Chapter 4 Vector-Valued Functions

Chapter 4 

Vector-Valued Functions 

4.1 Introduction to Vector-Valued Functions 

Parametric Curves in 3-Space 

Recall that if f and g are well-behaved functions, then the pair of parametric equations 

x = f(t), y = g(t) (4.1) 

generatesacurve in2-spacethatistracedinaspecific directionastheparameter tincreases. 

We define this direction to be the orientation of the curve or the direction of increasing 

parameter, and we called the curve together with its orientation the graph of the parametric 

equations or the parametric curve represented by the equations. Analogously, if f, g and h 

are three well-behaved functions, then the parametric equations 

x = f(t), y = g(t), z = h(t) (4.2) 

generatesacurve in3-spacethatistracedinaspecificdirectionastincreases. Asin2-space, 

this direction is called the orientation or direction of increasing parameter, and the 

curve together with its orientation is called the graph of the parametric equations or the 

parametric curve represented by the equations. If no restrictions are state explicitly 

or are implied by the equation, then it will be understood that t varies over the interval 

(−∞,+∞). 

Example 4.1 Describe the parametric curve represented by the equations 

Solution ......... 

x = 1−t, y = 3t, z = 2t. 

Example 4.2 Describe the parametric curve represented by the equations 

where a and c are positive constants. 

x = acosθ, y = asinθ, z = ct. 

62

MA112 Section 750001: Prepared by Dr.Archara Pacheenburawana 63 

Solution As the parameter t increases, the value of z = ct also increases, so the point 

(x,y,z) moves upward. However, as t increases, the point (x,y,z) also moves in a path 

directly over the circle 

x = acosθ, y = asinθ 

in the xy-plane. The combination of these upward and circular motions produces a cork 

screw-shaped curve that wraps around a right circular cylinder of radius a centered on the 

z-axis. This curve is called a circular helix. 

z 

x 

y 

✠ 

Parametric Equations for Intersections of Surfaces 

Curve in 3-space often arise as intersections of surfaces. One method for finding parametric 

equations for the curve of intersection is to choose one of the variables as the parameter and 

use the two equations to express the remaining two variables in terms of that parameter. 

For example, suppose we want to find parametric equations of the intersection of the 

cylinder z = x 3 and y = x 2 . We choose x = t as the parameter and substitute this into the 

equations z = x 3 and y = x 2 , then we obtain the parametric equations 

This curve is called a twisted cubic. 

x = t, y = t 2 , z = t 3 (4.3)


Vector-Valued Functions 

The twisted cubic defined by the equations in (4.3) is the set of points of the form (t,t 2 ,t 3 ) 

for real values of t. If we view each of these points as a terminal point for a vector r whose 

initial point is at the origin, 

r = 〈x,y,z〉 = 〈t,t 2 ,t 3 〉 = ti+t 2 j+t 2 k 

then we obtain r as a function of the parameter t, that is, r = r(t). Since this function 

produces a vector, we say that r = r(t) defines r as a vector-valued function of a real 

variable, or more simply, a vector-valued function. 

If r(t) is a vector-valued function in 2-space, then for each allowable value of t the vector 

r = r(t) can be represented in terms of components as 

r = r(t) = 〈x(t),y(t)〉 = x(t)i+y(t)j 

The functions x(t) and y(t) are called the component functions or the components of 

r(t). Similarly, the component functions of a vector-valued function 

in 3-space are x(t), y(t) and z(t). 

r = r(t) = 〈x(t),y(t),z(t)〉 = x(t)i+y(t)j+z(t)k 

Example 4.3 The component functions of 

are 

r(t) = 〈t,t 2 ,t 3 〉 = ti+t 2 j+t 3 k 

x(t) = t, y(t) = t 2 , z(t) = t 3 

The domain of a vector-valued function r(t) is the set of allowable values for t. If r(t) 

is defined in terms of component functions and the domain is not specified explicitly, then 

the domain is the intersection of the natural domains of the component functions; this is 

called the natural domain of r(t). 

Example 4.4 Find the natural domain of 


r(t) = 〈ln|t−1|,e t , √ t〉 = (ln|t−1|)i+e t j+ √ tk 

Graphs of Vector-Valued Functions 

If r(t) is a vector-valued function in 2-space or 3-space, then we define the graph of r(t) to 

be the parametric curve described by the component functions for r(t). 

For example, if 

r(t) = 〈1−t,3t,2t〉 = (1−t)i+3tj+2tk (4.4) 

then the graph of r = r(t) is the graph of the parametric equations 

x = 1−t, y = 3t, z = 2t 

Thus, the graph of (4.4) is the line in Example 4.1. 

✠

• 


Example 4.5 Describe the graph of the vector-valued function 


r(t) = 〈cost,sint,t〉 = costi+sintj+tk 

If a parametric curve is viewed as the graph of a vector-valued function, then we can 

also imagine the graph to be traced by the tip of the moving vector. For example, if the 

curve C in 3-space is the graph of 

r(t) = x(t)i+y(t)j+z(t)k 

and if we position r(t) so its initial point is at the origin, then its terminal point will fall 

on the curve C. 

z 

(x(t),y(t),z(t)) 

r(t) 

C 

y 

x 

Thus, when r(t) is positioned with its initial point at the origin, its terminal point will trace 

out the curve C as the parameter t varies, in which case we call r(t) the radius vector or 

the position vector for C. 

Example 4.6 Sketch the graph and a radius vector of 

(a) r(t) = costi+sintj, 0 ≤ t ≤ 2π 

(b) r(t) = costi+sintj+2k, 0 ≤ t ≤ 2π 


Vector Form of a Line Segment 

Recall that if r 0 is a vector in 2-space or 3-space with its initial points at the origin, then 

the line that passes through the terminal point of r 0 and is parallel to the vector v can be 

expressed in vector form as 

r = r 0 +tv 

In particular, if r 0 and r 1 are vectors in 2-space or 3-space with their initial points at 

the origin, then the line that passes through the terminal points of these vectors can be 

expressed in vector form as 

r = r 0 +t(r 1 −r 0 ) or r = (1−t)r 0 +tr 1 (4.5)

• 

• 

• 

• 


r 0 

t(r 1 −r 0 ) 

r 

O 

r 1 

r = (1−t)r 0 +tr 1 

It is common to call (4.5) the two-point vector form of a line. It is understood 

in (4.5) that t varies from −∞ to +∞. However, if we restrict t to vary over the interval 

0 ≤ t ≤ 1, then r will vary from r 0 to r 1 . Thus, the equation 

r = (1−t)r 0 +tr 1 (0 ≤ t ≤ 1) (4.6) 

represents the line segment in 2-space or 3-space that is traced from r 0 to r 1 . 

4.2 Calculus of Vector-Valued Functions 

In this section we will define limits, derivative, and integral of vector-valued functions. 

Limits and Continuity 

Our first goal in this section is to develop a notion of what it means for a vector-valued 

function r(t) in 2-space or 3-space to approach a limiting vector L ar t approaches a number 

a. That is, we want to define 

limr(t) = L (4.7) 

t→a 

Definition 4.1 Let r(t) be a vector-valued function that is defined for all t in some open 

interval containing the number a, except that r(t) need not be defined at a. We will write 

limr(t) = L 

t→a 

if and only if 

Theorem 4.1 

lim‖r(t)−L‖ = 0 

t→a 

(a) If r(t) = 〈x(t),y(t)〉 = x(t)i+y(t)j, then 

limr(t) = 

t→a 

〈 

lim 

t→a 

x(t),lim 

t→a 

y(t) 

〉 

= lim 

t→a 

x(t)i+lim 

t→a 

y(t)j 

provided the limits of the component functions exist. Conversely, the limits of the 

component functions exist provided r(t) approaches a limiting vector as t approaches 

a.


(b) If r(t) = 〈x(t),y(t),z(t)〉 = x(t)i+y(t)j+z(t)k, then 

〈 

〉 

limr(t) = limx(t),limy(t),limz(t) 

t→a t→a t→a t→a 

= lim 

t→a 

x(t)i+lim 

t→a 

y(t)j+lim 

t→a 

z(t)k 

provided the limits of the component functions exist. Conversely, the limits of the 

component functions exist provided r(t) approaches a limiting vector as t approaches 

a. 

Example 4.7 Find lim 

t→0 

( 

(t 2 +1)i+5costj+sintk ) . 


Example 4.8 Find lim 

t→0 

〈t 2 ,e t ,−2cosπt〉. 


Motivated by the definition of continuity for real-valued functions, we define a vectorvalued 

function r(t) to be continuous at t = a if 

limr(t) = r(a) (4.8) 

t→a 

That is, r(a) is defined, the limit of r(t) as t → a exists, and the two are equal. As in the 

case for real-valued functions, we say that r(t) is continuous on an interval I if it is 

continuous at each point of I. It follows from Theorem 4.1 that a vector-valued function is 

continuous at t = a if and only if its component functions are continuous at t = a. 

Derivatives 

The derivative of a vector-valued function is defined by a limit similar to that for the 

derivative of a real-values function. 

Definition 4.2 If r(t) is a vector-valued function, we define the derivative of r with 

respect to t to be the vector-valued function r ′ given by 

r ′ (t) = lim 

h→0 

r(t+h)−r(t) 

h 

(4.9) 

The domain of r consists of all values of t in the domain of r(t) for which the limit exists. 

The function r(t) is differentiable at t if the limit in (4.9) exists. All of the standard 

notations for derivatives continue to apply. For example, the derivative of r(t) can be 

expressed as 

d 

dt [r(t)], dr 

dt , r′ (t), or r ′

• 


Geometric Interpretation of the Derivative. 

Suppose that C is the graph of a vector-valued function r(t) in 2-space or 3-space and that 

r ′ (t) exists and is nonzero for a given value of t. If the vector r ′ (t) is positioned with its 

initial point as the terminal point of the radius vector r(t), then r ′ (t) is tangent to C and 

points in the direction of increasing parameter. 

y 

r ′ (t) 

r(t) 

C 

x 

Theorem 4.2 If r(t) is a vector-valued function, then r is differentiable at t if and only if 

each of its component functions is differentiable at t, in which case the component functions 

of r ′ (t) are the derivatives of the corresponding component functions of r(t). 

Example 4.9 Find the derivative of r(t) = sin(t 2 )i+e cost j+tlntk. 


Derivative Rules 

Many of the rules for differentiating real-valued functions have analogs in the context of 

differentiating vector-valued functions. We state some of these in the following theorem. 

Theorem 4.3 (Rules of Differentiation). Let r(t), r 1 (t), and r 2 (t) be vector-valued 

functions that are all in 2-space or all in 3-space, and let f(t) be a real-valued function, k a 

scalar, and c a constant vector (that is, a vector whose value does not depend on t). Then 

the following rules of differentiation hold: 

(a) 

(b) 

(c) 

(d) 

(c) 

d 

dt [c] = 0 

d 

dt [kr(t)] = k d dt [r(t)] 

d 

dt [r 1(t)+r 2 (t)] = d dt [r 1(t)]+ d dt [r 2(t)] 

d 

dt [r 1(t)−r 2 (t)] = d dt [r 1(t)]− d dt [r 2(t)] 

d 

dt [f(t)r(t)] = f(t) d dt [r(t)]+r(t) d dt [f(t)]

• 


Tangent Lines to Graphs of Vector-Valued Functions 

Definition 4.3 Let P be a point on the graph of a vector-valued function r(t), and let r(t 0 ) 

be the radius vector from the origin to P. 

y 

r ′ (t 0 ) 

r(t 0 ) 

P 

Tangent line 

x 

If r ′ (t 0 ) exists and r ′ (t 0 ) ≠ 0, then we call r ′ (t 0 ) a tangent vector to the graph of r(t) at 

r(t 0 ), and we call the line through P that is parallel to the tangent vector the tangent line 

to the graph of r(t) at r(t 0 ). 

Let r 0 = r(t 0 ) and v 0 = r ′ (t 0 ). Then the tangent line to the graph of r(t) at r 0 is given 

by the vector equation 

r = r 0 +tv 0 (4.10) 

Example 4.10 Find parametric equations of the tangent line to the circular helix 

x = cost, y = sint, z = t 

where t = t 0 , and use that result to find parametric equations for the tangent line at the 

point where t = π. 


Example 4.11 Let 

and 

r 1 (t) = (tan −1 t)i+(sint)j+t 2 k 

r 2 (t) = (t 2 −t)i+(2t−2)j+(lnt)k 

The graphs of r 1 (t) and r 2 (t) intersect at the origin. Find the degree measure of the acute 

angle between the tangent lines to the graphs of r 1 (t) and r 2 (t) at the origin. 

Solution .........


Derivatives of Dot and Cross Products 

The following rules provide a method for differentiating dot products in 2-space and 3-space 

and cross product in 3-space. 

d 

dt [r 1(t)·r 2 (t)] = r 1 (t)· dr 2 

dt + dr 1 

dt ·r 2(t) (4.11) 

d 

dt [r 1(t)×r 2 (t)] = r 1 (t)× dr 2 

dt + dr 1 

dt ×r 2(t) (4.12) 

Theorem 4.4 If r(t) is a differentiable vector-valued function in 2-space or 3-space and 

‖r(t)‖ is constant for all t, then 

r(t)·r ′ (t) = 0 (4.13) 

that is, r(t) and r ′ (t) are orthogonal vectors for all t. 

Definite Integrals of Vector-Valued Functions 

If r(t) is a vector-valued function that is continuous on the interval a ≤ t ≤ b, then we 

define the definite integral of r(t) over this interval as a limit of Riemann sums, that is, 

In general, we have 

∫ b 

a 

∫ b 

a 

(∫ b 

r(t)dt = 

∫ b 

a 

r(t)dt = 

(∫ b 

r(t)dt = 

a 

a 

lim 

max △t k →0 

n∑ 

r(t ∗ k)△t k (4.14) 

k=1 

) (∫ b 

) 

x(t)dt i+ y(t)dt j 2-space (4.15) 

a 

) (∫ b 

) (∫ b 

) 

x(t)dt i+ y(t)dt j+ z(t)dt k 3-space (4.16) 

a 

a 

Example 4.12 Let r(t) = t 2 i+e t j−(2cosπt)k. Evaluate 


∫ 1 

0 

r(t)dt. 

Rules of Integration 

As with differentiating, many of the rules for integrating real-values functions have analogs 

for vector-values functions. 

Theorem 4.5 (Rulesof Integration). Letr(t), r 1 (t), and r 2 (t) be vector-valued functions 

in 2-space or 3-space that are continuous on the interval a ≤ t ≤ b, and let k be a scalar. 

Then the following rules of integration hold:


(a) 

(b) 

(c) 

∫ b 

a 

∫ b 

a 

∫ b 

a 

∫ b 

kr(t)dt = k r(t)dt 

a 

[r 1 (t)+r 2 (t)]dt = 

[r 1 (t)−r 2 (t)]dt = 

∫ b 

a 

∫ b 

a 

r 1 (t)dt+ 

r 1 (t)dt− 

∫ b 

a 

∫ b 

a 

r 2 (t)dt 

r 2 (t)dt 

Antiderivatives of Vector-Valued Functions 

An antiderivative for a vector-valued function r(t) is a vector-valued function R(t) such 

that 

R ′ (t) = r(t) (4.17) 

We express Equation(4.17) using integral notation as 

∫ 

r(t)dt = R(t)+C (4.18) 

where C represents an arbitrary constant vector. Note that the vector R(t) + C is also 

called the indefinite integral of r(t). 

