Math 214, Calculus III Final Exam Solutions 1. (25 points) Suppose ...

Math 214, Calculus III
Final Exam Solutions
1. (25 points) Suppose we know that for a certain function f(x, y),
f(3, 4) = 25, f x (3, 4) = 6, f y (3, 4) = 8, and f(4, 5) = 41.
(a) Find a linear function L(x, y) that approximates f as well as possible near (3, 4).
Solution: The linear function L that best approximates f near (3, 4) is the linearization,
L(x, y) = f(3, 4)+[f x (3, 4)](x−3)+[f y (3, 4)](y−4) = [25]+[6](x−3)+[8](y−4) = 6x+8y−25.
That is, z = 6x + 8y − 25 is the equation of the tangent plane to z = f(x, y) at the point
(3, 4, 25).
(b) Use L to estimate f(2.9, 3.9), f(3.1, 4.1) and f(4, 5).
Solution: Using the linear function L(x, y) = 6x + 8y − 25 above as an approximation
of f(x, y), we estimate
f(2.9, 3.9) ≈ L(2.9, 3.9) = 25 + 6([2.9] − 3) + 8([3.9] − 4) = 25 − 0.6 − 0.8 = 23.6,
f(3.1, 4.1) ≈ L(3.1, 4.1) = 25 + 6([3.1] − 3) + 8([4.1] − 4) = 25 + 0.6 + 0.8 = 26.4,
f(4, 5) ≈ L(4, 5) = 25 + 6([4] − 3) + 8([5] − 4) = 25 + 6 + 8 = 39.
(c) Could f itself be a linear function? Why or why not?
Solution: If f is a linear function, then we must have f = L, since L is the linear
function that is the best approximation of f. However, f(4, 5) = 41 while L(4, 5) = 39
implies f ≠ L. Thus, we conclude that f is not a linear function.
2. (25 points) Consider the function
where C is some constant.
f(x, y) = x 2 + y 2 + Cxy
(a) Show that (0, 0) is a critical point of f.
Solution: The point (0, 0) will be a critical point of f provided f x (0, 0) = f y (0, 0) = 0.
So consider
f x (x, y) = 2x + Cy and f y (x, y) = 2y + Cx.
Then, no matter what the value of C happens to be, we see that
f x (0, 0) = 2[0] + C[0] = 0 and f y (0, 0) = 2[0] + C[0] = 0.
Therefore, (0, 0) is a critical point of f.
(b) For what values of C, if any, does f have a local minimum at (0, 0)?
Solution: To determine whether f has a local max, min or saddle point at any of its
critical points, we employ the Second Derivative Test. So, to get things started, we
compute the second partial derivatives of f:
f xx (x, y) = 2, f xy (x, y) = C = f yx (x, y), and f yy (x, y) = 2.
Thus the discriminant of f is
D(x, y) = f xx (x, y)f yy (x, y) − f xy (x, y)f yx (x, y) = [2][2] − [C][C] = 4 − C 2

Math 214, Calculus III
Final Exam Solutions
for all points (x, y). Recall that if D(x, y) > 0, then all of the concavities of f at (x, y)
agree (i.e., the graph of f is either concave up in all directions at (x, y) or it is concave
down in all directions at (x, y)); if D(x, y) < 0, then there are at least two directions
in one of which the graph of f is concave up and in the other it is concave down (i.e.,
the concavities don’t all agree at (x, y)); if D(x, y) = 0, then the Second Derivative Test
doesn’t know what to do.
Let’s first address the possibility of D = 0: i.e., C 2 = 4. Then C = ±2, so
f(x, y) = x 2 + y 2 + 2xy = (x + y) 2 or f(x, y) = x 2 + y 2 − 2xy = (x − y) 2 .
The graph of either of these functions is a parabolic cylinder whose vertex is along the
line x+y = 0 (when C = 2) of along the line x−y = 0 (when C = −2) and the parabolas
are all opening up. So, as (0, 0) is on the vertex line for these parabolic cylinders and
the parabolas in the cylinder all open up, (0, 0) is a local minimum of f when C = ±2.
