16.3.Ex28-29

16.3.Ex28-29 - S E C T I O N 16.3 Triple Integrals (ET...

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SECTION 16.3 Triple Integrals (ET Section 15.3) 919 This gives the following inequalities of S : S : 0 z 4 , 0 x r 1 z 4 The upper surface z = 4 y 2 can be described by y = 4 z , hence the limits of y are 2 x y 4 z .We obtain the following iterated integral: ZZZ W xyz 2 dV = ZZ S Ã Z 4 z 2 x 2 dy ± dA = Z 4 0 Z q 1 z 4 0 Z 4 z 2 x 2 dydx dz 28. Let W be the region bounded by y + z = 2, 2 x = y , x = 0, and z = 0 (Figure 16). Express and evaluate the triple integral of xe z by projecting W onto the: (a) xy -plane (b) yz -plane (c) xz -plane y z x 1 2 2 Upper face y + z = 2 2 x = y FIGURE 16 SOLUTION (a) y z x 1 2 2 Upper face y + z = 2 2 x = y The upper face z = 2 y intersects the ±rst quadrant of the -plane ( z = 0) in the line y = 2. Therefore the projection of W onto the -plane is the triangle D de±ned by D : 0 x 1 , 2 x y 2 y x 2 1 2 1 0 y = 2 x D Therefore, W is the following region: 0 x 1 , 2 x y 2 , 0 z 2 y We obtain the following iterated integral: W z = Z 1 0 Z 2 2 x Z 2 y 0 z dzdydx = Z 1 0 Z 2 2 x z ¯ ¯ ¯ ¯ 2 y z = 0 dydx = Z 1 0 Z 2 2 x x ³ e 2 y 1 ´
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920 CHAPTER 16 MULTIPLE INTEGRATION (ET CHAPTER 15) = Z 1 0 x ³ e 2 y y ´ ¯ ¯ ¯ ¯ 2 y = 2 x dx = Z 1 0 x ³ e 0 2 ³ e 2 2 x 2 x ´ = Z 1 0 x ³ e 2 2 x + 2 x 3 ´ = e 2 Z 1 0 xe 2 x + Z 1 0 2 x 2 Z 1 0 3 xdx = e 2 Z 1 0 2 x + 2 x 3 3 ¯ ¯ ¯ ¯ 1 0 3 x 2 2 ¯ ¯ ¯ ¯ 1 0 = e 2 Z 1 0 2 x 5 6 Using integration by parts, we have Z 1 0 2 x =− 1 2 2 x ¯ ¯ ¯ ¯ 1 0 + 1 2 Z 1 0 e 2 x = µ 1 2 x 1 4 ± e 2 x ¯ ¯ ¯ ¯ 1 0 = 1 4 3 4 e
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This note was uploaded on 04/13/2008 for the course MATH 32B taught by Professor Rogawski during the Winter '08 term at UCLA.

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16.3.Ex28-29 - S E C T I O N 16.3 Triple Integrals (ET...

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