Since differentiation of vector-valued functions can be performed componentwise, it follows 

that antidifferentiation can be done this way as well. 

∫ 

Example 4.13 Evaluate the indefinite integral (2ti+3t 2 j+sin2tk)dt. 


Mostofthefamiliarintegrationpropertieshavevector counterparts. Forexample, vector 

differentiation and integration are inverse operations in the sense that 

[∫ 

d 

dt 

] 

r(t)dt = r(t) and 

∫ 

r ′ (t)dt = r(t)+C (4.19) 

Moreover, if R(t) is an antiderivative of r(t) on an interval containing t = a and t = b, then 

we have the following vector form of the Fundamental Theorem of Calculus: 

∫ b 

a 

] b ∫ 

r(t)dt = R(t) = 

a 

] b 

r(t)dt = R(b)−R(a) (4.20) 

a 

Example 4.14 Evaluate the definite integral 


∫ 1 

0 

( 

sinπti+(6t 2 +4t)j ) dt. 

Example 4.15 Find r(t) given that r ′ (t) = 〈3,2t〉 and r(1) = 〈2,5〉. 

Solution .........


4.3 Change of Parameter; Arc Length 

We observed in earlier sections that a curve in 2-space or 3-space can be represented parametrically 

in more than one way. Sometimes it will be desirable to change the parameter for 

a parametric curve to a different parameter that is better suited for the problem at hand. 

Smooth Parametrizations 

A curve represented by r(t) is smoothly parametrized by r(t), that r(t) is a smooth 

function of t if r ′ (t) is continuous and r ′ (t) ≠ 0 for any allowable value of t. Geometrically, 

this means that the a smooth parametrized curve can have no abrupt changes in direction 

as the parameter increases. 

Example 4.16 Determine whether the following vector-valued functions are smooth. 

(a) r(t) = acosti+asintj+ctk (a > 0,c > 0) 

(b) r(t) = t 2 i+t 3 j 


Arc Length from the Vector Viewpoint 

Recall that the arc length L of a parametric curve 

is given by the formula 

x = x(t), y = y(t) (a ≤ t ≤ b) (4.21) 

L = 

∫ b 

a 

√ (dx ) 2 

+ 

dt 

Analogously, the arc length L of a parametric curve 

in 3-space is given by the formula 

( ) 2 dy 

dt (4.22) 

dt 

x = x(t), y = y(t), z = z(t) (a ≤ t ≤ b) (4.23) 

L = 

∫ b 

a 

√ (dx ) 2 

+ 

dt 

( ) 2 dy 

+ 

dt 

( ) 2 dz 

dt (4.24) 

dt 

Formulas (4.22) and (4.24) have vector forms that we can obtain by letting 

r(t) = x(t)i+y(t)j or r(t) = x(t)i+y(t)j+z(t)k 

It follows that 

dr 

dt = dx dy 

i+ 

dt dt j or dr 

dt = dx dy dz 

i+ j+ 

dt dt dt k


and hence 

√ (dx ) dr 

2 ∥dt∥ = + 

dt 

( ) 2 dy 

or 

dt 

√ (dx ) dr 

2 ∥dt∥ = + 

dt 

( ) 2 dy 

+ 

dt 

( ) 2 dz 

dt 

Substituting these expressions in (4.22) and (4.24) leads us to the following theorem. 

Theorem 4.6 If C is the graph in 2-space or 3-space of a smooth vector-valued function 

r(t), then its arc length L from t = a to t = b is 

L = 

Example 4.17 Find the arc length of the vector-valued function 

from t = 0 to t = π. 


∫ b 

a 

dr 

∥dt∥ dt (4.25) 

r(t) = costi+sintj+tk 

Example 4.18 Find the arc length of the vector-valued function 


Arc Length as a Parameter 

r(t) = 〈2t,lnt,t 2 〉 for 1 ≤ t ≤ e. 

For many purpose the best parameter to use for representing a curve in 2-space or 3-space 

parametricallyisthelengthofarcmeasured alongthecurve fromsomefixed reference point. 

This can be done as follows: 

Using Arc Length as a Parameter 

Step 1. Select an arbitrary point on the curve C to serve as a reference point. 

Step 2. Starting from the reference point, choose one direction along the curve to be the 

positive direction and the other to be the negative direction. 

Step 3. If P is a point on the curve, let s be the signed arc length along C from the 

reference point to P, where s is positive if P is in the positive direction from the 

reference point and s is negative if P is in the negative direction.

• 

• 

• 

• 

• 

• 

• 


s = 3 

s = 2 

s = 1 

Reference point 

s = −1 

s = −2 

s = −3 

C 

Let us now treat s as a variable. As the value of s changes, the corresponding point 

P moves along C and the coordinate of P become functions of s. Thus, in 2-space the 

coordinate of P are (x(s),y(s)), and in 3-space they are (x(s),y(s),z(s)). Therefore in 

2-space or 3-space the curve C is given by the parametric equations 

x = x(s), y = y(s), or x = x(s), y = y(s), z = z(s) 

A parametric representation of a curve with arc length as the parameter is called an arc 

length parametrization of the curve. 

Example 4.19 Find the arc length parametrization of the circle x 2 + y 2 = a 2 with counterclockwise 

orientation and (a,0) as the reference point. 


Change of Parameter 

A change of parameter in a vector-valued function r(t) is a substitution t = g(τ) that 

produce a new vector-valued function r(g(τ)) have the same graph as r(t), but possibly 

traced differently as the parameter τ increases. 

Example 4.20 Find a change of parameter t = g(τ) for the circle 

such that 

r(t) = costi+sintj (0 ≤ t ≤ 2π) 

(a) the circle is traced counterclockwise as τ increases over the interval [0,1]; 

(b) the circle is traced clockwise as τ increases over the interval [0,1]. 


Theorem 4.7 (Chain Rule) Let r(t) be a vector-valued function in 2-space or 3-space 

that is differentiable with respect to t. If t = g(τ) is a change of parameter in which g is 

differentiable with respect to τ, then r((g(τ)) is differentiable with respect to τ and 

dr 

dτ = dr dt 

dt dτ 

(4.26)

• 

• 


A change of parameter t = g(τ) in which r(g(τ)) is smooth is called a smooth change 

of parameter. Smooth change of parameter fall into two categories—those for which 

dt/dτ > 0forallτ (calledpositive changes of parameter)andthoseforwhichdt/dτ < 0 

for all τ (called negative changes of parameter). A positive changes of parameter 

preserves theorientationofaparametriccurve, andanegativechangesofparameterreverses 

it. 

Finding Arc Length Parametrizations 

Theorem 4.8 Let C be the graph of a smooth vector-valued function r(t) in 2-space or 3- 

space, and let r 0 (t) be any point on C. Then the following formula defines a positive change 

of parameter from t to s, where s is an arc length parameter having r 0 (t) as its reference 

point: 

∫ t 

∥ ∥∥∥ dr 

s = 

t 0 

du∥ du (4.27) 

y 

r(t 0 ) 

s 

r(t) 

C 

x 

When needed, Formula (4.27) can be expressed in component form as 

s = 

s = 

∫ t 

√ (dx ) 2 

+ 

t 0 

du 

∫ t 

√ (dx ) 2 

+ 

t 0 

du 

( ) 2 dy 

+ 

du 

( ) 2 dy 

du 2-space (4.28) 

du 

( ) 2 dz 

du 3-space (4.29) 

du 

Example 4.21 Find the arc length parametrization of the circular helix 

r = costi+sintj+tk 

that has reference point r(0) = (1,0,0) and has the same orientation as the given helix. 


Example 4.22 A bug walks along the trunk of a tree following a path modeled by the 

circular helix in Example 4.21. The bug starts at the reference point (1,0,0) and walks up 

the helix for a distance of 10 units. What are the bug’s final coordinates



Example 4.23 Find the arc length parametrization of the line 

x = 2t+1, y = 3t−2 

that has the same orientation as the given line and uses (1,−2) as the reference point. 


Properties of Arc Length Parametrizations 

Theorem 4.9 

(a) If C is the graph of a smooth vector-valued function r(t) in 2-space or 3-space, where 

t is a general parameter, and if s is the arc length parameter for C defined by Formula 

(4.27), then for every value of t the tangent vector has length 

dr 

∥dt∥ = ds 

(4.30) 

dt 

(b) If C is the graph of a smooth vector-valued function r(t) in 2-space or 3-space, where 

s is an arc length parameter, then for every value of s the tangent vector to C has 

length 

dr 

∥ds∥ = 1 (4.31) 

(c) If C is the graph of a smooth vector-valued function r(t) in 2-space or 3-space, and 

if ‖dr/dt‖ = 1 for every value of t, then for any value of t 0 in the domain of r, the 

parameter s = t−t 0 is an arc length parameter that has its reference point at the point 

on C where t = t 0 . 

The component forms of Formulas (4.30) and (4.31) are as follow: 

∥ √ (dx ) 

ds ∥∥∥ 

dt = dr 

2 dt∥ = + 

dt 

∥ √ (dx ) 

ds ∥∥∥ 

dt = dr 

2 dt∥ = + 

dt 

√ (dx ) dr 

2 ∥ds∥ = + 

ds 

√ (dx ) dr 

2 ∥ds∥ = + 

ds 

( ) 2 dy 

+ 

ds 

( ) 2 dy 

+ 

dt 

( ) 2 dy 

2-space (4.32) 

dt 

( ) 2 dz 

3-space (4.33) 

dt 

( ) 2 dy 

= 1 2-space (4.34) 

ds 

( ) 2 dz 

= 1 3-space (4.35) 

ds


4.4 Unit Tangent and Normal Vectors 

Unit Tangent Vectors 

Recall that if C is a graph of a smooth vector-valued function r(t) in 2-space or 3-space, 

then the vector r ′ (t) is nonzero, tangent to C, and points in the direction of increasing 

parameter. Thus, by normalizing r ′ (t) we obtain a unit vector 

T(t) = r′ (t) 

‖r ′ (t)‖ 

(4.36) 

that is tangent to C and points in the direction of increasing parameter. We call T(t) the 

unit tangent vector to C at t. 

y 

C 

T(t) 

r(t) 

P 

x 

Example 4.24 Find the unit tangent vector to the graph of r(t) = t 2 i + t 3 j at the point 

where t = 2. 


Example 4.25 Find the unit tangent vector to the curve determined by 


Unit Normal Vectors 

r(t) = 〈t 2 +1,t〉. 

Recall from Theorem 4.4 that if a vector-valued function r(t) has constant norm, then r(t) 

and r ′ (t) are orthogonal vectors. In particular, T(t) has constant norm 1, so T(t) and T ′ (t) 

are orthogonal vectors. This implies that T ′ (t) is perpendicular to the tangent line to C at 

t, so we say that T ′ (t) is normal to C at t. It follows that if T ′ (t) ≠ 0, and if we normalize 

T ′ (t), then we obtain a unit vector 

N(t) = T′ (t) 

‖T ′ (t)‖ 

(4.37)

• 

• 


that is normal to C and points in the same direction as T ′ (t). We call N(t) the principal 

unit normal vector to C at t, or more simply, the unit normal vector. Observe that 

the unit normal vector is defined only at points where T ′ (t) ≠ 0. 

In 2-space there are two unit vectors that are orthogonal to T(t), and in 3-space there 

are infinitely such vectors. 

y 

C 

z 

T(t) 

C 

x 

T(t) 

y 

There are two unit vectors 

orthogonal to T(t). 

x 

There are infinitely many unit 

vectors orthogonal to T(t). 

Example 4.26 Find T(t) and N(t) for the circular helix 

where a > 0. 


x = acost, y = asint, z = ct 

Example 4.27 Find the unit tangent and principal unit normal vectors to the curve defined 

by r(t) = 〈t 2 ,t〉. 


Computing T and N for Curves Parametrized by Arc Length 

In the case where r(s) ia parametrized by arc length, the procedures for computing the 

unit tangent vector T(s) and the unit normal vector N(s) are simpler than in the general 

case. For example, we showed in Theorem 4.9 that if s is an arc length parameter, then 

‖r ′ (s)‖ = 1. Thus, Formula (4.36) for the unit tangent vector simplifies to 

T(s) = r ′ (s) (4.38) 

and consequently Formula (4.37) for the unit normal vector simplifies to 

N(s) = r′′ (s) 

‖r ′′ (s)‖ 

(4.39) 

Example 4.28 The circle of radius a with counterclockwise orientation and centered at the 

origin can be represented by the vector-valued function 

r = acosti+asintj (0 ≤ t ≤ 2π) 

Parametrize this circle by arc length and find T(s) and N(s). 

Solution .........


Partial Derivatives 

5.1 Functions of Two or More Variables 

Notation and Terminology 

The terminology and notation for functions of two or more variables is similar to that for 

functions of one variable. For example, the expression 

z = f(x,y) 

means that z is a function of x and y in the sense that a unique value of the dependent 

variable z is determined by specifying values for the independent variables x and y. Similarly, 

w = f(x,y,z) 

expresses w as a function x, y, and z, and 

u = f(x 1 ,x 2 ,...,x n ) 

expresses u as a function of x 1 ,x 2 ,...,x n . 

As with functions of one variable, the independent variables of a function of two or more 

variables may be restricted to lie in some set D, which we call the domain of f. Sometimes 

the domain will be determined by physical restrictions on the variables. If the function is 

defined by a formula and if there are no physical restrictions or other restrictions stated 

explicitly, then it is understand that the domain consists of all points for which the formula 

yields a real value for the dependent variable. We call this the natural domain of the 

function. 

Definition 5.1 A function f of two variables, x and y, is a rule that assigns a unique 

real number f(x,y) to each point (x,y) in some set D in the xy-plane. 

Definition 5.2 A function f of three variables, x, y, and z, is a rule that assigns 

a unique real number f(x,y,z) to each point (x,y,z) in some set D in three dimensional 

space. 

79



f(x,y) = 3x 2√ y −1 

Find f(1,4), f(0,9), f(t 2 ,t), f(ab,9b), and the natural domain of f. 


Example 5.2 Sketch the natural domain of the function (a) f(x,y) = ln(x 2 −y) and (b) 

g(x,y) = 2x 

y −x 2. 



f(x,y,z) = √ 1−x 2 −y 2 −z 2 

Find f ( 0, 1 2 ,−1 2) 

and the natural domain of f. 


Graphs of Functions of Two Variables 

Recall that for a function f of one variable, the graph of f(x) in the xy-plane was defined 

to be the graph of the equation y = f(x). Similarly, if f is a function of two variables, we 

define the graph of f(x,y) in xyz-space to be the graph of the equation z = f(x,y). In 

general, such a graph will be a surface in 3-space. 

Example 5.4 In each part, describe the graph of the function in an xyz-coordinate system. 

(a) f(x,y) = 1−x− 1 2 y (b) f(x,y) = √ 1−x 2 −y 2 

(c) f(x,y) = − √ x 2 +y 2 


5.2 Limits and Continuity 

For a function of one variable there are two one-sided limits at a points x 0 , namely, 

lim f(x) and lim f(x) 

x→x + 0 x→x − 0 

reflecting the fact that there are only two directions fromwhich x can approach x 0 , the right 

or the left. For function of two or three variables the situation is more complicated because 

there are infinitely many different curves along which one point can approach another. Our 

first objective is to define the limit of f(x,y) as (x,y) approaches a point (x 0 ,y 0 ) along the 

curve C (and similarly for functions of three variables). 

