18.3 Ex 17 - 23

18.3 Ex 17 - 23 - S E C T I O N 18.3 y x 2 y2 = 9 D 3 x...

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SECTION 18.3 Divergence Theorem (ET Section 17.3) 1255 3 x 2 + y 2 = 9 x y D We convert the integral to polar coordinates: x = r cos θ , y = r sin , 0 r 3 , 0 2 π ZZ S F · d S = Z 2 0 Z 3 0 2 ³ 9 r 2 ± rdrd = 4 Z 3 0 ( 9 r r 3 ) dr = 4 Ã 9 r 2 2 r 4 4 ¯ ¯ ¯ ¯ 3 0 ! = 81 16. F = ² e z 2 , sin ( x 2 z ), p x 2 + 9 y 2 ® , S is the region x 2 + y 2 z 8 x 2 y 2 . SOLUTION We compute the divergence of F : div ( F ) = x ( e z 2 ) + y sin ( x 2 z ) + z q x 2 + y 2 = 0 The Divergence Theorem implies that S F · d S = ZZZ W div ( F ) dV = 0 17. Let W be the region in Figure 16 bounded by the cylinder x 2 + y 2 = 9, the plane z = x + 1, and the xy -plane. Use the Divergence Theorem to compute the ±ux of F = h z , x , y + 2 z i through the boundary of W . x y z FIGURE 16 We compute the divergence of F = h z , x , y + 2 z i : div ( F ) = x ( z ) + y ( x ) + z ( y + 2 z ) = 2 By the Divergence Theorem we have S F · d S = W div ( F ) = W 2 To compute the triple integral, we identify the projection D of the region on the -plane. D is the region in the plane enclosed by the circle x 2 + y 2 = 9 and the line 0 = x + 1or x =− 1. We obtain the following integral: S F · d S = W 2 = D Z x + 1 0 2 dzdx dy = D 2 z ¯ ¯ ¯ ¯ x + 1 z = 0 dx dy = D ( 2 x + 2 ) We compute the double integral as the difference of two integrals: the integral over the disk D 2 of radius 3, and the integral over the part D 1 of the disk, shown in the ²gure.
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1256 CHAPTER 18 FUNDAMENTAL THEOREMS OF VECTOR ANALYSIS (ET CHAPTER 17) x y 3 D 1 We obtain ZZ S F · d S = D 2 ( 2 x + 2 ) dx dy D 1 ( 2 x + 2 ) = D 2 2 xdxdy + D 2 2 D 1 ( 2 x + 2 ) x y 3 D 2 The frst integral is zero due to symmetry. The second integral is twice the area oF D 2 ,thatis,2 · π · 3 2 = 18 . ThereFore, S F · d S = 18 D 1 ( 2 x + 2 ) We compute the double integral over the upper part oF D 1 . Due to symmetry, this integral is equal to halF oF the integral over D 1 . x y 3 1 q We describe the region in polar coordinates: 1 . 91 θ , 1 cos r 3 Then D 1 ( 2 x + 2 ) = 2 Z 1 . 91 Z 3 1 / cos ( 2 r cos + 2 ) rdrd = Z 1 . 91 Z 3 1 / cos ( 4 r 2 cos + 4 r ) dr d = Z 1 . 91 4 r 3 cos 3 + 2 r 2 ¯ ¯ ¯ ¯ 3 r = 1 cos d = Z 1 . 91 µ 36 cos + 18 + 4 3 cos cos 3 2 cos 2 d = Z 1 . 91 µ 36 cos + 18 2 3 · 1 cos 2 d = 36 sin + 18 2 3 tan ¯ ¯ ¯ ¯ 1 . 91
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SECTION 18.3 Divergence Theorem (ET Section 17.3) 1257 = 18 π
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This homework help 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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18.3 Ex 17 - 23 - S E C T I O N 18.3 y x 2 y2 = 9 D 3 x...

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