Computing the centroid of a three-dimensional solid

Question

I am trying to find the centroid of the solid bounded above by $x^2 +y^2 +z^2 = 12$ and below by $z = x^2 + y^2$. I am having some troubles evaluating one of the sub-integrals, specifically, the moment with respect to $y$, that is, $M_{xz}$. I perform this integral in terms of cylindrical coordinates (i.e. $x = r$ $cos (\theta)$, $y = r$ $sin (\theta)$, $z = z$). The integral is your basic constant $k$ multiplied by the Jacobian for cylindrical coordinates, $r$ $dz dr d\theta$.

I found the mass but the problem I am having is in $M_{xz}$. I eventually get to:

$\int_0^{\sqrt{12}} r^2 \sqrt{12-r^2} dr$,

but I am not sure how to perform the integration.

I know you can't do a simple u-substitution.

Any suggestions?

Answer 1

Hint: Set $r = \sqrt{12}\cos\theta$.

Answer 2

$r=\sqrt{12}\cos\theta \implies dr = -\sqrt{12}\sin\theta d\theta$

$$ \sqrt{12}\int_{0}^{\sqrt {12}} r^{2}\left(\sqrt{1-\left(\frac{r}{\sqrt{12}}\right)^2}\right)dr = -36\int_{\frac{\pi}{2}}^{0}4\cos^2\theta \sin^2\theta d\theta = 36 \int_{0}^{\pi/2}\sin^{2}{(2\theta)}d\theta = \\ = 18\int_{0}^{\pi/2}\left(1-\cos4\theta\right)d\theta = 9\pi $$