If C is a smooth parametric curve in 2-space or 3-space that is represented by the 

equations 

x = x(t), y = y(t) or x = x(t), y = y(t), z = z(t)


and if x 0 = x(t 0 ),y 0 = y(t 0 ), and z 0 = z(t 0 ), then the limits 

lim 

(x,y) → (x 0 ,y 0 ) 

(along C) 

f(x,y) and 

lim 

(x,y,z) → (x 0 ,y 0 ,z 0 ) 

(along C) 

f(x,y,z) 

are defined by 

lim 

(x,y) → (x 0 ,y 0 ) 

(along C) 

f(x,y) = lim 

t→t0 

f(x(t),y(t)) (5.1) 

lim 

(x,y,z) → (x 0 ,y 0 ,z 0 ) 

(along C) 

f(x,y,z) = lim 

t→t0 

f(x(t),y(t),z(t)) (5.2) 

Example 5.5 Find the limit of the function 

as (x,y) → (0,0) along 

f(x,y) = − xy 

x 2 +y 2 

(a) the x-axis (b) the y-axis (c) the line y = x 

(d) the line y = −x (e) the parabola y = x 2 


Relationships between General Limits and Limit Along Smooth 

Curve 

Theorem 5.1 

(a) If f(x,y) → L as (x,y) → (x 0 ,y 0 ), then f(x,y) → L as (x,y) → (x 0 ,y 0 ) along any 

smooth curve. 

(b) If the limit of f(x,y) fail to exit as (x,y) → (x 0 ,y 0 ) along some smooth curve, or if 

f(x,y) has different limit as (x,y) → (x 0 ,y 0 ) along two different smooth curves, then 

the limit of f(x,y) does not exist as (x,y) → (x 0 ,y 0 ). 

Example 5.6 The limit 

lim − xy 

(x,y)→(0,0) x 2 +y 2 

does not exist because in Example 5.5 we found two different smooth curves along which this 

limit had different values. Specifically, 

lim 

(x,y) → (0,0) 

(along y = 0) 

− xy 

x 2 +y 2 = 0 and 

lim 

(x,y) → (0,0) 

(along y = x) 

− xy 

x 2 +y 2 = −1 2 . 

✠


Continuity 

Definition 5.3 A function f(x,y) is said to be continuous at (x 0 ,y 0 ) if f(x 0 ,y 0 ) is 

defined and if 

lim 

(x,y)→(x 0 ,y 0 ) f(x,y) = f(x 0,y 0 ) 

In addition, if f is continuous at every point in an open set D, then we say that f is 

continuous on D, and if f is continuous at every point in the xy-plane, then we say that 

f is continuous everywhere. 

Theorem 5.2 

(a) If g(x) is continuous at x 0 and h(y) is continuous at y 0 , then f(x,y) = g(x)h(y) is 

continuous at (x 0 ,y 0 ). 

(b) If h(x,y) is continuous at (x 0 ,y 0 ) and g(u) is continuous at u = h(x 0 ,y 0 ), then the 

composition f(x,y) = g(h(x,y)) is continuous at (x 0 ,y 0 ). 

(c) If f(x,y) is continuous at (x 0 ,y 0 ) and if x(t) and y(t) is continuous at t 0 with x(t 0 ) = 

x 0 and y(t 0 ) = y 0 , then the composition f(x(t),y(t)) is continuous at t 0 . 

Example 5.7 Show that the functions f(x,y) = 3x 2 y 5 and f(x,y) = sin(3x 2 y 5 ) are continuous 

everywhere. 


Recognizing continuous Functions 

• A composition of continuous functions is continuous. 

• A sum, difference, or product of continuous functions is continuous. 

• A quotient of continuous functions is continuous, except where the denominator is 

zero. 

Example 5.8 Evaluate 


lim 

(x,y)→(−1,2) 

Example 5.9 Since the function 

xy 

x 2 +y 2. 

f(x,y) = x3 y 2 

1−xy 

is a quotient of continuous functions, it is continuous except where 1−xy = 0. Thus f(x,y) 

is continuous everywhere except on the hyperbola xy = 1. 

✠


5.3 Partial Derivatives 

Partial Derivatives of Functions of Two Variables 

If z = f(x,y), then one can inquire how the value of z changes if y is held fixed and x is 

allowed to vary, or if x is held fixed and y is allowed to vary. For example, the ideal gas 

law in physics states that under appropriate conditions the pressure exerted by a gas is a 

function of the volume of the gas and its temperature. Thus, a physicist studying gases 

might be interested in the rate of change of the pressure if the volume is held fixed and the 

temperature is allowed to vary, or if the temperature is held fixed and the volume is allowed 

to vary. We now define a derivative that describes such rates of change. 

Definition 5.4 If z = f(x,y) and (x 0 ,y 0 ) is a point in the domain of f, then the partial 

derivative of f with respect to x at (x 0 ,y 0 ) [ also called the partial derivative of 

z with respect to x at (x 0 ,y 0 ) ] is the derivative at x 0 of the function that results when 

y = y 0 is held fixed and x is allowed to vary. This partial derivative is denoted by f x (x 0 ,y 0 ) 

and is given by 

f x (x 0 ,y 0 ) = d 

dx [f(x,y 0)] ∣ = lim 

x=x0 

∆x→0 

f(x 0 +∆x,y 0 )−f(x 0 ,y 0 ) 

∆x 

(5.3) 

Similarly, the partial derivative of f with respect to y at (x 0 ,y 0 ) [ also called the 

partial derivative of z with respect to y at (x 0 ,y 0 ) ] is the derivative at y 0 of the 

function that results when x = x 0 is held fixed and y is allowed to vary. This partial 

derivative is denoted by f y (x 0 ,y 0 ) and is given by 

f y (x 0 ,y 0 ) = d 

dy [f(x 0,y)] ∣ = lim 

y=y0 

∆y→0 

f(x 0 ,y 0 +∆y)−f(x 0 ,y 0 ) 

∆y 

Example 5.10 Find f x (1,0) and f y (2,−1) for the function f(x,y) = 3x 2 +x 3 y +4y 2 . 


The Partial Derivative Functions 

(5.4) 

Formulas (5.3) and (5.4) define the partial derivatives of a function at a specific point 

(x 0 ,y 0 ). However, often it will be desirable to omit the subscripts and think of the partial 

derivatives as a functions of the variables x and y. These functions are 

and 

f(x+∆x,y)−f(x,y) 

f x (x,y) = lim 

∆x→0 ∆x 

f(x,y +∆y)−f(x,y) 

f y (x,y) = lim 

∆y→0 ∆y 

The following example gives an alternative way of performing the computations in Example 

5.10.


Example 5.11 Find f x (x,y) and f y (x,y) for f(x,y) = 3x 2 + x 3 y + 4y 2 , and use those 

partial derivatives to compute f x (1,0) and f y (2,−1). 


Partial Derivative Notation 

If z = f(x,y), then the partial derivatives f x and f y are also denoted by the symbols 

∂f 

∂x , ∂z 

∂x 

and 

∂f 

∂y , ∂z 

∂y 

Some typical notations for the partial derivatives of z = f(x,y) at a point (x 0 ,y 0 ) are 

∂f 

∂z 

∂x∣ , 

∂f 

∂f 

x=x0 ,y=y 0 

∂x∣ (x0 ,y 0 ), 

∂x∣ (x0 ,y 0 ), 

∂x (x ∂z 

0,y 0 ), 

∂x (x 0,y 0 ) 

Example 5.12 Find ∂z 

∂x 


and 

∂z 

∂y if z = exy + x y . 

Partial Derivatives Viewed as Rates of Change and Slopes 

Recall that if y = f(x), then the value of f ′ (x 0 ) can be interpreted either as the rate of 

change of y with respect to x at x 0 or as the slope of the tangent line to the graph of f at 

x 0 . Partial derivatives have analogous interpretations. 

Suppose that C 1 is the intersection of the surface z = f(x,y) with the plane y = y 0 and 

that C 2 is its intersection with the plane x = x 0 . Thus, f x (x,y 0 ) can be interpreted as the 

rate of change of z with respect to x along the curve C 1 , and f y (x 0 ,y) can be interpreted as 

the rate of change of z with respect to y along the curve C 2 . In particular, f x (x 0 ,y 0 ) is the 

rate of change of z with respect to x along the curve C 1 at the point (x 0 ,y 0 ), and f y (x 0 ,y 0 ) 

is the rate of change of z with respect to y along the curve C 2 at the point (x 0 ,y 0 ). 

Example 5.13 For a real gas, van der Waals’s equation states that 

( ) 

P + n2 a 

(V −nb) = nRT 

V 2 

Here, P is the pressure of the gas, V is the volume of the gas, T is the temperature (in 

degrees Kelvin), n is the number of moles of gas, R is the universal gas constant and a and 

b are constants. Compute and interpret P V and T P 


Geometrically, f x (x 0 ,y 0 ) can be viewed as the slope of the tangent line to the curve C 1 

at the point (x 0 ,y 0 ), and f y (x 0 ,y 0 ) can be viewed as the slope of the tangent line to the 

curve C 2 at the point (x 0 ,y 0 ). We will call f x (x 0 ,y 0 ) the slope of the surface in the 

x-direction at (x 0 ,y 0 ) and f y (x 0 ,y 0 ) the slope of the surface in the y-direction at 

(x 0 ,y 0 ).


Example 5.14 Let f(x,y) = x 2 y +5y 3 . 

(a) Find the slope of the surface z = f(x,y) in the x-direction at the point (1,−2). 

(b) Find the slope of the surface z = f(x,y) in the y-direction at the point (1,−2). 


Implicit Partial Differentiation 

( 

Example 5.15 Find the slope of the sphere x 2 +y 2 +z 2 = 1 in the y-direction at the points 

2 , 1, ) ( 2 

3 3 3 and 2 , ) 1 

3 3 ,−2 3 . 


Example 5.16 Suppose that D = √ x 2 +y 2 is the length of the diagonal of a rectangle 

whose sides have lengths x and y that are allowed to vary. Find the formula for the rate of 

change of D with respect to x if x varies with y held constant, and use this formula to find 

the rate of change of D with respect to x at the point where x = 3 and y = 4. 


Partial Derivatives of Functions with More Than Two Variables 

For a function f(x,y,z) of three variables, there are three partial derivatives: 

f x (x,y,z), f y (x,y,z), f z (x,y,z) 

The partial derivative f x is calculated by holding y and z constant and differentiating with 

respect to x. For f y the variables x and z are held constant, and for f z the variables x and 

y are held constant. If a dependent variable 

w = f(x,y,z) 

is used, then the three partial derivatives of f can be denoted by 

∂w 

∂x , ∂w ∂w 

, and 

∂y ∂z 

Example 5.17 If f(x,y,z) = x 3 y 2 z 4 +2xy +z, then 

f x (x,y,z) = 

f y (x,y,z) = 

f z (x,y,z) = 

f z (−1,1,2) =


Example 5.18 If f(ρ,θ,φ) = ρ 2 cosφsinθ, then 

f ρ (ρ,θ,φ) = 

f θ (ρ,θ,φ) = 

f φ (ρ,θ,φ) = 

In general, if f(v 1 ,v 2 ,...,v n ) is a function of n variables, there are n partial derivatives 

of f, each of which is obtained by holding n −1 of the variables fixed and differentiating 

the function f with respect to the remaining variable. If w = f(v 1 ,v 2 ,...,v n ), then these 

partial derivatives are denoted by 

∂w 

∂v 1 

, 

∂w 

,..., ∂w 

∂v 2 ∂v n 

where ∂w/∂v i is obtained by holding all variables except v i fixed and differentiating with 

respect to v i . 

Example 5.19 Find 

for i = 1,2,...,n. 

[√ ] 

∂ 

x 2 1 +x 2 2 +···+x 

∂x 2 n 

i 


Higher-Order Partial Derivatives 

Suppose that f is a function of two variables x and y. Since the partial derivatives ∂f/∂x 

and ∂f/∂y are also functions of x and y, these functions may themselves have partial 

derivatives. This gives rise to four possible second-order partial derivatives of f, which are 

defined by 

∂ 2 f 

∂x = ∂ ( ) ∂f ∂ 2 f 

= f 2 xx 

∂x ∂x ∂y = ∂ ( ) ∂f 

= f 2 yy 

∂y ∂y 

Differentiate twice 

with respect to x. 

∂ 2 f 

∂y∂x = ∂ ( ) ∂f 

∂y ∂x 

Differentiate first with 

respect to x and then 

with respect to y. 

= f xy 

∂ 2 f 

∂x∂y = ∂ 

∂x 

Differentiate twice 

with respect to y. 

( ) ∂f 

= f yx 

∂y 

Differentiate first with 

respect to y and then 

with respect to x. 

The last two cases are called the mixed second-order partial derivatives or the mixed 

second partials. Also, the derivatives ∂f/∂x and ∂f/∂x are often called the first-order 

partial derivatives when it is necessary to distinguish them from higher-order partial 

derivatives.


Example 5.20 Find the second-order partial derivatives of f(x,y) = x 2 y −y 3 +lnx. 


Third-order, fourth-order, and higher-order partial derivatives can be obtained by successive 

differentiation. Some possibilities are 

∂ 3 f 

∂x = ∂ ( ) ∂ 2 f 

∂ 4 f 

= f 3 ∂x ∂x 2 xxx 

∂y = ∂ ( ) ∂ 3 f 

= f 4 ∂y ∂y 3 yyyy 

∂ 3 f 

∂y 2 ∂x = ∂ ( ) ∂ 2 f ∂ 4 f 

= f xyy 

∂y ∂y∂x ∂y 2 ∂x = ∂ ( ) ∂ 3 f 

= f 2 ∂y ∂y∂x 2 xxyy 

Example 5.21 Let f(x,y) = cos(xy)−x 3 +y 4 . Find f xyy . 


Equality of Mixed Partials 

Theorem 5.3 Let f be a function of two variables. If f xy and f yx are continuous on some 

open disk, then f xy = f yx on that disk. 

5.4 Differentiability, Differentials, and Local Linearity 

Differentiability 

Definition 5.5 A function f of two variables is said to be differentiable at (x 0 ,y 0 ) provided 

f x (x 0 ,y 0 ) and f y (x 0 ,y 0 ) both exist and 

∆f −f x (x 0 ,y 0 )∆x−f y (x 0 ,y 0 )∆y 

lim √ = 0 (5.5) 

(∆x,∆y)→(0,0) (∆x)2 +(∆y) 2 

NoteForafunctionf(x,y),thesymbol ∆f, calledtheincrement off, denotes thechange 

in the value of f(x,y) that results when (x,y) varies from some initial position (x 0 ,y 0 ) to 

some new position (x 0 +∆x,y 0 +∆y); thus 

∆f = f(x 0 +∆x,y 0 +∆y)−f(x 0 ,y 0 ) 

Example 5.22 Use Definition 5.5 to prove that f(x,y) = x 2 +y 2 is differentiable at (0,0). 


Definition 5.6 A function f of three variables is said to be differentiable at (x 0 ,y 0 ,z 0 ) 

provided f x (x 0 ,y 0 ,z 0 ), f y (x 0 ,y 0 ,z 0 ), and f z (x 0 ,y 0 ,z 0 ) exist and 

∆f −f x (x 0 ,y 0 ,z 0 )∆x−f y (x 0 ,y 0 ,z 0 )∆y −f z (x 0 ,y 0 ,z 0 )∆z 

lim √ = 0 (5.6) 

(∆x,∆y,∆z)→(0,0) (∆x)2 +(∆y) 2 +(∆z) 2 

If a function f of two variables is differentiable at each point of the region R in the 

xy-plane, then we say that f is differentiable on R; and if f is differentiable at every 

point in the xy-plane, then we say that f is differentiable everywhere. For a function 

f of three variables we have corresponding conventions.