Now consider the situation when the Second Derivative Test will be conclusive, i.e.,
D ≠ 0. In order for f to have a minimum at (0, 0), first of all the discriminant must be
positive, so 4 − C 2 > 0 which implies that C 2 < 4 so −2 < C < 2. Then, so long as
−2 < C < 2, (0, 0) will be a minimum of f provided that f xx (0, 0) > 0, since this implies
that the graph of f is concave up in all directions at (0, 0). But f xx (0, 0) = 2 > 0, so
the graph of f is concave up and (0, 0) is a minimum (with f(0, 0) = 0 as the minimum
value of f). Therefore, f has a local minimum at the point (0, 0) if
−2 ≤ C ≤ 2.
(c) For what values of C, if any, does f have a local maximum at (0, 0)?
Solution: From the analysis in part (b), for f to have any chance to have a local
maximum at (0, 0), we need D(0, 0) > 0, so −2 < C < 2, to ensure that all of the
concavities of the graph of f agree at (0, 0) and we need f xx (0, 0) < 0 to imply that
these concavities are all down to conclude that f has a local max at (0, 0). However,
f xx (0, 0) = 2 > 0 is not less than zero. Therefore, there is
for which f has a local maximum at (0, 0).
no value of C
(d) For what values of C, if any, does f have a saddle point at (0, 0)?
Solution: Appealing to our work in part (b) again, we know that f will have a
saddle point at (0, 0) so long as the discriminant is negative, as this will imply that
the concavities in two different directions disagree with one another. Thus, we want
D(0, 0) = 4 − C 2 < 0, i.e., 4 < C 2 . This inequality has the solution
C < −2 or C > 2
and either of these options will imply that f has a saddle point at (0, 0).
3. (25 points) A flat circular plate has the shape of the region x 2 + y 2 ≤ 1. The plate, including
the boundary where x 2 + y 2 = 1, is heated so that the temperature at the point (x, y) is
T (x, y) = x 2 + 2y 2 − x.
Find the temperatures at the hottest and coldest points on the plate.

Math 214, Calculus III
Final Exam Solutions
Solution: This is an extreme value problem with constraints, i.e., that we’re looking for the
absolute maximum and absolute minimum values of the function T (x, y) = x 2 + 2y 2 − x on
the region x 2 + y 2 ≤ 1. So, to begin, we look for critical points of T in the interior of the
plate, x 2 + y 2 < 1. Thus, consider
T x (x, y) = 2x − 1 = 0 =⇒ x = 1/2, T y (x, y) = 4y = 0 =⇒ y = 0
Therefore, (1/2, 0) is the only critical point of the function T (x, y) and (1/2, 0) is within the
interior of the plate, as (1/2) 2 + (0) 2 = 1/4 < 1. Moreover, we have
T xx = 2, T xy = 0 = T yx , T yy = 4 =⇒ D = [2][4] − [0] 2 = 8 > 0
implies that (1/2, 0) is a local minimum of the function T since D > 0 implies all of the
concavities agree at (1/2, 0) and T xx (1/2, 0) = 2 > 0 implies these concavities are all up,
meaning that (1/2, 0, −1/4) is a local minimum point.
Now we consider the boundary of the plate and search for extreme values subject to the
constraint x 2 + y 2 = 1, i.e., y 2 = 1 − x 2 , so that we may write T in terms of just the variable
x as
T (x, ± √ 1 − x 2 ) = x 2 + 2(1 − x 2 ) − x = −x 2 − x + 2.
Then T ′ (x) = −2x − 1, which is zero when x = −1/2, so y = ± √ 1 − (−1/2) 2 = ± √ 3/4.
That is, T has “critical points” on the boundary, x 2 + y 2 = 1, at the points (−1/2, √ 3/4)
and (−1/2, − √ 3/4), at which points T has the values
T (−1/2, ± √ 3/4) = (−1/2) 2 + 2(± √ 3/4) 2 − (−1/2) = 1/4 + 2(3/4) + 1/2 = 9/4.
So, from our analysis, the only possible points where absolute extreme values of T can occur
are at the one interior point (1/2, 0), where T (1/2, 0) = −1/4, and at the two boundary
points (−1/2, ± √ 3/4), where T (−1/2, ± √ 3/4) = 9/4. Since 9/4 > −1/4, we conclude that
the temperature at the hottest point(s) on the plate is 9/4 = 2.25, which occurs at the two
boundary points (−1/2, ± √ 3/4), and at the coldest point on the plate, i.e., at (1/2, 0), the
temperature is T (1/2, 0) = −1/4 = −0.25.
