# Center of mass for $V=\{{(x,y,z)|x^2+y^2\leq 2x\sin(z), 0\leq z\leq \pi}\}$

I'm asked in an exercise to calculate the center of mass for $V=\{{(x,y,z)|x^2+y^2\leq 2x\sin(z), 0\leq z\leq \pi}\}$, but I'm having trouble doing this "the normal way" - the integral(s) I get seems unsolvable (I've tried calculating the integral in two orders: $dxdydz$ and $dydxdz$, both ended up in something unsolvable).

(The density of the body is the same for every point)

Help would be appreciated :)!

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Have you considered spherical coordinates? $x^2+y^2$ and $x\sin(z)$ seem like they might translate into reasonable bounds in terms of spherical coordinates. – Arturo Magidin May 22 '12 at 5:58
Thanks for your comment! We have not yet learned spherical coordinates... so I don't know/can't use them! – ro44 May 22 '12 at 6:01
@ro44 Are you the same person? math.stackexchange.com/users/31847/ro44 – user17762 May 22 '12 at 6:01
Yeah, that's me. I deleted my browser cookies so I guess it forgot who I was. – ro44 May 22 '12 at 6:02

## 1 Answer

I've managed to solve the question. The idea is to notice that for a set $z$, $x^2+y^2\leq 2xsin(z)$ is a circle, and thus we can use the equations for the area and center of mass of a circle to calculate the internal double integral. The final answer is: $(8/(3\pi), 0, \pi/2)$

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