16.4 Ex 56 - 60

16.4 Ex 56 - 60 - 974 C H A P T E R 16 M U LTI P L E I N T...

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974 CHAPTER 16 MULTIPLE INTEGRATION (ET CHAPTER 15) z π 4 π 4 = 0 = The spherical inequalities for W are thus W : 0 θ 2 π , 0 φ 4 , 0 ρ 4cos We obtain the following integral: ZZZ W 1 dV = Z 2 0 Z / 4 0 Z 0 2 sin d d d = Z 2 0 Z / 4 0 3 sin 3 ¯ ¯ ¯ ¯ = 0 d d = Z 2 0 Z / 4 0 64 cos 3 sin 3 d d = Ã Z 2 0 64 3 d Z / 4 0 cos 3 sin d ! = 128 3 Z / 4 0 cos 3 sin d We compute the integral using the substitution u = cos , du =− sin d : W 1 = 128 3 Z 1 / 2 1 u 3 ( ) = 128 3 Z 1 1 / 2 u 3 = 128 3 · u 4 4 ¯ ¯ ¯ ¯ 1 1 / 2 = 8 56. f ( x , y , z ) = ; x 2 + y 2 + z 2 4, z 1 x 0 SOLUTION 1 2 2 z y x W is the region inside the sphere of radius 2, below the plane z = 1 and above and below the right xy -plane. The equation of the sphere x 2 + y 2 + z 2 = 4 in spherical coordinates is = 2, and the equation of the plane z = 1is, cos = 1 = 1 cos (1) To evaluate the triple integral we let W 1 be the region inside the sphere above the plane z = 1for x 0, and W 2 be the region enclosed by the sphere, for x 0. D x y 2 Thus, W f ( x , y , z ) = W 2 f ( x , y , z ) W 1 f ( x , y , z )
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SECTION 16.4 Integration in Polar, Cylindrical, and Spherical Coordinates (ET Section 15.4) 975 Setting ρ = 2 in (1) gives 2 = 1 cos φ cos = 1 2 = π 3 W 1 y x z p 3 Therefore, we obtain the following deFnition: W 1 :− 2 θ 2 , 0 3 , 1 cos 2 W 2 2 2 , 0 , 0 2 We compute the integral over W 1 : ZZZ W 1 f ( x , y , z ) dV = Z / 2 / 2 Z / 3 0 Z 2 1 / cos · 2 sin d d d = Z / 2 / 2 Z / 3 0 Z 2 1 / cos 3 sin d d d = Z / 2 / 2 Z / 3 0 4 sin 4 ¯ ¯ ¯ ¯ 2 = 1 cos d d = Z / 2 / 2 Z / 3 0 ± 4sin sin 4cos 4 d d = à Z / 2 / 2 d Z / 3 0 ± sin 4 ! d = Z / 3 0 d Z / 3 0 sin 4 d (2) We compute the second integral using the substitution u = cos , du =− sin d , and the Frst using an integration formula. We get Z / 3 0 4s in d ¯ ¯ ¯ ¯ / 3 0 4 ± 1 2 1 = 2 Z / 3 0 sin 4 d = Z 1 / 2 1 1 4 u 4 ( ) = Z 1 1 / 2 u 4 4 = u 3 12 ¯ ¯ ¯ ¯ 1 1 / 2 = 7 12 Substituting the integrals in (2) we get W 1 f ( x , y , z ) = 2 7 12 = 17 12 (3) We compute the integral over W 2 : W 2 f ( x , y , z ) = Z / 2 / 2 Z 0 Z 2 0 · 2 sin d d d = Z / 2 / 2 Z 0 Z 2 0 3 sin d d d = à Z / 2 / 2 1 d ! ±Z 0 sin d à Z 2 0 3 d ! = · ± cos ¯ ¯ ¯ ¯ 0 à 4 4 ¯ ¯ ¯ ¯ 2 0 !
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16.4 Ex 56 - 60 - 974 C H A P T E R 16 M U LTI P L E I N T...

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