4. (25 points) Consider the curve r(t) = 〈cos(t) + t sin(t), sin(t) − t cos(t)〉 defined for t > 0.
(a) Find the velocity vector, v(t), for this curve.
Solution: The velocity vector is v(t) = r ′ (t), so it is
v(t) = 〈− sin(t) + t cos(t) + sin(t), cos(t) − t(− sin(t)) − cos(t)〉 = 〈t cos(t), t sin(t)〉.
(b) Find the unit tangent vector, T(t), for this curve.
Solution: The unit tangent vector is
T(t) = 1
|v(t)| v(t) = 1
√ 〈t cos t, t sin t〉
(t cos t) 2 + (t sin t) 2
1
= √
t 2 (cos 2 t + sin 2 t) 〈t cos t, t sin t〉 = 1 〈t cos t, t sin t〉
t
= 〈cos t, sin t〉.

Math 214, Calculus III
Final Exam Solutions
(c) Find the principal normal vector, N(t), for this curve.
Solution: The principal normal vector is
N(t) = 1
|T ′ (t)| T′ (t) =
1
1
√ 〈− sin t, cos t〉 = √ 〈− sin t, cos t〉 = 〈− sin t, cos t〉.
(− sin t) 2 + (cos t) 2 1
(d) Use your previous results together with the equation
κN = dT
ds = T′ (t)
|r ′ (t)|
to find the curvature κ for this curve.
Solution: From part (a), we found that r ′ (t) = v(t) = 〈t cos(t), t sin(t)〉. In (b),
we found T(t) = 〈cos t, sin t〉, so that T ′ (t) = 〈− sin t, cos t〉. In part (c), we found
N(t) = 〈− sin t, cos t〉 = T ′ (t). Thus,
κ = |κN(t)| =
T ′ √ √
(t)
∣|r ′ (t)| ∣ = |T′ (t)| (− sin t)
|r ′ (t)| = 2 + (cos t)
√ 2 1
(t cos t) 2 + (t sin t) = √ = 1 2 t 2 t .
5. (25 points) The graph y = f(x) in the xy-plane automatically has the parametrization x =
x, y = f(x), and the vector formula r(x) = 〈x, f(x)〉. Using this formula, if f is a twicedifferentiable
function of x, we can show
κ(x) =
|f ′′ (x)|
[1 + (f ′ (x)) 2 ] 3/2 .
(a) Use the formula for the curvature κ(x) above to find the curvature for the parabola
y = x 2 .
Solution: To find the curvature of the parabola y = x 2 , so that f(x) = x 2 is our
function, we should find f ′ (x) = 2x and f ′′ (x) = 2. Then we have
κ(x) =
|f ′′ (x)|
[1 + (f ′ (x)) 2 ] 3/2 = |2|
[1 + (2x) 2 ] 3/2 = 2
[1 + 4x 2 ] 3/2 .
(b) We find the total curvature of the portion of a smooth curve that runs from s = s 0 to
s = s 1 (s 1 > s 0 ) by integrating κ from s 0 to s 1 . If the curve has some other parameter,
say t, then the total curvature is
K =
∫ s1
s 0
κ ds =
∫ t1
t 0
κ ds
dt dt = ∫ t1
t 0
κ|v(t)| dt,
where t 0 and t 1 correspond to s 0 and s 1 .
Find the total curvature of the parabola y = x 2 , −∞ < x < ∞.
Solution: The parametrization of the parabola is, as described in the introduction to
this problem, given by
r(x) = 〈x, f(x)〉 = 〈x, x 2 〉.

