qz05sol - m = Z C k ( 1-x ) ds = Z 3 / 2 / 2 k ( 1-3cos t )...

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Math 53 Section 002 Quiz IV Solution 1. Evaluate the triple integral ZZZ H z 3 p x 2 + y 2 + z 2 dV , where H is the solid hemi- sphere that lies above the xy -plane and has center the origin and radius 1. Solution. The solid hemisphere H can be described in spherical coordinates as H = { ( ρ , θ , φ ) | 0 ρ 1 , 0 θ 2 π , 0 φ π / 2 } . So we may compute the integral in polar coordinates ZZZ H z 3 p x 2 + y 2 + z 2 dV = Z π / 2 0 Z 2 π 0 Z 1 0 ( ρ cos φ ) 3 ρ · ρ 2 sin φ d ρ d θ d φ = Z π / 2 0 cos 3 φ sin φ d φ Z 2 π 0 d θ Z 1 0 ρ 6 d ρ = ± Z 0 1 u 3 ( - du ) ² h 2 π i ± 1 7 ρ 2 ³ ³ ³ 1 0 ² = 2 π 7 ± 1 4 u 4 ² 0 1 = 2 π 7 · 1 4 = π 14 . 2. A thin wire is bent into the shape of a semicircle x 2 + y 2 = 9, x 0. If the linear density at any point is proportional to its distance from the line x = 1, please find the center of mass. Solution. To find the center of mass, we need to compute the mass of the wire first. The semicircle can be parametrized by means of the equations C : x = 3cos t , y = 3sin t , π / 2 t 3 π / 2 . The linear density is ρ ( x , y ) = k ( 1 - x ) = k ( 1 - 3cos t ) , where k is a constant. Hence the mass of the wire is
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Unformatted text preview: m = Z C k ( 1-x ) ds = Z 3 / 2 / 2 k ( 1-3cos t ) s x t 2 + y t 2 dt = Z 3 / 2 / 2 k ( 1-3cos t ) q (-sin t ) 2 +( cos t ) 2 dt = k Z 3 / 2 / 2 ( 1-3cos t ) dt = k ( t-3sin t ) 3 / 2 / 2 = k ( + 6 ) . 1 By symmetry we see that y = 0. For x we have x = 1 m Z C x ( x , y ) ds = 1 k ( + 6 ) Z C xk ( 1-x ) ds = 1 + 6 Z 3 / 2 / 2 3cos t ( 1-3cos t ) dt = 1 + 6 Z 3 / 2 / 2 3cos t-9 2 ( 1 + cos2 t ) dt = 1 + 6 3sin t-9 2 t-9 4 sin2 t 3 / 2 / 2 = 1 + 6 -6-9 2 =-9 + 12 2 + 12 . So the center of mass is -9 + 12 2 + 12 , . 2...
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qz05sol - m = Z C k ( 1-x ) ds = Z 3 / 2 / 2 k ( 1-3cos t )...

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