Find the center of mass of that part of the sphere
$x^2+y^2+z^2 \le a^2$
having $x,y,z \ge 0$ (that is, the part in the first octant)
With density given by $\rho(x,y,z)=(x^2+y^2+z^2)^{3/2}$
It should be solved using a triple integral.
I get either $\frac{2a}3$ or $\frac{6a}7$, but I don't know whether any of these are correct. I used spherical coordinates. The problem is that the density function has a power of $\frac{3}{2}$.
Best Answer
First, we solve it for the unit sphere, since the solution is just scaled up by $a$.
Secondly, we observe that if we have a single octant, with center of mass at $(u, u, u)$, then if we combine the four positive-$z$ octants (say), then the center of mass will be at $(0, 0, u)$, by symmetry. So if we find the center of mass of the positive-$z$ hemisphere, we thereby obtain the center of mass of any octant.
We note that the center of mass of a unit hemispherical shell is at coordinate
$$ \frac{\int_{\theta=0}^{\pi/2} \cos\theta\sin\theta \, d\theta} {\int_{\theta=0}^{\pi/2} \cos\theta \,d\theta} = \frac{1}{2} $$
so that of a hemispherical shell of radius $r$ is at coordinate $\frac{r}{2}$.
In the sphere in question, we have shells of area proportional to $r^2$, multiplied by a density proportional to $r^3$, so the coordinate of the center of mass of the hemisphere is
$$ \frac{\int_{r=0}^1 \frac{r}{2} r^5 \, dr}{\int_{r=0}^1 r^5 \, dr} = \frac{\frac{1}{14}}{\frac{1}{6}} = \frac{3}{7} $$
For a uniform sphere, it would be
$$ \frac{\int_{r=0}^1 \frac{r}{2} r^2 \, dr}{\int_{r=0}^1 r^2 \, dr} = \frac{\frac{1}{8}}{\frac{1}{3}} = \frac{3}{8} $$
thus confirming the values obtained by Ron Gordon. It is mildly curious to me that that high a density gradient has such a small effect.