Differentiability and Continuity 

Theorem 5.4 If a function is differentiable at a point, then it is continuous at that point. 

Theorem 5.5 If all first-order partial derivatives of f exist and are continuous at a point, 

then f is differentiable at that point. 

For example, consider the function 

f(x,y,z) = x+yz 

Since f x (x,y,z) = 1,f y (x,y,z) = z, and f z (x,y,z) = y are defined and continuous everywhere, 

we conclude that f is differentiable everywhere. 

Differentials 

If z = f(x,y) is differentiable at a point (x 0 ,y 0 ), we let 

dz = f x (x 0 ,y 0 )dx+f y (x 0 ,y 0 )dy (5.7) 

denote a new function with dependent variable dz and independent variables dx and dy. 

We refer to this function (also denoted df) as the total differential of z at (x 0 ,y 0 ) or as 

the total differential of f at (x 0 ,y 0 ). Similarly, for a function w = f(x,y,z) of three 

variables we have the total differential of w at (x 0 ,y 0 ,z 0 ), 

dw = f x (x 0 ,y 0 ,z 0 )dx+f y (x 0 ,y 0 ,z 0 )dy +f z (x 0 ,y 0 ,z 0 )dz (5.8) 

which is also referred toas the total differential of f at (x 0 ,y 0 ,z 0 ). It is common practice 

to omit the subscripts and write Equations (5.7) and (5.8) as 

and 

dz = f x (x,y)dx+f y (x,y)dy (5.9) 

dw = f x (x,y,z)dx+f y (x,y,z)dy+f z (x,y,z)dz (5.10) 

In the two-variable case, the approximation 

can be written in the form 

∆f ≈ f x (x 0 ,y 0 )∆x+f y (x 0 ,y 0 )∆y 

∆f ≈ df (5.11) 

for dx = ∆x and dy = ∆y. Equivalently, we can write approximation (5.11) as 

∆z ≈ dz (5.12) 

Example 5.23 Use a total differential to approximate the change in z = xy 2 from it value 

at (0.5,1.0) to its value at (0.503,1.004). Compare this estimate with the actual change in 

z. 

Solution .........


Local Linear Approximations 

When f(x,y) is differentiable at (x 0 ,y 0 ) we get 

L(x,y) = f(x 0 ,y 0 )+f x (x 0 ,y 0 )(x−x 0 )+f y (x 0 ,y 0 )(y −y 0 ) 

and refer to L(x,y) as the local linear approximation to f at (x 0 ,y 0 ). 

Example 5.24 Let L(x,y) denote the local linear approximation to f(x,y) = √ x 2 +y 2 at 

the point (3,4). Compare the error in approximating 

f(3.04,3.98) = √ (3.04) 2 +(3.98) 2 

by L(3.04,3.98) with the distance between the points (3,4) and (3.04,3.98). 


is 

Forafunctionf(x,y,z)thatisdifferentiableat(x 0 ,y 0 ,z 0 ),thelocallinearapproximation 

L(x,y,z) = f(x 0 ,y 0 ,z 0 )+f x (x 0 ,y 0 ,z 0 )(x−x 0 ) 

+f y (x 0 ,y 0 ,z 0 )(y −y 0 )+f z (x 0 ,y 0 ,z 0 )(z −z 0 ) 

5.5 The Chain Rule 

The Chain Rule for Derivatives 

Theorem 5.6 (Chain Rule for Derivatives) If x = x(t) and y = y(t) are differentiable 

at t, and if z = f(x,y) is differentiable at the point (x,y) = (x(t),y(t)), then 

z = f(x(t),y(t)) is differentiable at t and 

dz 

dt = ∂z dx 

∂x dt + ∂z dy 

∂y dt 

(5.13) 

where the ordinary derivatives are evaluated at t and the partial derivatives are evaluated 

at (x,y). 

If each of the functions x = x(t), y = y(t), and z = z(t) is differentiable at t, and if w = 

f(x,y,z) is differentiable at the point (x,y,z) = (x(t),y(t),z(t)), then w = f(x(t),y(t),z(t)) 

is differentiable at t and 

dw 

dt = ∂w dx 

∂x dt + ∂w dy 

∂y dt + ∂w dz 

(5.14) 

∂z dt 

where the ordinary derivatives are evaluated at t and the partial derivatives are evaluated 

at (x,y,z). 

Formula (5.13) can be represented schematically by a “tree diagram” that is constructed 

as follows.


z 

∂z 

∂x 

∂z 

∂y 

x 

dx 

dt 

y 

dy 

dt 

t 

dz 

dt = ∂z dx 

∂x dt + ∂z dy 

∂y dt 

t 

Example 5.25 Suppose that 

Use the chain rule to find dz/dt. 



z = x 2 e y , x(t) = t 2 −1, y(t) = sint 

z = √ xy +y, x = cosθ, y = sinθ 

Use the chain rule to find dz/dθ when θ = π/2. 



w = √ x 2 +y 2 +z 2 , x = cosθ, y = sinθ, z = tanθ 

Use the chain rule to find dw/dθ. 


The Chain Rule for Partial Derivatives 

Theorem 5.7 (Chain Rule for Partial Derivatives) If x = x(u,v) and y = y(u,v) 

have first-order derivatives at the point (u,v), and if z = f(x,y) is differentiable at the point 

(x,y) = (x(u,v),y(u,v)), then z = f(x(u,v),y(u,v)) has first-order partial derivatives at 

the point (u,v) given by 

∂z 

∂u = ∂z ∂x 

∂x∂u + ∂z ∂y 

∂y∂u 

and 

∂z 

∂v = ∂z ∂x 

∂x∂v + ∂z ∂y 

∂y∂v 

If each function x = x(u,v), y = y(u,v), and z = z(u,v) have first-order derivatives at 

the point (u,v), and if the function w = f(x,y,z) is differentiable at the point (x,y,z) =


(x(u,v),y(u,v),z(u,v)), then the function w = f(x(u,v),y(u,v),z(u,v)) has first-order 

partial derivatives at the point (u,v) given by 

∂w 

∂u = ∂w ∂x 

∂x ∂u + ∂w ∂y 

∂y ∂u + ∂w ∂z 

∂z ∂u 

and 

∂w 

∂v = ∂w ∂x 

∂x ∂v + ∂w ∂y 

∂y ∂v + ∂w ∂z 

∂z ∂v 

The following figure shows tree diagrams for the formulas in Theorem 5.7. 

z 

z 

∂z 

∂x 

∂z 

∂y 

∂z 

∂x 

∂z 

∂y 

x 

y 

x 

y 

∂x 

∂u 

∂x 

∂v 

∂y 

∂u 

∂y 

∂v 

∂x 

∂u 

∂x 

∂v 

∂y 

∂u 

∂y 

∂v 

u v u v 

∂z 

∂u = ∂z ∂x 

∂x∂u + ∂z ∂y 

∂y∂u 

u v u v 

∂z 

∂v = ∂z ∂x 

∂x∂v + ∂z ∂y 

∂y∂v 

w 

∂w 

∂x 

∂w 

∂y 

∂w 

∂z 

x y z 

∂x 

∂u 

∂x 

∂v 

∂y 

∂u 

∂y 

∂v 

∂z 

∂u 

∂z 

∂v 

u v u v u v 

∂w 

∂u = ∂w ∂x 

∂x ∂u + ∂w ∂y 

∂y ∂u + ∂w ∂z 

∂z ∂u 

∂w 

∂v = ∂w ∂x 

∂x ∂v + ∂w ∂y 

∂y ∂v + ∂w ∂z 

∂z ∂v 

Example 5.28 Given that 

z = e xy , x(u,v) = 3usinv, y(u,v) = 4v 2 u 

find ∂z/∂u and ∂z/∂v using the chain rule. 

Solution .........



w = e xyz , x = 3u+v, y = 3u−v, z = u 2 v 

Use appropriate forms of the chain rule to find ∂w/∂u and ∂w/∂v. 


Other Versions of the Chain Rule 

Although we will not prove it, the chain rule extends to functions w = f(v 1 ,v 2 ,...,v n ) of 

n variables. For example, if each v i is a function of t, i = 1,2,...,n, the relevant formula is 

dw 

dt = ∂w dv 1 

∂v 1 dt + ∂w dv 2 ∂w dv n 

+···+ 

∂v 2 dt ∂v n dt 

(5.15) 

There are infinitely many variations of the chain rule, depending on the number of 

variablesandthechoice ofindependent anddependent variables. Agoodworking procedure 

is to use tree diagrams to derive new versions of the chain rule as needed. 

Example 5.30 Suppose that w = x 2 +y 2 +z 2 and 

x = ρsinφcosθ, y = ρsinφsinθ, z = ρcosφ 

Use appropriate forms of the chain rule to find ∂w/∂ρ and ∂w/∂θ. 



w = xy +yz, y = sinx, z = e x 

Use appropriate form of the chain rule to find dw/dx. 


Implicit Differentiation 

Consider the special case where z = f(x,y) is a function of x and y and y is a differentiable 

function of x. Equation (5.13) then becomes 

dz 

dx = ∂f dx 

∂xdx + ∂f dy 

∂y dx = ∂f 

∂x + ∂f dy 

∂y dx 

(5.16) 

This result can be used to find derivatives of functions that are defined implicitly. For 

example, suppose that the equation 

f(x,y) = c (5.17)


defines y implicitly as a differentiable function of x and we are interested in finding dy/dx. 

Differentiating both sides of (5.17) with respect to x and applying (5.16) yields 

Thus, if ∂f/∂y ≠ 0, we obtain 

In summary, we have the following result. 

∂f 

∂x + ∂f dy 

∂y dx = 0 

dy 

dx = −∂f/∂x ∂f/∂y 

Theorem 5.8 If the equation f(x,y) = c defines y implicitly as a differentiable function 

of x, and if ∂f/∂y ≠ 0, then 

dy 

dx = −∂f/∂x 

(5.18) 

∂f/∂y 

Example 5.32 Given that 

x 3 +y 2 x−3 = 0 

find dy/dx using (5.18), and check the result using implicit differentiation. 


Theorem 5.9 If the equation f(x,y,z) = c defines z implicitly as a differentiable function 

of x and y, and if ∂f/∂z ≠ 0, then 

dz 

dx = −∂f/∂x ∂f/∂z 

and 

dz 

dy = −∂f/∂y ∂f/∂z 

Example 

( 

5.33 Consider the sphere x 2 +y 2 +z 2 = 1. Find ∂z/∂x and ∂z/∂y at the point 

2 , 1 , ) 2 

3 3 3 . 


5.6 Directional Derivatives and Gradients 

Directional Derivatives 

Definition 5.7 If f(x,y) is a function of x and y, and if u = u 1 i + u 2 j is a unit vector, 

then the directional derivative of f in the direction of u at (x 0 ,y 0 ) is denoted by 

D u f(x 0 ,y 0 ) and is defined by 

D u f(x 0 ,y 0 ) = d ds [f(x 0 +su 1 ,y 0 +su 2 )] s=0 (5.19) 

provided this derivative exists.


Geometrically, D u f(x 0 ,y 0 ) can be interpreted as the slope of the surface z = f(x,y) 

in the direction of u at the point (x 0 ,y 0 ,f(x 0 ,y 0 )). Usually the value of D u f(x 0 ,y 0 ) will 

depend on both the point (x 0 ,y 0 ) and the direction u. Thus, at a fixed point the slope of 

the surface may vary with the direction. 

Example 5.34 Let f(x,y) = xy. Find and interpret D u f(1,2) for the unit vector u = 

√ 

3 

2 i+ 1 2 j. 


The definition of a directional derivative for a function f(x,y,z) of three variables is 

similar to Definition 5.7. 

Definition 5.8 If u = u 1 i+u 2 j+u 3 k is a unit vector, and if f(x,y,z) is a function of x, 

y, and z, then the directional derivative of f in the direction of u at (x 0 ,y 0 ,z 0 ) is 

denoted by D u f(x 0 ,y 0 ,z 0 ) and is defined by 

provided this derivative exists. 

D u f(x 0 ,y 0 ,z 0 ) = d ds [f(x 0 +su 1 ,y 0 +su 2 ,z 0 +su 3 )] s=0 (5.20) 

For a function that is differentiable at a point, directional derivatives exist in every 

direction from the point and can be computed directly in terms of the first-order partial 

derivatives of the function. 

Theorem 5.10 

(a) If f(x,y) is differentiable at (x 0 ,y 0 ) and if u = u 1 i + u 2 j is a unit vector, then the 

directional derivative D u f(x 0 ,y 0 ) exists and is given by 

D u f(x 0 ,y 0 ) = f x (x 0 ,y 0 )u 1 +f y (x 0 ,y 0 )u 2 (5.21) 

(b) If f(x,y,z) is differentiable at (x 0 ,y 0 ,z 0 ) and if u = u 1 i+u 2 j+u 3 k is a unit vector, 

then the directional derivative D u f(x 0 ,y 0 ,z 0 ) exists and is given by 

D u f(x 0 ,y 0 ,z 0 ) = f x (x 0 ,y 0 ,x 0 )u 1 +f y (x 0 ,y 0 ,z 0 )u 2 +f z (x 0 ,y 0 ,z 0 )u 3 (5.22) 

Example 5.35 For f(x,y) = x 2 y−4y 3 , compute D u f(2,1) where u is a unit vector in the 

direction from (2,1) to (4,0). 


Recall that a unit vector u in the xy-plane can be expressed as 

u = cosφi+sinφj 

where φ is the angle from the positive x-axis to u. Thus, Formula (5.21) can also be 

expressed as 

D u f(x 0 ,y 0 ) = f x (x 0 ,y 0 )cosφ+f y (x 0 ,y 0 )sinφ (5.23)


Example 5.36 Find the directional derivative of f(x,y) = e xy at (−2,0) in the direction 

of the unit vector that makes an angle of π/3 with the positive x-axis. 


Example 5.37 Find the directional derivative of f(x,y,z) = x 2 y−yz 3 +z at (1,−2,0) in 

the direction of the vector a = 2i+j−2k. 


The Gradient 

Definition 5.9 

(a) If f is a function of x and y, then the gradient of f is defined by 

∇f(x,y) = f x (x,y)i+f y (x,y)j (5.24) 

(b) If f is a function of x, y, and z, then the gradient of f is defined by 

∇f(x,y,z) = f x (x,y,z)i+f y (x,y,z)j+f k (x,y,z)k (5.25) 

Formulas (5.21) and (5.22) can now be written as 

D u f(x 0 ,y 0 ) = ∇f(x 0 ,y 0 )·u (5.26) 

and 

D u f(x 0 ,y 0 ,z 0 ) = ∇f(x 0 ,y 0 ,z 0 )·u (5.27) 

respectively. For example, using Formula (5.27) our solution to Example 5.37 would take 

the form 

D u f(1,−2,0) = ∇f(1,−2,0)·u 

( 2 

= (−4i+j+k)· 

3 i+ 1 3 j− 2 ) 

3 k ( 2 

= (−4) + 

3) 

1 3 − 2 3 = −3 

Formula (5.26) can be interpreted to mean that the slope of the surface z = f(x,y) at the 

point (x 0 ,y 0 ) in the direction of u is the dot product of the gradient with u.


