I am trying to find the center of mass of a semiellipsoid using cylindrical coordinates.
$$ \frac{r^2}{a^2}+\frac{z^2}{b^2} \leq 1 $$ with $z < 0$ and density = 1.
I know that the center of mass is
$$ \overline{CM} = \frac{1}{M} \int \int \int \pmatrix{ r \cos(\theta) \\ r \sin (\theta) \\ z } f(x)\, {\rm d} r\, {\rm d}\theta \,{\rm d}z $$
I am having trouble figuring out the bounds.. can anyone help?
You have the equation that gives the limit for $r$ for each $z$
$$ r = a \sqrt{ 1 - \frac{z^2}{b^2} } $$
So the volume integral is
$$ V = \int_0^b \int_0^{a \sqrt{ 1 - \frac{z^2}{b^2} }} \int_0^{2\pi} r\, {\rm d}\theta\,{\rm d}r\,{\rm d}z = \tfrac{2}{3} \pi a ^2 b$$
and so the mass is
$$ M = \int_0^b \int_0^{a \sqrt{ 1 - \frac{z^2}{b^2} }} \int_0^{2\pi} \rho\, r\, {\rm d}\theta\,{\rm d}r\,{\rm d}z = \rho \tfrac{2}{3} \pi a ^2 b$$
Each interior point has coordinates $\overrightarrow{\rm pos} = \pmatrix{r \cos \theta \\ r \sin \theta \\ z} $ and so the center of mass is
$$ \overrightarrow{\rm cm} = \frac{1}{M} \int_0^b \int_0^{a \sqrt{ 1 - \frac{z^2}{b^2} }} \int_0^{2\pi} \rho\,\pmatrix{r \cos \theta \\ r \sin \theta \\ z} r\, {\rm d}\theta\,{\rm d}r\,{\rm d}z = \pmatrix{0\\ 0 \\ \tfrac{3}{8} b} $$
Finally, the mass moment of inertia tensor about the origin is
$$\overline{I}_0 = \int_0^b \int_0^{a \sqrt{ 1 - \frac{z^2}{b^2} }} \int_0^{2\pi} \rho\,\pmatrix{r^2 \sin^2 \theta + z^2 & -r^2 \sin \theta \cos \theta & -r z \cos \theta \\ -r^2 \sin \theta \cos \theta & r^2 \cos^2 \theta + z^2 & -r z \sin \theta \\ -r z \cos \theta & - r z \sin \theta & r^2 } r\, {\rm d}\theta\,{\rm d}r\,{\rm d}z $$
which after some simplifications becomes
$$ \overline{I}_0 = M \int_0^b \int_0^{a \sqrt{ 1 - \frac{z^2}{b^2} }} \pmatrix{ \frac{ 3 r (2 z^2+r^2)}{2 a^2 b} & & \\ & \frac{3 r (2 z^2+r^2)}{2 a^2 b} & \\ & & \frac{3 r^2}{a^2 b} } \,{\rm d}r\,{\rm d}z $$
and
$$ \overline{I}_0 = M \pmatrix{ \frac{a^2+b^2}{5} & & \\ & \frac{a^2+b^2}{5} & \\ & & \frac{2 a^2}{5} } $$
Finally, you want the MMOI tensor about the center of mass
$$ \overline{I}_{\rm cm} = \overline{I}_{0} - M \pmatrix{ \frac{9 b^2}{64} & & \\ & \frac{9 b^2}{64} & \\ & & 0 } $$
Verification. From the above, you can see that when $ a = \frac{\sqrt{19}}{8} b$ all the MMOI values are equal to each other. I wanted to try it out in CAD, and indeed this is true