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16.3.Ex28-29

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

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S E C T I O N 16.3 Triple Integrals (ET Section 15.3) 919 This gives the following inequalities of S : S : 0 z 4 , 0 x 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: W xyz 2 dV = S 4 z 2 x xyz 2 dy d A = 4 0 1 z 4 0 4 z 2 x xyz 2 dy dx 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 first quadrant of the xy -plane ( z = 0) in the line y = 2. Therefore the projection of W onto the xy -plane is the triangle D defined 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 xe z dV = 1 0 2 2 x 2 y 0 xe z dz dy dx = 1 0 2 2 x xe z 2 y z = 0 dy dx = 1 0 2 2 x x e 2 y 1 dy dx

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920 C H A P T E R 16 MULTIPLE INTEGRATION (ET CHAPTER 15) = 1 0 x e 2 y y 2 y = 2 x dx = 1 0 x e 0 2 e 2 2 x 2 x dx = 1 0 x e 2 2 x + 2 x 3 dx = e 2 1 0 xe 2 x dx + 1 0 2 x 2 dx 1 0 3 x dx = e 2 1 0 xe 2 x dx + 2 x 3 3 1 0 3 x 2 2 1 0 = e 2 1 0 xe 2 x dx 5 6 Using integration by parts, we have 1 0 xe 2 x dx = − 1 2 xe 2 x 1 0 + 1 2 1 0 e 2 x dx = 1 2 x 1 4 e 2
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16.3.Ex28-29 - S E C T I O N 16.3 Triple Integrals(ET...

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