Math 214, Calculus III
Final Exam Solutions
Using this together with the formula for the total curvature K and our answer for κ(x)
in part (a) above, we find
K =
=
=
=
∫ ∞
−∞
∫ ∞
−∞
∫ ∞
−∞
∫ ∞
−∞
∫ 0
κ(x)|v(x)| dx
[
]
2
[1 + 4x 2 ] 3/2 |〈1, 2x〉| dx
[
]
2 √(1)
[1 + 4x 2 ] 3/2 2 + (2x) 2 dx
2
1 + 4x 2 dx
∫
2
∞
=
−∞ 1 + 4x 2 dx + 0
[∫ 0
= lim
a→−∞
= lim
a→−∞
= lim
a→−∞
= lim
a→−∞
a
[∫ 0
a
[∫ 0
x=a
[∫ 0
x=a
[∫ 0
= lim
a→−∞
x=a
2
1 + 4x 2 dx
] [∫
2
b
1 + 4x 2 dx + lim
b→∞ 0
] [∫
2 dx
b
1 + (2x) 2 + lim
b→∞ 0
] [∫
du
b
1 + u 2 + lim
b→∞ x=0
sec 2 ]
θ dθ
1 + (tan θ) 2 + lim
]
dθ
[∫ b
+ lim
b→∞ x=0
b→∞
[∫ b
]
dθ
]
2
1 + 4x 2 dx
]
2 dx
1 + (2x) 2
]
du
1 + u 2
x=0
sec 2 ]
θ dθ
1 + (tan θ) 2
= lim
a→−∞ [θ]0 x=a + lim
b→∞ [θ]b x=0
[
= lim tan −1 (u) ] 0
a→−∞
x=a + lim
[
tan −1 (u) ] b
b→∞
x=0
[
= lim tan −1 (2x) ] 0
a→−∞
a + lim
[
tan −1 (2x) ] 0
b→∞
a
[
= lim tan −1 (0) − tan −1 (2a) ] [
+ lim tan −1 (2b) − tan −1 (0) ]
a→−∞
b→∞
[ (
= 0 − − π )] [( π
) ]
+ − 0 = π.
2 2
6. (25 points) Suppose f(x, y, z) is differentiable in a region whose interior contains a smooth
curve
C : r(t) = 〈g(t), h(t), k(t)〉.
Consider the function f(r(t)) = f (g(t), h(t), k(t)) of the variable t.
(a) Use the Chain Rule to find df
dt .
Solution: By the Chain Rule, we have
(b) Show that df
dt = ∇f · r′ (t).
df
dt = ∂f dx
∂x dt + ∂f dy
∂y dt + ∂f dz
∂z dt = [f x]g ′ (t) + [f y ]h ′ (t) + [f z ]k ′ (t).

Math 214, Calculus III
Final Exam Solutions
Solution: The gradient of f, ∇f is
∇f = 〈f x , f y , f z 〉
while
Hence,
r ′ (t) = 〈g ′ (t), h ′ (t), k ′ (t)〉.
∇f · r ′ (t) = 〈f x , f y , f z 〉 · 〈g ′ (t), h ′ (t), k ′ (t)〉 = [f x ]g ′ (t) + [f y ]h ′ (t) + [f z ]k ′ (t) = df
dt .
(c) Suppose P 0 is a point on C where f has a local maximum or a local minimum relative
to its values on C. Use part (b) to explain why ∇f is orthogonal to C at P 0 .
Solution: As a function of t, if f(r(t)) has either a local max or min at P 0 , then we
must have df
df
= 0. However, by part (b),
dt dt = ∇f · r′ (t), so ∇f · r ′ (t) = 0 at P 0 . But
two vectors are orthogonal if and only if their dot product is zero, so we conclude that
∇f is orthogonal to r ′ (t) at P 0 . Yet, r ′ (t) is a tangent vector to the curve C at the point
P 0 , so ∇f is orthogonal to the curve C at P 0 as claimed.
7. (25 points) Find the average value of f(x, y) = x 2 + y 2 over each of the following regions:
(a) the square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1.
Solution: The average value of f over a region D is defined to be f ave =
Thus, as the area of the square A : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 is 1, we have
f ave = 1 ∫∫
∫ 1 ∫ 1
(x 2 + y 2 ) dA = (x 2 + y 2 ) dy dx =
[1] A
x=0 y=0
∫ 1
[
= x 2 + 1 ] [ 1
dx =
x=0 3 3 x3 + 1 ] 1
3 x = 1
x=0
3 + 1 3 = 2 3 .
∫ 1
x=0
1 ∫∫
area(D)
D
f(x, y) dA.
[
x 2 y + 1 3 y3 ] 1
y=0
(b) the square 0 ≤ x ≤ a, 0 ≤ y ≤ a, where a > 0.