Properties of the Gradient 

Theorem 5.11 Let f be a function of either two variables or three variables, and let P 

denote the point P(x 0 ,y 0 ) or P(x 0 ,y 0 ,z 0 ), respectively. Assume that f is differentiable at 

P. 

(a) If ∇f = 0 at P, then all directional derivatives of f at P are zero. 

(b) If ∇f ≠ 0 at P, then amongall possibledirectional derivativesof f at P, the derivative 

in the direction of ∇f at P has the largest value. The value of this largest directional 

derivative is ‖∇f‖ at P. 

(c) If ∇f ≠ 0 at P, then amongall possibledirectional derivativesof f at P, the derivative 

in the direction opposite to that of ∇f at P has the smallest value. The value of this 

smallest directional derivative is −‖∇f‖ at P. 

Example 5.38 Let f(x,y) = x 2 e y . Find the maximum value of a directional derivative at 

(−2,0), and find the unit vector in the direction in which the maximum value occurs. 

Solution .........

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Multiple Integrals 

6.1 Double Integrals 

Volume 

Recall that the definite integral of a function of one variable 

∫ b 

n∑ 

f(x)dx = lim f(x ∗ k )∆x k = lim 

max∆x k →0 

n→+∞ 

a 

k=1 

n∑ 

f(x ∗ k )∆x k (6.1) 

arose from the problem of finding areas under curves. Integrals of functions of two variables 

arise from the problem of finding volumes under surfaces. 

The Volume Problem. Given a function f of two variables that is continuous 

and nonnegative on a region R in the xy-plane, find the volume of the solid 

enclosed between the surface z = f(x,y) and the region R. 

The procedure for finding the volume V of the solid will be similar to the limiting process 

used for finding areas. We proceed as follows: 

• Using lines parallel to the coordinate axes, divide the rectangle enclosing the region R 

intosubrectangles, andexcludefromconsiderationallthosesubrectanglesthatcontain 

any points outside of R. This leaves only rectangles that are subset of R. Assume 

that there are n such rectangles, and denote the area of the kth such rectangle by 

∆A k . 

y 

k=1 

(x ∗ k ,y∗ k ) 

Area ∆A k 

x 

97


• Choose any arbitrary point in each subrectangle, and denote the point in the kth 

subrectangle by (x ∗ k ,y∗ k ). The product f(x∗ k ,y∗ k )∆A k is the volume of a rectangular 

parallelepiped with base area ∆A k and height f(x ∗ k ,y∗ k ), so the sum 

n∑ 

f(x ∗ k ,y∗ k )∆A k 

k=1 

can be view as an approximation to the volume V of the entire solid. 

• If we repeat the above process with more and more subdivisions in such a way that 

both the lengths and the widths of the subrectangles approach zero, then the exact 

volume of the solid will be 

V = lim 

n→+∞ 

n∑ 

f(x ∗ k,yk)∆A ∗ k 

k=1 

Definition 6.1 (Volume Under a Surface). If f is a function of two variables that is 

continuous and nonnegative on a region R in the xy-plane, then the volume of the solid 

enclosed between the surface z = f(x,y) and the region R is defined by 

V = lim 

n→+∞ 

n∑ 

f(x ∗ k ,y∗ k )∆A k (6.2) 

k=1 

Here, n → +∞ indicates the process of increasing the number of subrectangles of the rectangle 

enclosing R in such a way that both the lengths and the widths of the subrectangles 

approach zero. 

Definition of a Double Integral 

Thesumsin(6.2)arecalledRiemann sums, andthelimitoftheRiemannsumsisdenoted 

by 

∫∫ 

n∑ 

f(x,y)dA = lim f(x ∗ k,yk)∆A ∗ k (6.3) 

n→+∞ 

R 

which is called the double integral of f(x,y) over R. 

If f is continuous and nonnegative on a region R, then the volume formula in (6.2) can 

be expressed as 

∫∫ 

V = f(x,y)dA (6.4) 

R 

If f has both positive and negative values on R, then a positive value for the double integral 

of f over R means that there is move volume above R than below, a negative value for the 

double integral means that there is more volume below R than above, and a value of zero 

means that the volume above R is the same as the volume below R. 

k=1


Evaluating Double Integrals 

Except in the simplest cases, it is impractical to obtain the value of a double integral from 

thelimitin(6.3). However, wewillnowshowhowtoevaluatedoubleintegralsbycalculating 

two successive single integrals. For now we will consider the case where R is a rectangle; in 

the next section we will consider double integrals over more complicated regions. 

Thepartial derivatives ofafunctionf(x,y)arecalculatedby holding oneofthevariables 

fixed and differentiating with respect to the other variable. Let us consider the reverse of 

this process, partial integration. The symbols 

∫ b 

a 

f(x,y)dx and 

∫ d 

c 

f(x,y)dy 

denote partial definite integrals; the first integral, called the partial definite integral 

with respect to x, is evaluated by holding y fixed and integrating with respect to x, and 

the second integral, called the partial definite integral with respect to y, is evaluated 

by holding x fixed and integrating with respect to y. 

Example 6.1 

∫ 1 

0 

∫ 1 

0 

xy 2 dx = 

xy 2 dy = 

Apartial definite integral with respect toxisafunctionof y andhence canbeintegrated 

with respect to y; similarly, a partial definite integral with respect to y can be integrated 

with respect to x. This two-stage integration process is called iterated (or repeated) 

integration. We introduce the following notation: 

∫ d ∫ b 

c 

a 

∫ b 

∫ d 

a 

c 

f(x,y)dxdy = 

f(x,y)dydx= 

These integrals are called iterated integrals. 

∫ d 

c 

∫ b 

a 

[∫ b 

a 

[∫ d 

c 

] 

f(x,y)dx dy (6.5) 

] 

f(x,y)dy dx (6.6) 


(a) 

∫ 3 ∫ 4 

1 2 


(40−2xy)dydx 

(b) 

∫ 4 ∫ 3 

2 1 

(40−2xy)dxdy 

It is no accident that both parts of Example 6.2 produced the same answer. The conclusion 

that the double integral of f(x,y) over R has the same value as either of the two 

possible iterated integrals is true even when f is negative at some points in R.


Theorem 6.1 (Fubini’s Theorem) Let R be the rectangle defined by the inequalities 

a ≤ x ≤ b, c ≤ y ≤ d 

If f(x,y) is continuous on this rectangle, then 

∫∫ 

R 

f(x,y)dA = 

∫ d ∫ b 

c 

a 

f(x,y)dxdy = 

∫ b ∫ d 

a 

c 

f(x,y)dydx 

Example 6.3 Use a double integral to find the volume of the solid that is bounded above 

by the plane z = 4−x−y and below by the rectangle R = [0,1]×[0,2]. 

z 

4 

z = 4−x−y 

1 

(1,2) 

2 

y 

x 


Example 6.4 Evaluate the double integral 

∫∫ 

(6x 2 +4xy 3 )dA 

over the rectangle R = {(x,y) : 0 ≤ x ≤ 2,1 ≤ y ≤ 4}. 


Properties of Double Integrals 

R 

The following properties are analogs of the single integral properties. 

∫∫ ∫∫ 

cf(x,y)dA = c f(x,y)dA (6.7) 

R 

R 

∫∫ 

∫∫ 

[f(x,y)+g(x,y)]dA = 

∫∫ 

f(x,y)dA+ 

g(x,y)dA (6.8) 

R 

∫∫ 

∫∫ 

[f(x,y)−g(x,y)]dA = 

R 

R 

∫∫ 

f(x,y)dA− 

g(x,y)dA (6.9) 

R 

R 

R


It is evident intuitive that if f(x,y) is nonnegative on the region R, then subdividing R 

into two regions R 1 and R 2 has the effect of subdividing the solid between R and z = f(x,y) 

into two solids, the sum of whose volumes is the volumes of the entries solid. This suggests 

the following result. 

∫∫ 

R 

∫∫ ∫∫ 

[f(x,y)+g(x,y)]dA= f(x,y)dA+ f(x,y)dA (6.10) 

R 1 R 2 

6.2 Double Integrals Over Nonrectangular Regions 

Iterated Integrals with Nonconstant Limits of Integration 

Later in this section we will see that double integrals over nonrectangular regions can be 

evaluated as iterated integrals of the following types: 

∫ b ∫ g2 (x) 

a 

g 1 (x) 

∫ d 

∫ h2 (y) 

c 

h 1 (y) 

f(x,y)dydx = 

f(x,y)dxdy = 

We begin with the following example. 


∫ b 

a 

∫ d 

c 

[ ∫ ] 

g2 (x) 

f(x,y)dy dx (6.11) 

g 1 (x) 

[ ∫ ] 

h2 (y) 

f(x,y)dx dy (6.12) 

h 1 (y) 

(a) 

∫ 1 ∫ x 2 

0 −x 


y 2 xdydx 

(b) 

∫ π/3 ∫ cosy 

0 0 

xsinydxdy 

Double Integrals Over Nonrectangular Regions 

We will now limit our study of double integrals to two basic types of regions, which we will 

call type I and type II; they are defined as follows. 

Definition 6.2 

(a) A type I region is bounded on the left and right by vertical lines x = a and x = b 

and is bounded below and above by continuous curves y = g 1 (x) and y = g 2 (x), where 

g 1 (x) ≤ g 2 (x) for a ≤ x ≤ b. 

(b) A type II region is bounded below and above by horizontal lines y = c and y = d 

and is bounded on the left and right by continuous curves x = h 1 (y) and x = h 2 (y), 

where h 1 (y) ≤ h 2 (y) for c ≤ y ≤ d.


y 

y 

y = g 2 (x) 

d 

x = h 1 (y) 

y = g 1 (x) 

a b x 

A type I region 

c 

A type II region 

x = h 2 (y) 

x 

The following theorem will enable us to evaluate double integrals over type I and type 

II regions using iterated integrals. 

Theorem 6.2 

(a) If R is a type I region on which f(x,y) is continuous, then 

∫∫ 

R 

f(x,y)dA = 

∫ b ∫ g2 (x) 

a 

g 1 (x) 

(b) If R is a type II region on which f(x,y) is continuous, then 

∫∫ 

R 

f(x,y)dA = 

∫ d ∫ h2 (y) 

c 

h 1 (y) 

f(x,y)dydx (6.13) 

f(x,y)dxdy (6.14) 

Setting up Limits of Integration for Evaluating Double Integrals 

To apply [ Theorem 6.2, it is helpful to start ] with a two-dimensional sketch of the region 

R. It is not necessary to graph f(x,y). For a type I region, the limits of integration in 

Formula (6.13) can be obtained as follows: 

Determining Limits of Integration: Type I Region 

Step 1. Since x is held fixed for the first integration, we draw a vertical line through the 

region R at an arbitrary fixed value x. This line crosses the boundary of R twice. 

The lower point of intersection is on the curve y = g 1 (x) and the higher point is on 

the curve y = g 2 (x). These two intersection determine the lower and upper y-limits 

of integration in Formula (6.13). 

Step 2. Imagine moving the line drawn in Step 1 first to the left and then to the right. The 

leftmost position where the line intersects the region R is x = a, and the rightmost 

position where the line intersects the region R is x = b. This yields the limits for the 

x-integration in Formula (6.13).


y 

y 

y = g 2 (x) 

y = g 1 (x) 

a b x 

a b x 


∫∫ 

(4e x2 −5siny)dA 

over the region R enclosed between y = x, y = 0, and x = 4. 


R 


∫∫ 

(x+2y)dA 

over the region R enclosed between the graphs of y = 2x 2 and y = 1+x 2 . 


R 

If R is a type II region, then the limits of integration in (6.14)can be obtained asfollows: 

Determining Limits of Integration: Type II Region 

Step 1. Since y is held fixed for the first integration, we draw a horizontal line through the 

region R at a fixed value y. This line crosses the boundary of R twice. The leftmost 

point of intersection is on the curve x = h 1 (y) and the rightmost point is on the curve 

x = h 2 (y). These intersections determine the x-limits of integration in (6.14). 

Step 2. Imagine moving the line drawn in Step 1 first down and then up. The lowest 

position where the line intersects the region R is y = c, and the highest position 

where the line intersects the region R is y = d. This yields the limits for the y-limit 

of integration in (6.14).


y 

y 

d 

d 

x = h 1 (y) 

c 

x = h 2 (y) 

x 

c 

x 


∫∫ 

xydA 

over the region R enclosed between the graphs of y = x−1 and y 2 = 2x+6. 



∫∫ 

R 

R 

(2x−y 2 )dA 

over the triangular region R enclosed between the lines y = −x+1, y = x+1, and y = 3. 


Example 6.10 Use a double integral to find the volume of the tetrahedron bounded by the 

coordinate planes and the plane z = 4−4x−2y. 


Example 6.11 Find the volume of the solid bounded by the cylinder x 2 +y 2 = 4 and the 

plane y +z = 4 and z = 0. 


Reversing the Order of Integration 

Sometimes the evaluation of an iterated integral can be simplified by reversing the order of 

integration. The next example illustrates how this is done. 

Example 6.12 Since there is no elementary antiderivative of e x2 , the integral 

∫ 2 ∫ 1 

0 

y/2 

e x2 dxdy 

can not be evaluated by performing the x-intersection first. Evaluate this integral by expressing 

it as an equivalent iterated integral with the order of integration reversed. 

Solution .........


Area Calculated as a Double Integral 

Although double integrals arose in the context of calculating volumes, they can also be used 

to calculated areas. That is, 

∫∫ 

area of R = 

∫∫ 

1dA = 

dA (6.15) 

R 

R 

Example 6.13 Use a double integral to find the area of the region R enclosed between the 

parabola y = 1 2 x2 and the line y = 2x. 


6.3 Double Integrals in Polar Coordinates 

Simple Polar Regions 

Some double integrals are easier to evaluate if the region of integration is expressed in polar 

coordinates. This is usually if the region is bounded by a cardioid, a rose curve, a spiral, 

or, more generally, by any curve whose equation is simpler in polar coordinates than in 

rectangular coordinates. 

Figure(a) shows a region R in a polar coordinate system that is enclosed between two 

rays, θ = α and θ = β, and two polar curves, r = r 1 (θ) and r = r 2 (θ). If the functions 

r 1 (θ) and r 2 (θ) are continuous and their graphs do not cross, then the region R is called a 

simple polar region. If r 1 (θ) is identically zero, then the boundary r = r 1 (θ) reduces to a 

point (the origin), and the region has the general shape shown in Figure(b). If, in addition, 

β = α+2π, then therays coincide, andthe regionhas the general shape shown inFigure(c). 

θ = β 

r = r 2 (θ) 

R 

r = r 1 (θ) 

θ = β 

r = r 2 (θ) 

R 

r = r 2 (θ) 

0 

β 

α 

R 

θ = α 

0 

R 

θ = α 

0 

RR 

β = α+2π 

(a) 

(b) 

(c) 

Definition 6.3 A simple polar region in a polar coordinate is a region that is enclosed 

between two rays, θ = α and θ = β, and two continuous polar curves, r = r 1 (θ) and r = 

r 2 (θ), where the equations of the rays and the polar curves satisfy the following conditions: 

(i) α ≤ β (ii) β −α ≤ 2π (iii) 0 ≤ r 1 (θ) ≤ r 2 (θ)

• 


A polar rectangle is a simple polar region for which the bounding polar curves are 

circular arcs. For example, Figure below shows the polar rectangle R given by 

2 ≤ r ≤ 4, 

π 

12 ≤ θ ≤ π 3 

θ = π 3 

r = 2 

R 

r = 4 

θ = π 12 

o 

Double Integrals in Polar Coordinates 

The Volume Problem in Polar Coordinates. Given a function 

f(r,θ) that is continuous and nonnegative on a simple polar region R 

in the xy-plane, find the volume of the solid that is enclosed between 

the region R and the surface whose equation in cylindrical 

coordinates is z = f(r,θ). 

