Multivariate Calculus - Bruce E. Shapiro
Multivariate Calculus - Bruce E. Shapiro Multivariate Calculus - Bruce E. Shapiro
160 LECTURE 19. SURFACE AREA WITH DOUBLE INTEGRALS Example 19.4 Find the area of the surface z = x 2 /4 + 4 that is cut off by the planes x = 0, x = 1, y = 0, y = 2. Solution. Differentiating f x = x/2, f y = 0; hence the area is √ A = 1 + fx 2 + fy 2 dA R ∫ 1 ∫ √ 2 = 1 + x2 0 0 4 dydx ∫ √ 2 ∫ 4 + x 2 2 = dydx 0 4 0 = 1 ∫ 2 2 (2) √ 4 + x 2 dx = ∫ 2 0 0 √ 4 + x 2 dx From integral 44 in the back of the book, ∫ √x 2 + a 2 dx = x 2 √ x 2 + a 2 + a2 2 ln ∣ ∣∣x + √ x 2 + a 2 ∣ ∣∣ Setting a = 2, ∫ √x 2 + 4 dx = x 2 √ x 2 + 4 + 2 ln |x + √ x 2 + 4| Therefore A = = = ∫ 2 0 √ 4 + x 2 dx [ x √ x 2 2 + 4 + 2 ln |x + √ ∣∣ x 2 2 + 4|]∣ 0 [ ] [ 2 √ 0 √ √ ] 4 + 4 + 2 ln |2 + 4 + 4| − 0 + 4 + 2 ln |0 + 0 2√ 2 2 + 4| = √ 8 + 2 ln |2 + √ 8| − 2 ln 2 = 2 √ 2 + 2 ln(1 + √ 2) Revised December 6, 2006. Math 250, Fall 2006
Lecture 20 Triple Integrals Triple Integrals in Cartesian Coordinates We can easily extend our definition of a double integral to a triple integral. Suppose that f(x, y, z) : V ∪ R 3 ↦→ R. Then we extend our Riemann sum so that it covers the volume V with small boxes V j of dimension The volume ∆V j of the ith box is then ∆x j × ∆y j × ∆z j ∆V j = ∆x j ∆y j ∆z j Then the Riemann Sum representing ∭ V f(x, y, z)dV is n∑ V ≈ f(x j , y j , z j )∆V i j=1 and the triple integral of f over V is V f(x, y, z)dV = lim n→∞,∆V j →0 j=1 n∑ f(x j , y j , z j )∆V i We can also make general definitions of x-simple and y-simple sets. These definitions are exactly the same as they were previously but with an added dimension in the domain. Example 20.1 Find ∫ 2 ∫ 3 ∫ 2x−y 0 −2 0 dzdydx 161
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Lecture 20<br />
Triple Integrals<br />
Triple Integrals in Cartesian Coordinates<br />
We can easily extend our definition of a double integral to a triple integral. Suppose<br />
that f(x, y, z) : V ∪ R 3 ↦→ R. Then we extend our Riemann sum so that it covers<br />
the volume V with small boxes V j of dimension<br />
The volume ∆V j of the ith box is then<br />
∆x j × ∆y j × ∆z j<br />
∆V j = ∆x j ∆y j ∆z j<br />
Then the Riemann Sum representing ∭ V<br />
f(x, y, z)dV is<br />
n∑<br />
V ≈ f(x j , y j , z j )∆V i<br />
j=1<br />
and the triple integral of f over V is<br />
<br />
V<br />
f(x, y, z)dV =<br />
lim<br />
n→∞,∆V j →0<br />
j=1<br />
n∑<br />
f(x j , y j , z j )∆V i<br />
We can also make general definitions of x-simple and y-simple sets. These definitions<br />
are exactly the same as they were previously but with an added dimension in the<br />
domain.<br />
Example 20.1 Find<br />
∫ 2 ∫ 3 ∫ 2x−y<br />
0 −2 0<br />
dzdydx<br />
161