I am looking for exact or even approximate formulas for the moment of inertia of a hollow spheroid (oblate and prolate.) I have find formulae for a hollow spherical shell and for a filled ellipsoid but not for a hollowed spheroid.
I was wondering if someone here can be of help.
Thanks
a
It looks a bit messy. Suppose we parameterize the top surface of an oblate spheroid as $$\vec r=\langle x,y,z\rangle=\langle r\cos\theta,r\sin\theta,\frac ba\sqrt{a^2-r^2}\rangle$$ Then $$d\vec r=\langle\cos\theta,\sin\theta,\frac{-br}{a\sqrt{a^2-r^2}}\rangle\,dr+\langle-r\sin\theta,r\cos\theta,0\rangle\,d\theta$$ $$\begin{align}d^2\vec A&=\pm\langle\cos\theta,\sin\theta,\frac{-br}{a\sqrt{a^2-r^2}}\rangle\,dr\times\langle-r\sin\theta,r\cos\theta,0\rangle\,d\theta\\ &=\pm\langle\frac{br^2\cos\theta}{a\sqrt{a^2-r^2}},\frac{br^2\sin\theta}{a\sqrt{a^2-r^2}},r\rangle\,dr\,d\theta\end{align}$$ $$d^2A=\left|\left|d^2\vec A\right|\right|=\sqrt{\frac{b^2r^2}{a^2(a^2-r^2)}+1}\,r\,dr\,d\theta=\sqrt\frac{a^4-(a^2-b^2)r^2}{a^2(a^2-r^2)}\,r\,dr\,d\theta$$ Now that we have the scalar areal element, let's find the area. $$\begin{align}A&=\int\int d^2A=2\int_0^{2\pi}\int_0^a\sqrt\frac{a^4-(a^2-b^2)r^2}{a^2(a^2-r^2)}\,r\,dr\,d\theta\\ &=4\pi\int_0^1\frac a2\frac{\sqrt{a^2-(a^2-b^2)u}}{\sqrt{1-u}}du\\ &=2\pi a\sqrt{a^2-b^2}\int_0^1\frac{\left(\frac{a^2}{a^2-b^2}-u\right)du}{\sqrt{\left(\frac{2a^2-b^2}{2(a^2-b^2)}-u\right)^2-\frac{b^4}{4(a^2-b^2)^2}}}\\ &=2\pi a\sqrt{a^2-b^2}\int_{\frac{b^2}{2(a^2-b^2)}}^{\frac{2a^2-b^2}{2(a^2-b^2)}}\frac{\left(\frac{b^2}{2(a^2-b^2)}+v\right)dv}{\sqrt{v^2-\frac{b^4}{4(a^2-b^2)^2}}}\\ &=\frac{\pi ab^2}{\sqrt{a^2-b^2}}\int_1^{\frac{2a^2}{b^2}-1}\frac{(1+w)dw}{\sqrt{w^2-1}}=\frac{\pi ab^2}{\sqrt{a^2-b^2}}\int_0^{\cosh^{-1}\left(\frac{2a^2}{b^2}-1\right)}(1+\cosh\phi)d\phi\\ &=\frac{\pi ab^2}{\sqrt{a^2-b^2}}\left[\phi+\sinh\phi\right]_0^{\cosh^{-1}\left(\frac{2a^2}{b^2}-1\right)}\\ &=\frac{\pi ab^2}{\sqrt{a^2-b^2}}\left[\ln\left(\frac{2a^2}{b^2}-1+\frac{2a}{b^2}\sqrt{a^2-b^2}\right)+\frac{2a}{b^2}\sqrt{a^2-b^2}\right]\end{align}$$ Where we have made the substitutions $r^2=a^2u$, $u=\frac{2a^2-b^2}{2(a^2-b^2)}-v$, $v=\frac{b^2}{2(a^2-b^2)}w$, and $w=\cosh\phi$. Oh my, that was exhausting and we only have the area so far. The products of inertia are all zero by symmetry, and $$I_{zz}=\int\int