To motivate a formula for the volume V of the solid, we will use a limit process similar to 

that used to obtain Formula (6.2) of Section15.1, except that here we will use circular arcs 

and rays to subdivide the region R into polar rectangles. 

Area ∆A k 

(r ⋆ k ,θ⋆ k ) 

0 

Assumethattherearensuchpolarrectangles, anddenotetheareaofthekthpolarrectangle 

by ∆A k . Let (rk ⋆,θ⋆ k ) be any point in this polar rectangle. The product f(r⋆ k ,θ⋆ k )∆A k is the 

volume of a solid with base area ∆A k and height f(rk ⋆,θ⋆ k ), so the sum 

n∑ 

f(rk ⋆ ,θ⋆ k )∆A k 

can be viewed as an approximation to the volume V of the entire solid. 

k=1


If we now increase the number of subdivisions in such a way that the dimensions of the 

polar rectangles approach zero, then the exact volume of the solid is 

V = lim 

n→+∞ 

n∑ 

f(rk ⋆ ,θ⋆ k )∆A k (6.16) 

k=1 

If f(r,θ) is continuous on R and has both positive and negative values, then the limit 

lim 

n→+∞ 

n∑ 

f(rk ⋆ ,θ⋆ k )∆A k (6.17) 

k=1 

represent the net signed volume between the region R and the surface z = f(r,θ). The 

sums in (6.17) are called polar Riemann sums, and the limit of the polar Riemann sums 

is denoted by 

∫∫ 

n∑ 

f(r,θ)dA = lim f(rk ⋆ n→+∞ 

,θ⋆ k )∆A k (6.18) 

R 

which is called the polar double integral of f(r,θ) over R. If f(r,θ) is continuous and 

nonnegative on R, then the volume formula (6.16) can be expressed as 

∫∫ 

V = 

R 

k=1 

f(r,θ)dA (6.19) 

Evaluating Polar Double Integrals 

Theorem 6.3 If R is a simple polar region whose boundaries are the rays θ = α and θ = β 

and the curves r = r 1 (θ) and r = r 2 (θ) shown in Figure below, 

θ = β 

r = r 2 (θ) 

r = r 1 (θ) 

θ 

θ = α 

and if f(r,θ) is continuous on R, then 

0 

∫∫ 

R 

f(r,θ)dA = 

∫ β 

∫ r2 (θ) 

α r 1 (θ) 

f(r,θ)rdrdθ (6.20)

• 

• 


Determining Limits of Integration for a Polar Double Integral: 

Simple Polar Region 

Step 1. Since θ is held fixed for the first integration, draw a radial line from the origin 

through the region R at a fixed angle θ. This line crosses the boundary of R at most 

twice. The innermost point of intersection is on the inner boundary curve r = r 1 (θ) 

and the outermost point is on the outer boundary curve r = r 2 (x). These intersection 

determine the r-limits of integration in (6.20). 

Step 2. Imagine rotating the radial line from Step 1 about the origin, thus sweeping out 

the region R. The least angle at which the radial line intersects the region R is θ = α 

and the greatest angle is θ = β. This determines the θ-limits of integration. 

θ = β 

r = r 2 (θ) 

θ = β 

r = r 1 (θ) 

θ 

θ = α 

θ 

θ = α 

0 

0 


∫∫ 

sinθdA 

R 

where R is the region in the first quadrant that is outside the circle r = 2 and inside the 

cardioid r = 2(1+cosθ) 


Example 6.15 The sphere of radius a centered at the origin is expressed in rectangular 

coordinates as x 2 +y 2 +z 2 = a 2 , and hence its equation in cylindrical coordinates is r 2 +z 2 = 

a 2 . Use this equation and a polar double integral to find the volume of the sphere. 


Finding Areas Using Polar Double Integrals 

Recall from Formula (6.15) of Section15.2 that the area of a region R in the xy-plane can 

be expressed as 

∫∫ ∫∫ 

area of R = 1dA = dA (6.21) 

R 

The argument used to derive this result can also be used to show that the formula applies 

to polar double integrals over regions in polar coordinates. 

R

• 


Example 6.16 Use a polar double integral to find the area enclosed by the three-petaled 

rose r = sin3θ 


Converting Double Integrals from Rectangular to Polar Coordinates 

Sometimes a double integral that is difficult to evaluate in rectangular coordinates can be 

evaluatedmoreeasilyinpolarcoordinatesbymakingthesubstitutionx = rcosθ, y = rsinθ 

andexpressing theregion ofintegration in polarform; that is, we rewrite the double integral 

in rectangular coordinates as 

∫∫ 

∫∫ 

f(x,y)dA = 

∫∫ 

f(rcosθ,rsinθ)dA = 

f(rcosθ,rsinθ)rdrdθ (6.22) 

R 

R 

R 

Example 6.17 Use polar coordinates to evaluate 


∫ 1 ∫ √ 1−x 2 

−1 0 

(x 2 +y 2 ) 3/2 dydx. 

6.4 Triple Integrals 

Definition of a Triple Integral 

A single integral of a function f(x) is defined over a finite closed interval on the x-axis, 

and a double integral of a function f(x,y) is defined over a finite closed region R in the 

xy-plane. Our goal in this section is to define what is meant by a triple integral of f(x,y,z) 

over a closed solid region G in an xyz-coordinate system. 

Todefinethetripleintegraloff(x,y,z)overG, wefirstdividetheboxinton“subboxes” 

by plane parallel to the coordinate planes. As shown in Figure below, we denote the volume 

ofthekthremainingsubboxby∆V k andthepointselected inthekthsubboxby(x ⋆ k ,y⋆ k ,z⋆ k ). 

z 

Volume = ∆V k 

x 

0 

y 

(x ⋆ k ,y⋆ k ,z⋆ k )


Next we form the product 

f(x ⋆ k,y ⋆ k,z ⋆ k)∆V k 

for each subbox, then add the products for all the subboxes to obtain the Riemann sum 

n∑ 

f(x ⋆ k ,y⋆ k ,z⋆ k )∆V k 

k=1 

Finally, we repeat this process with more and more subdivisions in such a way that the 

length, width, and height of each subbox approach zero, and n approaches +∞. The limit 

∫∫∫ 

G 

f(x,y,z)dV = lim 

n→+∞ 

n∑ 

f(x ⋆ k,yk,z ⋆ k)∆V ⋆ k (6.23) 

k=1 

is called the triple integral of f(x,y,z) over the region G. 

Properties of Triple Integrals 

Triple integrals enjoy many properties of single and double integrals: 

∫∫∫ ∫∫∫ 

cf(x,y,z)dV = c f(x,y,z)dV (c a constant) 

∫∫∫ 

G 

∫∫∫ 

[f(x,y,z)+g(x,y,z)]dV = 

G 

∫∫∫ 

f(x,y,z)dV + 

g(x,y,z)dV 

G 

∫∫∫ 

∫∫∫ 

[f(x,y,z)−g(x,y,z)]dV = 

G 

G 

∫∫∫ 

f(x,y,z)dV − 

g(x,y,z)dV 

G 

G 

G 

Moreover, if the region G is subdivided into two subregions G 1 and G 2 , then 

∫∫∫ ∫∫∫ ∫∫∫ 

f(x,y,z)dV = f(x,y,z)dV + f(x,y,z)dV 

G 1 G 2 

G 

G 

G 1 

G 2 

Evaluating Triple Integrals Over Rectangular Boxes 

Theorem 6.4 (Fubini’s Theorem) Let G be the rectangular box defined bythe inequalities 

a ≤ x ≤ b, c ≤ y ≤ d, k ≤ z ≤ l


If f is continuous on the region G, then 

∫∫∫ 

G 

f(x,y,z)dV = 

∫ b ∫ d ∫ l 

a 

c 

k 

f(x,y,z)dzdydx (6.24) 

Moreover, the iterated integral on the right can be replaces with any of the five other iterated 

integrals that result by altering the order of integration. 

Example 6.18 Evaluate the triple integral 

∫∫∫ 

xyz 2 dV 

G 

over the rectangular box G defined by the inequalities 0 ≤ x ≤ 1, −1 ≤ y ≤ 2, 0 ≤ z ≤ 3. 


Evaluating Triple Integrals Over More General Regions 

Nextwewillconsiderhowtripleintegralcanbeevaluatedoversolidsthatarenotrectangular 

boxes. We will assume that the solid G is bounded above by a surface z = g 2 (x,y) and 

below by a surface z = g 1 (x,y) and that the projection of the solid on the xy-plane is a 

type I or type II region R. 

z 

z = g 2 (x,y) 

G 

z = g 1 (x,y) 

x 

R 

y 

Inaddition, wewillassumethatg 1 (x,y)andg 2 (x,y)arecontinuousonRandthatg 1 (x,y) ≤ 

g 2 (x,y) on R. Geometrically, this means that the surfaces may touch but cannot cross. We 

call a solid of this type a simple xy-solid. 

Theorem 6.5 Let G be a simple xy-solid with upper surface z = g 2 (x,y) and lower surface 

z = g 1 (x,y), and let R be the projection of G on the xy-plane. If f(x,y,z) is continuous 

on G, then 

∫∫∫ ∫∫ [ ∫ ] 

g2 (x,y) 

f(x,y,z)dV = f(x,y,z)dz dA (6.25) 

G 

R 

g 1 (x,y)


Determining Limits of Integration: Simple xy-Solid 

Step 1. Find an equation z = g 2 (x,y) for the upper surface and an equation z = g 1 (x,y) 

for the lower surface of G. The functions g 1 (x,y) and g 2 (x,y) determine the lower 

and upper z-limits of integration. 

Step 2. Make a two-dimensional sketch of the projection R of the solid on the xy-plane. 

From this sketch determine the limits of integration for the double integral over R in 

(6.25). 

Example 6.19 Let G be the wedge in the first octant that is cut from the cylindrical solid 

y 2 +z 2 ≤ 1 by the planes y = x and x = 0. 

y = x 

z 

y 2 +z 2 = 1 

(z = √ 1−y 2 ) 

x = 0 

1 

y 

Evaluate 

x∫∫∫ 

zdV 



∫∫∫ 

G 

6xydV, where G is the tetrahedron by the planes x = 0, 

y = 0, z = 0, and 2x+y +z = 4. 

G 

z 

2x+y +z = 4 

x 

2 

R 

4 

y 


Volume Calculated as a Triple Integral 

In the case where f(x,y,z) = 1, Formula (6.23) yields 

∫∫∫ 

G 

dV = lim 

n→+∞ 

n∑ 

∆V k 

k=1

• 


which is the volume of G; that is, 

∫∫∫ 

volume of G = 

dV (6.26) 

Example 6.21 Use a triple integral to find the volume of the solid within the cylinder 

x 2 +y 2 = 9 and between the planes z = 1 and x+z = 5. 

z 

G 

x+z = 5 

x 2 +y 2 = 9 

z = 1 

x 

R 

x 2 +y 2 = 9 

y 


Example 6.22 Find the volume of the solid G enclosed between the paraboloids 


z = 5x 2 +5y 2 and z = 6−7x 2 −y 2 

Example 6.23 Find the volume of the solid G enclosed between the graph of y = 4−x 2 −z 2 

and the xz-plane. 


Integration in Other Orders 

In Formula (6.25) for integrating over a simple xy-solid, the z-integration was performed 

first. However, there are situations in which it is preferable to integrate is a different order. 

For a simple xz-solid it is usually best to integrate with respect to y first, and for yz-solid 

it is usually best to integrate with respect to x first: 

∫∫∫ ∫∫ [ ∫ ] 

g2 (x,z) 

f(x,y,z)dV = f(x,y,z)dy dA 

G 

simple xz-solid 

∫∫∫ 

G 

simple yz-solid 

R 

∫∫ 

f(x,y,z)dV = 

R 

g 1 (x,z) 

[ ∫ ] 

g2 (y,z) 

f(x,y,z)dx dA 

g 1 (y,z)


Example 6.24 In Example 6.19, we evaluated 

∫∫∫ 

zdV 

G 

over the wedge G by integrating first with respect to z. Evaluate this integral by integrating 

first with respect to x. 

Solution .........


Topics in Vector Calculus 

7.1 Vector Fields 

Definition 7.1 A vector field in a plane is a function that associates with each point P 

in the plane a unique vector F(P) parallel to the plane. Similarly, a vector field in 3-space 

is a function that associates with each point P in 3-space a unique vector F(P) in 3-space. 

If F(P) is a vector field in an xy-coordinate system, then the point P will have some 

coordinate (x,y) and the associated vector will have components that are functions of x 

and y. Thus, the vector field F(P) can be expressed as 

F(x,y) = f(x,y)i+g(x,y)j 

Similarly in 3-space with an xyz-coordinate system, a vector field F(P) can be expressed 

as 

F(x,y,z) = f(x,y,z)i+g(x,y,z)j+h(x,y,z)k 

Graphical Representations of Vector Fields 

A vector field in 2-space can be pictured geometrically by drawing representative field 

vectors F(x,y) at some well-chosen points in the xy-plane. 

115


A Compact Notation for Vector Fields 

Sometimes it is helpful to denote the vector fields F(x,y) and F(x,y,z) entirely in vector 

notation by identifying (x,y) with the radius vector r = xi+yj and (x,y,z) with the radius 

vector r = xi+yj+zk. With this notation a vector field in either 2-space or 3-space can 

be written as F(r). 

Gradient Fields 

An important class of vector fields arise from the process of finding gradients. Recall that 

if φ is a function of three variables, then the gradient of φ is defined as 

∇φ = ∂φ ∂φ ∂φ 

i+ j+ 

∂x ∂y ∂z k 

This formula defines a vector field in 3-space called the gradient field of φ. Similarly, 

the gradient field of a function of two variables defines a gradient field in 2-space. At each 

point in a gradient field where the gradient is nonzero, the vector points in the direction in 

which the rate of increase of φ is maximum. 

Example 7.1 Sketch the gradient field of φ(x,y) = x+y. 


Conservative Fields and Potential Functions 

IfF(r) isanarbitraryvector fieldin2-spaceor 3-space, we canaskwhether it isthe gradient 

field of some function φ, and if so, how we can find φ. This is an important problem in 

various applications. There is some terminology for such fields that we will introduce now. 

Definition 7.2 A vector field F in 2-space or 3-space is said to be conservative in a 

region if it is the gradient field for some function φ in that region; that is, if 

F = ∇φ 

The function φ is called a potential function for F in the region. 

In a later section we will discuss methods for finding potential functions for conservative 

vector fields. 

Divergence and Curl 

We will now define two important operations on vector fields in 3-space—the divergence 

and the curl of the field. These names originate in the study of fluid flow, in which case 

the divergence relates to the way in which fluid flows toward or away from a point and the 

curl relates to the rotational properties of the fluid at a point.