Solution: On the square B : 0 ≤ x ≤ a, 0 ≤ y ≤ a, whose area is area(B) = a 2 , we have
f ave = 1 ∫∫
[a 2 (x 2 + y 2 ) dA = 1 ∫ a ∫ a
] B
a 2 (x 2 + y 2 ) dy dx = 1 ∫ a
[
x=0 y=0
a 2 x 2 y + 1 ] a
x=0 3 y3 y=0
= 1 a 2 ∫ a
x=0
[ax 2 + a3
3
]
dx = 1 [ ] a a
a 2 3 x3 + a3
3 x = 1 [ a
4
0
a 2 3 + a4
3
]
= 2 3 a2 .
dx
dx
(c) the disc x 2 + y 2 ≤ a 2 , where a > 0.
Solution: On the disc D : x 2 + y 2 ≤ a 2 , whose area is area(D) = πa 2 , we have
f ave = 1 ∫∫
[πa 2 ]
= 1
πa 2 ∫ 2π
θ=0
(x 2 + y 2 ) dA = 1 ∫ 2π
D
πa 2 θ=0
[ ] 1
4 a4 dθ = 1 [ ] a
4 2π
πa 2 4 θ
θ=0
∫ a
r=0
(r 2 ) r dr dθ = 1
πa 2 ∫ 2π
= 1
πa 2 [ a
4
4 (2π) ]
= a2
2 .
θ=0
[ 1
4 r4 ] a
r=0
dθ

Math 214, Calculus III
Final Exam Solutions
8. (25 points) To celebrate the beginning of spring break, you treat yourself to an ice cream cone.
Upon examination, you recognize that the ice cream cone fills the region in space bounded
by the top half of the sphere
x 2 + y 2 + (z − 1) 2 = 1
above and by the cone
z = √ x 2 + y 2
below. Find the volume of your ice cream cone.
[Hint: First integrate with respect to z, and then convert to polar coordinates.]
Solution: The top half of the sphere x 2 + y 2 + (z − 1) 2 = 1 is obtained by solving for z
by taking the positive square root (the negative square root yields the bottom half of the
sphere), so the top of the ice cream cone is
(z − 1) 2 = 1 − x 2 − y 2 =⇒ z − 1 = √ 1 − x 2 − y 2 =⇒ z = 1 + √ 1 − x 2 − y 2 .
Now the top half of this sphere and the cone intersect when we have √ x 2 + y 2 = z =
1 + √ 1 − x 2 − y 2 , i.e.,
√
x 2 + y 2 − √ 1 − x 2 − y 2 = 1 =⇒ (x 2 + y 2 ) − 2 √ x 2 + y 2√ 1 − x 2 − y 2 + (1 − x 2 − y 2 ) = 1
=⇒ −2 √ x 2 + y 2√ 1 − x 2 − y 2 = 0
=⇒ x 2 + y 2 = 0 or x 2 + y 2 = 1.
The first option, x 2 + y 2 = 0, is where the bottom half of the sphere and the cone intersect at
the vertex of the cone, so we discard this option (as we are only concerned with the top half
of the cone, so that z ≥ 1), which means that the region in the xy-plane that is the shadow
of the ice cream cone is the disc D : x 2 + y 2 ≤ 1, which is given by 0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π in
polar coordinates. Therefore, the volume of the ice cream cone is
∫∫∫
V =
∫∫
=
=
=
=
D
∫ 2π ∫ 1
θ=0
∫ 2π
θ=0
∫ 2π
∫∫
dV =
D
[(1 + √ 1 − x 2 − y 2 )
−
r=0
[
1
∫
√
1+ 1−x 2 −y 2 ∫∫
√ dz dA =
z= x 2 +y 2 D
(√ )]
x 2 + y 2 dA
[1 + √ 1 − r 2 − √ r 2 ]
r dr dθ =
] 1
2 r2 − 1 (1 − r 2 ) 3/2
− 1 2 3/2 3 r3 dθ =
r=0
[ ] [ ] 1 1 2π
dθ =
θ=0 2 2 θ = π.
0
∫ 2π ∫ 1
[z] 1+√ 1−x 2 −y 2
z= √ x 2 +y 2 dA
θ=0 r=0
∫ 2π
[
r + r √ 1 − r 2 − r 2] dr dθ
θ=0
[( 1
2 − 0 − 1 3
So, along with your ice cream, you get to enjoy a little π. Happy “Pi Day.”
