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Find the Centroid of the Solid Region Bounded by the Graphs

Question 41

Multiple Choice

Find the centroid of the solid region bounded by the graphs of the equations. Use a computer algebra system to evaluate the triple integral. (Assume uniform density and find the center of mass.) z=9y2+1,z=0,x=2,x=2,y=0,y=1z = \frac { 9 } { y ^ { 2 } + 1 } , z = 0 , x = - 2 , x = 2 , y = 0 , y = 1


A) xˉ=2,yˉ=2ln(2) π,zˉ=2+π4π\bar { x } = 2 , \bar { y } = \frac { 2 \ln ( 2 ) } { \pi } , \bar { z } = \frac { 2 + \pi } { 4 \pi }
B) xˉ=0,yˉ=2ln(2) π,zˉ=9(2+π) 4π\bar { x } = 0 , \bar { y } = \frac { 2 \ln ( 2 ) } { \pi } , \bar { z } = \frac { 9 ( 2 + \pi ) } { 4 \pi }
C) xˉ=0,yˉ=ln(2) π,zˉ=9(2+π) 4π\bar { x } = 0 , \bar { y } = \frac { \ln ( 2 ) } { \pi } , \bar { z } = \frac { 9 ( 2 + \pi ) } { 4 \pi }
D) xˉ=0,yˉ=2ln(2) π,zˉ=9(4+π) 4π\bar { x } = 0 , \bar { y } = \frac { 2 \ln ( 2 ) } { \pi } , \bar { z } = \frac { 9 ( 4 + \pi ) } { 4 \pi }
E) xˉ=2,yˉ=(2+π) 94π,zˉ=2ln(2) π\bar { x } = 2 , \bar { y } = \frac { ( 2 + \pi ) 9 } { 4 \pi } , \bar { z } = \frac { 2 \ln ( 2 ) } { \pi }

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