MathJax TeX Test Page
The important thing to remember is that the total inertia is the sum of many moment of inertias of solid discs.
$$dI = \dfrac{1}{2}dm*r^2$$
Why is this true? Because the total moment of inertia is the sum of moment of inertias of solid disc, and the formula is $\frac{1}{2}mr^2$.
$$dm = \rho \text{ (density) } * \pi{}r^2 \text{ (area) } \mathrm{d}x \text{ (width) }$$
$$\rho = \dfrac{M}{\pi{}R^2h\frac{1}{3}} = \dfrac{3M}{\pi{}R^2h}$$
$$\mathrm{d}m = \dfrac{3M}{hR^2}*r^2\mathrm{d}x$$
$$\int_0^h \frac{1}{2}\left(\dfrac{3M}{hR^2}*r^2\right)* r^2\mathrm{d}x$$
$r$ is equal to $y$, which is equal to $\frac{R}{h}x$
$$\dfrac{3M}{2hR^2}\int_0^h y^4 \,\mathrm{d}x$$
Looking at the picture above, $y = \dfrac{R}{h}x$, so the integral is
$$\dfrac{3M}{2hR^2}\int_0^h \frac{R^4}{h^4}x^2 \,\mathrm{d}x$$
$$\dfrac{3MR^2}{2h^5}\int_0^h x^4 \,\mathrm{d}x$$
$$\dfrac{3MR^2}{2h^5} * \frac{h^5}{5} = \boxed{\frac{3MR^2}{10}}$$