Centroid of a solid semiellipsoid Assuming the result that the centroid of a solid hemisphere lies on the axis of symmetry 3/8 of the way from the base toward the top, show, by transforming the appropiate integrals, that the center of mass od the solid semiellipsoid $x^2/a^2+y^2/b^2 + z^2/c^2 < 1$, $z > 0$, lies on the $z$-axis 3/8 of the way from the base toward the top.
I don't know how to find the bounds of integration or if I have to change variables. I'd be very grateful to anyone who can help me.
 A: A demonstration of this can be obtained using simple geometric considerations, without using integrals. Remind that the volume of a $n$-ellipsoid is 
$$\frac{\pi^{n/2}}{\Gamma(n/2+1)}\prod_{k=1}^n r_k$$
where $r_1,r_2,r_3...r_n$ are the radii on the $n$ axes. 
Now consider the $4$-semiellipsoid $S_4$ generated by rotating a $3$-semiellipsoid $S_3$ around one of its axes. By construction, two of the four radii in $S_4$ are equal. Using the Pappus theorem, the volume of $S_4$ can be obtained  by multiplying the volume of $S_3$ to the distance traveled by the geometric centroid of $S_3$. So, calling $h$ the distance from the centroid to the center along the rotating radius $r_1$, we get
$$\frac{1}{2}\frac{\pi^2}{\Gamma(3)}r_1^2 r_2 r_3=\frac{2}{3} \pi r_1 r_2 r_3 \cdot \pi h$$
which can be simplified in
$$\frac{1}{4}r_1=\frac{2}{3} \cdot h$$
and then
$$h=\frac{3}{8}r_1$$
Note that this demonstration can be directly applied to the $3$-semiellipsoid  $x^2/a^2+y^2/b^2 + z^2/c^2 < 1$ described in the OP, whose radii are $a,b,c$.