r^2d^2A$$ Going throught the chain of substitutions, we find that $$\begin{align}r^2&=a^2u=a^2\left(\frac{2a^2-b^2}{2(a^2-b^2)}-v\right)=\frac{a^2b^2}{2(a^2-b^2)}\left(\frac{2a^2}{b^2}-1-w\right)\\ &=\frac{a^2b^2}{2(a^2-b^2)}\left(\frac{2a^2}{b^2}-1-\cosh\phi\right)\end{align}$$ So $$\begin{align}I_{zz}&=\frac{\pi a^3b^4}{2(a^2+b^2)^{3/2}}\int_0^{\cosh^{-1}\left(\frac{2a^2}{b^2}-1\right)}\left[-\cosh^2\phi+\frac{2(a^2-b^2)}{b^2}\cosh\phi+\frac{2a^2}{b^2}-1\right]d\phi\\ &=\frac{\pi a^3b^4}{2(a^2+b^2)^{3/2}}\left[-\frac12\sinh\phi\cosh\phi+\left(\frac{2a^2}{b^2}-1\right)\sinh\phi+\left(\frac{2a^2}{b^2}-\frac32\right)\phi\right]_0^{\cosh^{-1}\left(\frac{2a^2}{b^2}-1\right)}\\ &=\frac{\pi a^3b^4}{2(a^2+b^2)^{3/2}}\left[-\frac12\frac{2a}{b^2}\sqrt{a^2-b^2}\left(\frac{2a^2}{b^2}-1\right)+\frac{2(a^2-b^2)}{b^2}\frac{2a}{b^2}\sqrt{a^2-b^2}\right.\\ &\left.+\left(\frac{2a^2}{b^2}-\frac32\right)\ln\left(\frac{2a^2}{b^2}-1+\frac{2a}{b^2}\sqrt{a^2-b^2}\right)\right]\end{align}$$ If you think about it, $$\begin{align}I_{xx}&=I_{yy}=\int\int\left[z^2+y^2\right]d^2A=\int\int\left[\frac{b^2}{a^2}(a^2-r^2)+r^2\sin^2\theta\right]d^2A\\ &=b^2A+\left(\frac12-\frac{b^2}{a^2}\right)I_{zz}\end{align}$$ So, assuming my integrations for $A$ and $I_{zz}$ to be correct (which is quite a stretch) we also have the formulas for $I_{xx}$ and $I_{yy}$.
EDIT: I did a little checking and my result might actually be OK. For an oblate spheroid, the results may be cast as $$A=\frac{\pi ab^2}{\sqrt{a^2-b^2}}\left[\sinh^{-1}\left(\frac{2a}{b^2}\sqrt{a^2-b^2}\right)+\frac{2a}{b^2}\sqrt{a^2-b^2}\right]$$ $$\begin{align}I_{zz}&=\frac{\pi a^3b^4}{2(a^2-b^2)^{3/2}}\left[\left(\frac{2a^2}{b^2}-\frac32\right)\sinh^{-1}\left(\frac{2a}{b^2}\sqrt{a^2-b^2}\right)\right.\\ &\left.-\frac a{b^2}\left(\frac{2a^2}{b^2}-1\right)\sqrt{a^2-b^2}+\frac{4a}{b^4}(a^2-b^2)^{3/2}\right]\end{align}$$ While for the prolate spheroid we have $$A=\frac{\pi a^2b}{\sqrt{a^2-b^2}}\left[\sin^{-1}\left(\frac{2b}{a^2}\sqrt{a^2-b^2}\right)+\frac{2b}{a^2}\sqrt{a^2-b^2}\right]$$ $$\begin{align}I_{zz}&=\frac{\pi a^4b^3}{2(a^2-b^2)^{3/2}}\left[\left(\frac32-\frac{2b^2}{a^2}\right)\sin^{-1}\left(\frac{2b}{a^2}\sqrt{a^2-b^2}\right)\right.\\ &\left.-\frac b{a^2}\left(1-\frac{2b^2}{a^2}\right)\sqrt{a^2-b^2}+\frac{4b}{a^4}(a^2-b^2)^{3/2}\right]\end{align}$$