9. (25 points) Let F(x, y) = 〈3x 2 y 2 , 2x 3 y〉.
)
−
(0 − 1 )]
3 − 0
(a) Show that F is a conservative vector field.
Solution: First of all, we observe that F is defined on all of R 2 , which is open and simply
connected. So, to prove that F is conservative, it is enough to prove that P y = Q x , where
P = 3x 2 y 2 =⇒ P y = 6x 2 y
Q = 2x 3 y =⇒ Q x = 6x 2 y
dθ

Math 214, Calculus III
Final Exam Solutions
As we can see above, the vector field F = 〈P, Q〉 satisfies P y = Q x , so F is conservative.
(b) Find a potential function for F.
Solution: By the Fundamental Theorem for Line Integrals, we know that a potential
function for F exists, i.e., there is a scalar function f(x, y) with F = ∇f. In particular,
this means that f x = P , so
∫ ∫ [3x
f(x, y) = P dx = 2 y 2] dx = x 3 y 2 + C(y)
for some function C(y), which is constant with respect to x. From this form of f, we can
determine C(y) by comparing f y = 2x 3 y + C ′ (y) with Q = 2x 3 y. The equality f y = Q
then implies that C ′ (y) = 0, so that C(y) = K for some constant K (i.e., K is a number
and is not a function of either x or y). Thus
is the generic potential function of F.
f(x, y) = x 3 y 2 + K
(c) Let C be the curve parametrized by r(t) = 〈3t 2 + 1, 2t〉, 0 ≤ t ≤ 1. Find ∫ C F · dr.
Solution: By the Fundamental Theorem for Line Integrals,
∫
F · dr = f(r(1)) − f(r(0)) = f(4, 2) − f(1, 0) = [(4) 3 (2) 2 ] − [(1) 3 (0) 2 ] = 256.
C
10. (25 points) Evaluate the line integral ∮ C y2 dx+x 2 dy, where C is the circle x 2 +y 2 = 4 using:
(a) the parametrization r(t) = 〈2 cos(t), 2 sin(t)〉, 0 ≤ t ≤ 2π, of C.
Solution: The vector field F corresponding to the differential form y 2 dx + x 2 dy is
F(x, y) = 〈y 2 , x 2 〉, so ∮ C y2 dx + x 2 dy = ∮ C
F · dr. Computing the latter line integral
using the parametrization r(t), we obtain
∮
C
F · dr =
=
∫ 2π
0
∫ 2π
0
〈4 sin 2 t, 4 cos 2 t〉 · 〈−2 sin t, 2 cos t〉 dt =
∫ 2π
8(1 − cos 2 t)[− sin t] + 8(1 − sin 2 t)[cos t] dt
[ (
= 8 cos t − 1 ) (
3 cos3 t + 8 sin t − 1 )] 2π
3 sin3 t
0
[(
= 8 1 − 1 ) ] [(
+ (0 − 0) − 8 1 − 1 ) ]
+ (0 − 0)
3
3
0
−8 sin 3 t + 8 cos 3 t dt
= 0.
(b) Green’s Theorem.
Solution: As in part (a), we have ∮ C y2 dx + x 2 dy = ∮ C F · dr, where F(x, y) = 〈y2 , x 2 〉.
By Green’s Theorem, which applies to this problem since F is defined on all of R 2 (which
is an open, simply connected domain) and the circle C is both simple and closed, we

Math 214, Calculus III
Final Exam Solutions
have
∮ ∫∫
F · dr =
C
=
=
( ∂Q
∂x − ∂P ) ∫∫
dA =
∂y
R
∫ 2π ∫ 2
θ=0
∫ 2π
= 16
3
r=0
θ=0
∫ 2π
R
(2x − 2y) dA
(2r cos θ − 2r sin θ) r dr dθ =
[ ] 2 2
(cos θ − sin θ)
3 r3 dθ =
r=0
θ=0
(cos θ − sin θ) dθ = 16 3
∫ 2π
θ=0
∫ 2π ∫ 2
θ=0
r=0
2r 2 (cos θ − sin θ) dr dθ
[ ] 16
(cos θ − sin θ) dθ
3
[sin θ + cos θ]2π
θ=0 = 16 [(0 + 1) − (0 + 1)] = 0.
3