homework and exercises - 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.

Answer

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:

$$I = \int_{-R}^{R} \frac{M}{2 R} \cdot \left(R^2 - x^2\right)\ \text{d}x \\ = \frac{M}{2R}\left[{2R^3-\frac23 R^3}\right]\\ = \frac23 M R^2$$