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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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S E C T I O N 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 θ , 0 r 3 , 0 θ 2 π S F · d S = 2 π 0 3 0 2 9 r 2 r dr d θ = 4 π 3 0 ( 9 r r 3 ) dr = 4 π 9 r 2 2 r 4 4 3 0 = 81 π 16. F = e z 2 , sin ( x 2 z ), 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 x 2 + y 2 = 0 The Divergence Theorem implies that S F · d S = 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 flux of F = z , x , y + 2 z through the boundary of W . x y z FIGURE 16 SOLUTION We compute the divergence of F = z , x , y + 2 z : div ( F ) = x ( z ) + y ( x ) + z ( y + 2 z ) = 2 By the Divergence Theorem we have S F · d S = W div ( F ) dV = W 2 dV To compute the triple integral, we identify the projection D of the region on the xy -plane. D is the region in the xy plane enclosed by the circle x 2 + y 2 = 9 and the line 0 = x + 1 or x = − 1. We obtain the following integral: S F · d S = W 2 dV = D x + 1 0 2 dz dx dy = D 2 z x + 1 z = 0 dx dy = D ( 2 x + 2 ) dx dy 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 figure.
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1256 C H A P T E R 18 FUNDAMENTAL THEOREMS OF VECTOR ANALYSIS (ET CHAPTER 17) x y 3 D 1 We obtain S F · d S = D 2 ( 2 x + 2 ) dx dy D 1 ( 2 x + 2 ) dx dy = D 2 2 x dx dy + D 2 2 dx dy D 1 ( 2 x + 2 ) dx dy x y 3 D 2 The first integral is zero due to symmetry. The second integral is twice the area of D 2 , that is, 2 · π · 3 2 = 18 π . Therefore, S F · d S = 18 π D 1 ( 2 x + 2 ) dx dy 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 π , 1 cos θ r 3 Then D 1 ( 2 x + 2 ) dx dy = 2 π 1 . 91 3 1 / cos θ ( 2 r cos θ + 2 ) r dr d θ = π 1 . 91 3 1 / cos θ ( 4 r 2 cos θ + 4 r ) dr d θ = π 1 . 91 4 r 3 cos θ 3 + 2 r 2 3 r =
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