Definition 7.3 If F(x,y,z) = f(x,y,z)i + g(x,y,z)j + h(x,y,z)k, then we define the 

divergence of F, written divF, to be the function given by 

divF = ∂f 

∂x + ∂g 

∂y + ∂h 

∂z 

(7.1) 

Definition 7.4 If F(x,y,z) = f(x,y,z)i+g(x,y,z)j+h(x,y,z)k, then we define the curl 

of F, written curlF, to be the vector field given by 

curlF = 

( ∂h 

∂y − ∂g ) ( ∂f 

i+ 

∂z ∂z − ∂h ) ( ∂g 

j+ 

∂x ∂x − ∂f ) 

k (7.2) 

∂y 

Note that divF has scalar values, whereas curlF has vector values (i.e., curlF is itself 

a vector field). Moreover, for computational purposes it is useful to note that the formula 

for the curl can be expressed in the determinant form 

i j k 

∂ ∂ ∂ 

curlF = 

∂x ∂y ∂z 

∣ f g h ∣ 

Example 7.2 Find the divergence and the curl of the vector field 


F(x,y,z) = x 2 yi+2y 3 zj+3zk 

Example 7.3 Show that the divergence of the vector field 

is zero. 


The ∇ Operator 

F(x,y,z) = 

c 

(xi+yj+zk) 

(x 2 +y 2 +z 2 ) 

3/2 

(7.3) 

Thus far, the symbol ∇ that appears in the gradient expression ∇φ has not been given a 

meaning of its own. However, it is often convenient to view ∇ as an operator 

which when applied to φ(x,y,z) produces the gradient 

∇ = ∂ 

∂x i+ ∂ 

∂y j+ ∂ ∂z k (7.4) 

∇φ = ∂φ ∂φ ∂φ 

i+ j+ 

∂x ∂y ∂z k


We call (7.4) the del operator. This is analogous to the derivative operator d/dx, which 

when applied to f(x) produces the derivative f ′ (x). 

The del operator allows us to express the divergence of a vector field 

in dot product notation as 

F(x,y,z) = f(x,y,z)i+g(x,y,z)j+h(x,y,z)k (7.5) 

divF = ∇·F = ∂f 

∂x + ∂g 

∂y + ∂h 

∂z 

and the curl of this field in cross-product notation as 

i j k 

∂ ∂ ∂ 

curlF = ∇×F = 

∂x ∂y ∂z 

∣ f g h ∣ 

(7.6) 

(7.7) 

The Laplacian ∇ 2 

The operatorthat results by taking thedot product ofthe del operator with itself is denoted 

by ∇ 2 and is called the Laplacian operator. This operator has the form 

∇ 2 = ∇·∇ = ∂2 

∂x 2 + ∂2 

∂y 2 + ∂2 

∂z 2 (7.8) 

When applied to φ(x,y,z) the Laplacian operator produces the function 

∇ 2 φ = ∂2 φ 

∂x + ∂2 φ 

2 ∂y + ∂2 φ 

2 ∂z 2 

Note that ∇ 2 φ can also be expressed as div(∇φ). The equation ∇ 2 φ = 0 or, equivalently, 

is known as Laplace’s equation. 

7.2 Line Integral 

Line Integral 

∂ 2 φ 

∂x + ∂2 φ 

2 ∂y + ∂2 φ 

2 ∂z = 0 2 

The first goal of this section is to define what it means to integrate a function along a curve. 

To motivate the definition we will consider the problem of finding the mass of a very thin 

wire whose linear density function (mass per unit length) is known. We assume that we 

can model the wire by a smooth curve C between points P and Q in 3-space.

• 

• 

• 

• 

• 

• 

• 

• 


Q 

• 

• 

P 

Givenanypoint (x,y,z)onC, we letf(x,y,z)denotethecorresponding valueofthedensity 

function. To compute the mass of the wire, we process as follows: 

• Divide C into n very small sections using a succession of distinct partition points 

P = P 0 ,P 1 ,P 2 ,...,P n−1 ,P n = Q 

Let ∆M k be the mass of the kth section, and let ∆s k be the length of the arc between 

P k−1 and P k . 

P ∗ k (x∗ k ,y∗ k ,z∗ k ) 

P k 

P k−1 

P k 

• 

P k−1 

• 

Q 

• 

• 

• 

P = P 0 

• Choose anarbitrary sampling point Pk ∗(x∗ k ,y∗ k ,z∗ k ) onthe kth arc. If ∆s k is vary small, 

the value of f will not vary much along the kth section and we can approximate f 

along this section by the value f(x ∗ k ,y∗ k ,z∗ k ). It follows that the mass of the kth section 

can be approximated by 

∆M k ≈ f(x ∗ k ,y∗ k ,z∗ k )∆s k 

• The mass M of the entire wire can then be approximated by 

n∑ n∑ 

M = ∆M k ≈ f(x ∗ k ,y∗ k ,z∗ k )∆s k (7.9) 

k=1 

• We will use the expression max∆s k → 0 to indicate the process of increasing n in 

such a way that the lengths of all sections approach 0. Thus the exact value of M 

will be given by 

M = 

lim 

max∆s k →0 

k=1 

P 1 

P 2 

n∑ 

f(x ∗ k,yk,z ∗ k)∆s ∗ k (7.10) 

k=1


The limit in (7.10) is similar to the limit of Riemann sums used to define integral of a 

function over an interval. With this similarity in mind, we make the following definition. 

Definition 7.5 If C is a smooth curve in 2-space or 3-space, then the line integral of f 

with respect to s along C is 

or 

∫ 

C 

∫ 

C 

f(x,y)ds = lim 


f(x,y,z)ds = lim 


n∑ 

f(x ∗ k ,y∗ k )∆s k 2-space (7.11) 

k=1 

n∑ 

f(x ∗ k ,y∗ k ,z∗ k )∆s k 3-space (7.12) 

k=1 

provided this limit exists and does not depend on the choice of partition or on the choice of 

sample points. 

• If C is a curve in 3-space that model as a thin wire, and if f(x,y,z) is a linear density 

function of the wire, then it follows from (7.10) and Definition 7.5 that the mass M 

of the wire is given by 

∫ 

M = f(x,y,z)ds (7.13) 

C 

• If C is a smooth curve of arc length L, and f is identically, then it follows from 

Definition 7.5 that 

Evaluating Line Integral 

∫ 

C 

ds = lim 


n∑ 

k=1 

∆s k = lim 

max∆s k →0 L = L (7.14) 

Except in simple cases, it will not be feasible to evaluate a line integral directly from (7.11) 

or (7.12). However, we can show that it is possible to express a line integral as an ordinary 

definite integral. For example, suppose that C is a curve in the xy-plane that is smoothly 

parametrized by 

r(t) = x(t)i+y(t)j (a ≤ t ≤ b) 

Then 

∫ 

C 

f(x,y)ds = 

∫ b 

a 

f(x(t),y(t))‖r ′ (t)‖dt (7.15) 

Similarly, if C is a curve in 3-space that is smoothly parametrized by 

r(t) = x(t)i+y(t)j+z(t)k (a ≤ t ≤ b) 

Then 

∫ 

f(x,y,z)ds = 

∫ b 

C 

a 

f(x(t),y(t),z(t))‖r ′ (t)‖dt (7.16)


Example 7.4 Using the given parametrization, evaluate the line integral 

∫ 

(1+xy 2 )ds. 

(a) C : r(t) = ti+2tj (0 ≤ t ≤ 1) 

C 

(b) C : r(t) = (1−t)i+(2−2t)j (0 ≤ t ≤ 1) 

y 

y 

2 

(1,2) 

2 

(1,2) 

C 

C 

1 

(a) 

x 

1 

(b) 

x 


Note that the integrals in part (a) and (b) of Example 7.4 agree, even though the corresponding 

parametrizations of C have opposite orientations. This illustrates the important 

result that the value of a line integral of f along C does not depend on an orientation of C. 

Formula (7.15) has an alternative expression for a curve C in the xy-plane that is given 

by parametric equations 

x = x(t), y = y(t) (a ≤ t ≤ b) 

In this case, we write (7.15) in the expanded form 

∫ 

C 

f(x,y)ds = 

∫ b 

a 

√ (dx ) 2 

f(x(t),y(t)) + 

dt 

Similarly, if C is a curve in 3-space that is parametrized by 

Then we write (7.16) in the form 

∫ 

C 

f(x,y,z)ds = 

x = x(t), y = y(t), z = z(t) (a ≤ t ≤ b) 

∫ b 

a 

√ (dx ) 2 

f(x(t),y(t),z(t)) + 

dt 

( ) 2 dy 

dt (7.17) 

dt 

( ) 2 dy 

+ 

dt 

( ) 2 dz 

dt (7.18) 

dt 

Example 7.5 Evaluate the line integral ∫ C (xy−z3 )ds from (1,0,0) to (−1,0,π) along the 

helix C that is represented by parametric equations 

x = cost, y = sint, z = t (0 ≤ t ≤ π)


π 

(−1,0,π) 0 

y 

1 


z 

0 

(1,0,0) 0 

1 

x 

−1 

If δ(x,y) is the linear density function of a wire that is modeled by a smooth curve C in 

the xy-plane, then an argument similar to the derivative of Formula (7.13) shows that the 

mass of the wire is given by ∫ C δ(x,y)ds. 

Example 7.6 Suppose that a semicircular wire has the equation y = √ 25−x 2 and that its 

mass density is δ(x,y) = 15−y. 

y 

5 

y = √ 25−x 2 

Find the mass of the wire. 


−5 

5 

x 

In the special case where t is an arc length parameter, sat t = s, it follows from Formula 

(4.34) and (4.35) in Section4.3 that the radicals in (7.17) and (7.18) reduce to 1 and the 

equations simplify to 

∫ ∫ b 

f(x,y)ds = f(x(s),y(s))ds (7.19) 

and 

respectively. 

∫ 

C 

C 

f(x,y,z)ds = 

∫ b 

Line integrals with respect to x, y, and z 

a 

a 

f(x(s),y(s),z(s))ds (7.20) 

We now describe a second type of line integral in which we replace the “ds” in the integral 

by dx, dy, or dz. For example, suppose that f is a function defined on a smooth curve C 

in the xy-plane and that partition points of C are denoted by P k (x k ,y k ). Letting 

∆x k = x k −x k−1 

and ∆y k = y k −y k−1


we would like to define 

∫ 

∫ 

C 

C 

f(x,y)dx = lim 


f(x,y)dy = lim 


n∑ 

f(x ∗ k,yk)∆x ∗ k (7.21) 

k=1 

n∑ 

f(x ∗ k ,y∗ k )∆y k (7.22) 

However the values of ∆x k and ∆y k change sign if the order of the partition points along C 

is reversed. Therefore, in order to define the line integrals using Formulas (7.21) and (7.22), 

we must restrict ourselves to oriented curves C and to partition of C in which the partition 

points are ordered in the direction of the curve. We refer to (7.21) as the line integral 

of f with respect to x along C. Similarly, (7.22) defines the line integral of f with 

respect to y along C. If C is s smooth curve in 3 space, we can have line integrals of 

f with respect to x, y, and z along C. For example, 

∫ 

C 

f(x,y,z)dx = 

lim 


k=1 

n∑ 

f(x ∗ k,yk,z ∗ k)∆x ∗ k 

and so forth. 

The basic procedure for evaluating these line integrals is to find parametric equations 

for C, say 

x = x(t), y = y(t), z = z(t) (a ≤ t ≤ b) 

in which the orientation of C is in the direction of increasing t, and then express the 

integrand in terms of t. For example, 

∫ 

C 

f(x,y,z)dz = 

∫ b 

a 

k=1 

f(x(t),y(t),z(t))z ′ (t)dt 

Example 7.7 Evaluate ∫ 3xydy, where C is the line segment joining (0,0) and (1,2) with 

C 

the given orientation. 

(a) Oriented from (0,0) to (1,2). 

(b) Oriented from (1,2) to (0,0). 

y 

y 

2 

(1,2) 

2 

(1,2) 

C 

C 

1 

(a) 

x 

1 

(b) 

x



If C is a smooth oriented curve, we will let −C denote the oriented curve consisting of 

the same point as C but with the opposite orientation. 

B 

B 

C 

−C 

We then have 

∫ ∫ 

f(x,y)dx = − 

while 

−C 

C 

A 

f(x,y)dx and 

∫ 

−C 

A 

∫ 

∫ 

f(x,y)ds = 

−C 

C 

∫ 

g(x,y)dy = − g(x,y)dy (7.23) 

C 

f(x,y)ds (7.24) 

Similar identities hold for line integrals in 3-space. 

Frequently, the line integrals with respect to x and y occur in combination, in which 

case we will despense with one of the integral signs and write 

∫ 

∫ ∫ 

f(x,y)dx+g(x,y)dy = f(x,y)dx+ g(x,y)dy (7.25) 

C 

We will use a similar convention for combinations of line integrals with respect to x, y, and 

z along curves in 3-space. 

∫ 

Example 7.8 Evaluate 2xydx+(x 2 + y 2 )dy along the circular C given by x = cost, 

y = sint (0 ≤ t ≤ π/2). 

C 

y 

C 

C 

1 

C 


1 

x 

It can be shown that if f and g are continuous functions on C, then combinations of line 

integrals with respect to x and y can be evaluated together in a single step. For example, 

∫ 

∫ b 

f(x,y)dx+g(x,y)dy = [f(x(t),y(t))x ′ (t)+g(x(t),y(t))y ′ (t)]dt (7.26) 

C 

a 

Similar results hold for line integrals in 3-space.



y = sint (0 ≤ t ≤ π/2). 

∫ 

(3x 2 +y 2 )dx+2xydy along the circular C given by x = cost, 

C 

y 

1 

C 

1 

x 


It follows from (7.23) that 

∫ 

−C 

∫ 

f(x,y)dx+g(x,y)dy = − f(x,y)dx+g(x,y)dy (7.27) 

C 

so that reversing the orientation of C changes the sign of a line integral in which x and y 

occur in combination. Similarly, 

∫ 

−C 

f(x,y,z)dx+g(x,y,z)dy+h(x,y,z)dz 

∫ 

= − f(x,y,z)dx+g(x,y,z)dy+h(x,y,z)dz (7.28) 

Integrating a Vector Field Along a Curve 

C 

There is an alternative notation for line integrals with respect to x, y, and z that is particularly 

appropriate for dealing with problems involving vector fields. We will interpret dr 

as 

dr = dxi+dyj or dr = dxi+dyj+dzk 

depending on whether C is in 2-space or 3-space. For an oriented curve C in 2-space and a 

vector field 

F = f(x,y)i+g(x,y)j 

we will write 

∫ ∫ 

F·dr = 

C 

C 

∫ 

(f(x,y)i+g(x,y)j)·(dxi+dyj) = 

Similarly, for a curve C in 3-space and vector field 

F = f(x,y,z)i+g(x,y,z)j+h(x,y,z)k 

C 

f(x,y)dx+g(x,y)dy (7.29)


we will write 

∫ ∫ 

F·dr = (f(x,y,z)i+g(x,y,z)j+h(x,y,z)k)·(dxi+dyj+dzk) 

C 

∫C 

= f(x,y,z)dx+g(x,y,z)dy+h(x,y,z)dz 

C 

With these conventions, we are led to the following definition. 

(7.30) 

Definition 7.6 If F is a continuous vector field and C is a smooth oriented curve, then 

the line integral of F along C is 

∫ 

C 

F·dr (7.31) 

For example, suppose that C is an oriented curve in the plane given in vector form by 

r = r(t) = x(t)i+y(t)j (a ≤ t ≤ b) 

If we write 

then 

F(r(t)) = f(x(t),y(t))i+g(x(t),y(t))j 

∫ 

F·dr = 

∫ b 

C a 

F(r(t))·r ′ (t)dt (7.32) 

Example 7.10 Evaluate ∫ F·dr, where F(x,y) = cosxi+sinxj and where C is the given 

C 

oriented curve. 

