My Math Forum Center of Mass

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 September 3rd, 2017, 10:54 AM #1 Senior Member   Joined: Jan 2017 From: Toronto Posts: 118 Thanks: 2 Center of Mass Find the center of mass of an object that occupies the upper hemisphere of x^2 +y^2 +z^2 = 1 and has density x^2 + y^2. Official Answer: ( 0, 0, 3/8 ) $\displaystyle Total Mass = \int_{0}^{2\pi} \int_{0}^{\pi/2} \int_{0}^{1} \rho^4 {sin(\phi)}^3~d\rho d\phi d\theta ~=~ { 4 \pi } / 15$ $\displaystyle {Mass}^{xy} = \int_{0}^{2\pi} \int_{0}^{\pi/2} \int_{0}^{1} \rho^5 cos(\phi) {sin(\phi)}^3~d\rho d\phi d\theta ~=~ \pi / 12$ $\displaystyle Center~Of~mass = { ( \pi / 12) } / { ( { 4 \pi } / 15 ) } = 5/ 16$ (0, 0, 5 /16) What did I do wrong??
 September 3rd, 2017, 12:06 PM #2 Senior Member     Joined: Sep 2015 From: Southern California, USA Posts: 1,397 Thanks: 709 I get the answer you get in both cartesian and cylindrical coordinates. I think the book answer is incorrect.
September 3rd, 2017, 12:08 PM   #3
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I have been studying Surface Integration. What if the upper hemisphere x^2+y^2+z^2=1 is only about the 'surface'. The question may have been asking for the center of mass based on this 'surface'?

Quote:
 Originally Posted by romsek I get the answer you get in both cartesian and cylindrical coordinates. I think the book answer is incorrect.

September 3rd, 2017, 01:10 PM   #4
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Quote:
 Originally Posted by zollen I have been studying Surface Integration. What if the upper hemisphere x^2+y^2+z^2=1 is only about the 'surface'. The question may have been asking for the center of mass based on this 'surface'?
a surface has no mass

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