<s>[INST] Moment of inertia of a hollow sphere wrt the centre? I've been trying to compute the moment of inertia of a uniform hollow sphere (thin walled) wrt the centre, but I'm not quite sure what was wrong with my initial attempt (I've come to the correct answer now with a different method). Ok, here was my first method:
Consider a uniform hollow sphere of radius $R$ and mass $M$. On the hollow sphere, consider a concentric ring of radius $r$ and thickness $\text{d}x$. The mass of the ring is therefore $\text{d}m = \frac{M}{4\pi R^2}\cdot 2\pi r\cdot\text{d}x$. Now, use $r^2 = R^2 - x^2:$
$$\text{d}m = \frac{M}{4\pi R^2}\cdot 2\pi \left(R^2 - x^2 \right)^{1/2}\text{d}x$$
and the moment of inertia of a ring wrt the centre is $I = MR^2$, therefore:
$$\text{d}I = \text{d}m\cdot r^2 = \frac{M}{4\pi R^2}\cdot 2\pi\left(R^2 - x^2\right)^{3/2}\text{d}x $$
Integrating to get the total moment of inertia:
$$I = \int_{-R}^{R} \frac{M}{4\pi R^2} \cdot 2\pi\cdot \left(R^2 - x^2\right)^{3/2}\ \text{d}x = \frac{3MR^2 \pi}{16}$$
which obviously isn't correct as the real moment of inertia wrt the centre is $\frac{2MR^2}{3}$. 
What was wrong with this method? Was it how I constructed the element? Any help would be appreciated, thanks very much. 
 [/INST] The mass of the ring is wrong. The ring ends up at an angle, so its total width is not $dx$ but $\frac{dx}{sin\theta}$
You made what I believe was a typo when you wrote
$$\text{d}m = \frac{M}{4\pi R^2}\cdot 2\pi \left(R^2 - x^2 \right)\text{d}x$$
because based on what you wrote further down, you intended to write
$$\text{d}m = \frac{M}{4\pi R^2}\cdot 2\pi \sqrt{\left(R^2 - x^2 \right)}\text{d}x$$
This problem is much better done in polar coordinates - instead of $x$, use $\theta$. But the above is the basic reason why you went wrong.
In essence, $sin\theta=\frac{r}{R}$ so you could write
$$\text{d}m = \frac{M}{4\pi R^2}\cdot 2\pi \frac{r}{sin\theta} \ \text{d}x \\
= \frac{M}{4\pi R^2}\cdot 2\pi \frac{r}{\frac{r}{R}} \ \text{d}x\\
= \frac{M}{4\pi R^2}\cdot 2\pi R \ \text{d}x\\
= \frac{M}{2 R} \ \text{d}x$$
Now we can substitute this into the integral:
$$I = \int_{-R}^{R} \frac{M}{2 R} \cdot \left(R^2 - x^2\right)\ \text{d}x \\
= \frac{M}{2R}\left[{2R^3-\frac23 R^3}\right]\\
= \frac23 M R^2$$
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