(a) C : r(t) = − π i+tj (1 ≤ t ≤ 2) 

2 

(b) C : r(t) = ti+t 2 j (−1 ≤ t ≤ 2) 


If we let t denote an arc length parameter, say t = s, with T = r ′ (s) the unit tangent 

vector field along C, then 

∫ 

C 

which shows that 

F·dr = 

∫ b 

a 

F(r(s))·r ′ (t)ds = 

∫ 

C 

∫ 

F·dr = 

∫ b 

C 

a 

∫ 

F(r(s))·Tds = 

C 

F·Tds 

F·Tds (7.33) 

In words, the integral of a vector field along a curve has the same value as the integral of 

the tangential component of the vector field along the curve.


We can use (7.33) to interpret ∫ F·dr geometrically. If θ is the angle between F and 

C 

T at a point on C, then at this point 

F·T = ‖F‖‖T‖cosθ 

= ‖F‖cosθ Since ‖T‖ = 1 

Thus, 

−‖F‖ ≤ F·T ≤ ‖F‖ 

Example 7.11 Use (7.33) to evaluate ∫ F·dr where F(x,y) = −yi+xj and where C is 

C 

the given oriented curve. 

(a) C : x 2 +y 2 = 3 (0 ≤ x,y;oriented as Figure below) 

(b) C : r(t) = ti+2tj (0 ≤ t ≤ 1;see Figure below) 

y 

y 

√ 

3 

C 

√ 

3 

2 

C 

(a) 


x 

1 

(b) 

x 

Work as a Line Integral 

InSection11.3 we tooktheconcept ofworkastepfurther bydefining thework W performed 

by a constant force F moving a particle in a straight line from point P to point Q. We 

defined the work to be 

W = F·−→ PQ 

Our next goal isto define a moregeneral concept of work—the work performedby a variable 

force acting on a particle that moves along a curve path in 2-space or 3-space. 

Definition 7.7 Suppose that under the influence of a continuous force field F a particle 

moves along a smooth curve C and that C is oriented in the direction of motion of the 

particle. Then the work performed by the force field on the particle is 

∫ 

W = 

C 

F·dr (7.34)

• 

• 

• 

• 


Line Integrals Along Piecewise Smooth Curves 

Thus far, we have only considered line integrals along smooth curves. However, the notion 

of a line integral can be extended to curves formed from finitely many smooth curves C 1 , 

C 2 , ..., C n joined end to end. Such a curve is called piecewise smooth. 

C 4 

C 3 

C 2 

C 1 

C 1 

C 2 

C 1 

C 2 

C 3 

We define a line integral along a piecewise smooth curve C to be the sum of the integrals 

along the sections: ∫ ∫ ∫ ∫ 

= +···+ 

C 2 

C 

C 1 

+ 

Example 7.12 Evaluate ∫ 

x 2 ydx+xdy 

in a counterclockwise direction around the triangular path in Figure below. 

y 

C 

C n 

C 3 

B(1,2) 

• • 

C 2 

C 1 A(1,0) 

x 


7.3 Independence of Path; Conservative Vector Fields 

Work Integrals 

We saw in the last section that if F is a force field in 2-space or 3-space, then the work 

performed by the field on a particle moving along a parametric curve C from an initial point 

P to a final point Q is given by the integral 

∫ 

∫ 

F·dr or equivalently F·Tds 

C 

C


Accordingly, we call an integral of this type a work integral. Recall that a work integral 

can also be expressed in scalar form as 

∫ ∫ 

F·dr = f(x,y)dx+g(x,y)dy 2-space (7.35) 

∫ 

C 

C 

∫ 

F·dr = 

C 

where f, g, and h are the component functions of F. 

Independence of Path 

C 

f(x,y,z)dx+g(x,y,z)dy+h(x,y,z)dz 3-space (7.36) 

The parametric curve C in a work integral is called the path of integration. One of the 

important problems is to determine how the path of integration affects the work performed 

by a force field on a particle that moves from a fixed point P to a fixed point Q. We will 

show that if a force F is conservative (i.e., is the gradient of some potential function φ), then 

the work that the field performs on a particle that moves from P to Q does not depend on 

the particular path C that the particle follows. This is illustrated in the following example. 

Example 7.13 The force field F(x,y) = yi+xj is conservative since it is the gradient of 

φ(x,y) = xy. Thus, the preceding discussion suggests that the work performed by the field 

on a particle that moves from the points (0,0) to the point (1,1) should be the same along 

different paths. Confirm that the value of the work integral 

∫ 

F·dr 

is the same along the following path: 

(a) The line segment y = x from (0,0) to (1,1). 

(b) The parabola y = x 2 from (0,0) to (1,1). 

(c) The cubic y = x 3 from (0,0) to (1,1). 

C 

y 

(1,1) 

• 

x 

Solution .........


The Fundamental Theorem of Line Integrals 

Recall from the Fundamental Theorem of Calculus that if F is an antiderivative of f, then 

∫ b 

a 

f(x)dx = F(b)−f(a) 

The following result is the analog of that theorem for line integrals in 2-space. 

Theorem 7.1 (The Fundamental Theorem of Line Integrals). Suppose that 

F(x,y) = f(x,y)i+g(x,y)j 

is a conservative vector field in some open region D containing points (x 0 ,y 0 ) and (x 1 ,y 1 ) 

and that f(x,y) and g(x,y) are continuous in this region. If 

F(x,y) = ∇φ(x,y) 

and if C is any piecewise smooth parametric curve that start at (x 0 ,y 0 ), ends at (x 1 ,y 1 ), 

lies in the region D, then 

∫ 

C 

F(x,y)·dr = φ(x 1 ,y 1 )−φ(x 0 ,y 0 ) (7.37) 

or, equivalently, 

∫ 

C 

∇φ·dr = φ(x 1 ,y 1 )−φ(x 0 ,y 0 ) (7.38) 

Stated informally, this theorem shows that the values of a line integral of a conservative 

vector field along a piecewise smooth path is independent of the path; that is, the value 

of the integral depends on the endpoints and not on the actual path. Accordingly, for line 

integrals along paths in conservative vector fields, it is common to express (7.37) and (7.38) 

as 

Example 7.14 

∫ (x1 ,y 1 ) 

(x 0 ,y 0 ) 

F·dr = 

∫ (x1 ,y 1 ) 

(x 0 ,y 0 ) 

∇φ·dr = φ(x 1 ,y 1 )−φ(x 0 ,y 0 ) (7.39) 

(a) Confirm that the force field F(x,y) = yi + xj in Example 7.13 is conservative by 

showing that F(x,y) is the gradient of φ(x,y) = xy. 

(b) Use the Fundamental Theorem of Line Integrals to evaluate 


∫ (1,1) 

(0,0) 

F·dr.

• 

• 

• 

• 


Line Integrals Along Closed Paths 

Parametric curves that begin and end at the same point play an important role in the study 

of vector fields, so there is some special terminology associated with them. A parametric 

curve C that is represented by the vector-valued function r(t) for a ≤ t ≤ b is said to be 

closed if the initial point r(a) and the terminal point r(b) coincide; that is, r(a) = r(b). 

y 

C 

• r(a) = r(b) 

It follows from (7.39) that the line integral of a conservative vector field along a closed 

path C that begins and ends at (x 0 ,y 0 ) is zero. This is because the point (x 1 ,y 1 ) in (7.39) 

is the same as (x 0 ,y 0 ) and hence 

∫ 

F·dr = φ(x 1 ,y 1 )−φ(x 0 ,y 0 ) = 0 

C 

x 

Our next objective is to show that the converse of this result is also true. That is, we 

want to show that under appropriate conditions a vector field whose line integral is zero 

along all closed paths must be conservative. For this to be true we will need to require that 

the domain D of the vector field be connected, by which we mean that any two points in 

D can be joined by some piecewise smooth curve that lies entirely in D. Stated informally, 

D is connected if it does not consist of two or more separate pieces. 

D 

D 

Connected 

Not connected 

Theorem 7.2 If f(x,y) and g(x,y) are continuous on some open connected region D, then 

the following statements are equivalent (all true or all false): 

(a) F(x,y) = f(x,y)i+g(x,y)j is a conservative vector field on this region D. 

∫ 

(b) F·dr = 0 for every piecewise smooth closed curve C in D. 

C

• 

• 

• 


(c) 

∫ 

F·dr = 0 is independent of the path from any point P in D to any point Q in D 

C 

for every piecewise smooth curve C in D. 

A Test for Conservative Vector Fields 

Although Theorem 7.2 is an important characterization of conservative vector fields, it is 

not an effective computational tool because it is usually not possible to evaluate the line 

integral over all possible piecewise smooth curve in D, as required in parts (b) and (c). To 

develop a method for determining whether a vector field is conservative, we will need to 

introduce some new concepts about parametric curves and connected set. We will say that 

a parametric curve is simple if it does not intersect itself between its endpoints. A simple 

parametric curve may or may not be closed. 

r(b) 

r(a) = r(b) 

• 

r(b) 

C 

r(a) 

Not simple and 

not closed 

Closed but 

not simple 

• r(a) 

Simple but 

not closed 

• 

r(a) = r(b) 

Simple and 

closed 

In addition, we will say that a connected set D in 2-space is simply connected if no 

simple closed curve in D encloses points that are not in D. Stated informally, a connected 

set D is simply connected if it has no holds; a connected set with one or more holes is said 

to be multiply connected. 

The following theorem is the primary tool for determining whether a vector field in 

2-space is conservative. 

Theorem 7.3 (Conservative Field Test). If f(x,y) and g(x,y) are continuous and 

have continuous first derivatives on some open region D, and if the vector field F(x,y) = 

f(x,y)i+g(x,y)j is conservative on D, then 

∂f 

∂y = ∂g 

∂x 

(7.40) 

at each point in D. Conversely, if D is simply connected and (7.40) holds at each point in 

D, then F(x,y) = f(x,y)i+g(x,y)j is conservative. 

It is not hard to see why (7.40) must hold if F is conservative. For this purpose suppose 

that F = ∇φ, in which case we can express the functions f and g as 

∂φ 

∂x 

= f and 

∂φ 

∂y = g (7.41)


Thus, 

∂f 

∂y = ∂ ∂y 

( ) ∂φ 

= ∂2 φ 

∂x ∂y∂x and ∂g 

∂x = ∂ 

∂x 

( ) ∂φ 

= ∂2 φ 

∂y ∂x∂y 

But the mixed partial derivatives in these equations are equal, so (7.40) follows. 

Example 7.15 Use Theorem 7.3 to determine whether the vector field 

is conservative on some open set. 


F(x,y) = (y +x)i+(y −x)j 

Once it is established that a vector field is conservative, a potential function for the field 

can be obtained by first integrating either of the equations in (7.41). This is illustrated in 

the following example. 

Example 7.16 Let F(x,y) = 2xy 3 i+(1+3x 2 y 2 )j. 

(a) Show that F is a conservative vector field on the entire xy-plane. 

(b) Find φ by first integrating ∂φ/∂x. 

(c) Find φ by first integrating ∂φ/∂y. 


Example 7.17 Use the potential function obtained in Example 7.16 to evaluate the integral 


∫ (3,1) 

(1,4) 

2xy 3 dx+(1+3x 2 y 2 )dy 

Example 7.18 Let F(x,y) = e y i+xe y j denote the force field in the xy-plane. 

(a) Verify that the force field F is conservative on the entire xy-plane. 

(b) Find the work done by the field on the particle that moves from (1,0) to (−1,0) along 

the semicircular path C shown in Figure below. 

−1 

0 1 

Solution .........


Conservative Vector Fields in 3-Space 

All of the result in this section have analogs in 3-space: The Theorem 7.1 and 7.2 can be 

extended to vector fields in 3-space simply by adding a third variable and modifying the 

hypotheses appropriately. For example, in 3-space, Formula (7.37) becomes 

∫ 

F(x,y,z)·dr = φ(x 1 ,y 1 ,z 1 )−φ(x 0 ,y 0 ,z 0 ) (7.42) 

C 

Theorem 7.3 can also be extended to vector fields in 3-space. We can show that if 

F(x,y,z) = f(x,y,z)i+g(x,y,z)j+h(x,y,z)k is a conservative field, then 

∂f 

∂y = ∂g 

∂x , ∂f 

∂z = ∂h 

∂x , ∂g 

∂z = ∂h 

∂y 

(7.43) 

that is, curlF = 0. Conversely, a vector field satisfying these conditions on a suitably restrictedregionisconservativeontheregioniff, 

g,andharecontinuousandhavecontinuous 

first partial derivatives in the region. 

7.4 Green’s Theorem 

In this section we will discuss a remarkable and beautiful theorem that expresses a double 

integral over a plane region in terms of a line integral around its boundary. 

Green’s Theorem 

Theorem 7.4 (Green’s Theorem). Let R be a simply connected plane region whose 

boundary is a simple, closed, piecewise smooth curve C oriented counterclockwise. If f(x,y) 

and g(x,y) are continuous and have continuous first partial derivatives on some open set 

containing R, then 

∫ 

∫∫ ( ∂g 

f(x,y)dx+g(x,y)dy = 

∂x − ∂f ) 

dA (7.44) 

∂y 

C 

Example 7.19 Use Green’s Theorem to evaluate 

∫ 

x 2 ydx+xdy 

along the triangle path shown in Figure below. 

y 

C 

R 

(1,2) 

• 

1 

x



A Notation for Line Integrals Around Simple Closed Curves 

It is common practice to denote a line integral around a simple closed curve by an integral 

sign with a superimposed circle. With this notation Formula (7.44) would be written as 

∮ 

∫∫ ( ∂g 

f(x,y)dx+g(x,y)dy = 

∂x − ∂f ) 

dA (7.45) 

∂y 

C 

Finding Work Using Green’s Theorem 

It follows from Formula (7.30) of Section16.2 that the integral on the left side of (7.45) 

is the work performed by the force field F(x,y) = f(x,y)i+g(x,y)j on a particle moving 

counterclockwise around the simple closed curve C. In the case where this vector field is 

conservative, it follows from Theorem 7.2 that the integrand in the double integral on the 

right side of (7.45) is zero, so the work performed by the field is zero, as expected. For 

vector fields that are not conservative, it is often more efficient to calculate the work around 

simple closed curve by using Green’s Theorem than by parametrizing the curve. 

Example 7.20 Find the work done by the force field 

F(x,y) = (e x −y 3 )i+(cosy +x 3 )j 

on a particle that travels once around the unit circle x 2 + y 2 = 1 in the counterclockwise 

direction. 


Finding Areas Using Green’s Theorem 

Green’s Theorem leads to some useful new formulas for the area A of the region R that 

satisfies the conditions of the theorem. Two such formulas can be obtained as follows: 

∫∫ ∮ ∫∫ ∮ 

A = dA = xdy and A = dA = (−y)dx 

R 

C 

Set f(x,y) = 0 and 

g(x,y) = x in (7.44) 

R 

R 

C 

Set f(x,y) = −y and 

g(x,y) = 0 in (7.44) 

A third formula can be obtained by adding these two equations together. Thus, we have 

the following three formulas that express the area A of a region R in terms of line integrals 

around the boundary: 

∮ ∮ 

A = xdy = − ydx = 1 ∮ 

−ydx+xdy (7.46) 

C C 2 C 

Example 7.21 Use a line integral to find the area enclosed by the ellipse x2 

a + y2 

2 b = 1 2 

Solution .........

Chapter 4 Vector-Valued Functions

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