Shown in the fig. is a hollow ice-cream cone (it is open at the top). If its mass is M, radius of its top, R and height, H, then its moment of inertia about its axis is :

$\begin{array}{l}From the vertex, along curved surface go to y and take a strip of thickness dy\\ \text{}and radius x.\\ Mass of this strip = dm=\frac{mass on CSA}{CSA}\times area of strip\\ \Rightarrow dm=\frac{M}{\pi R\sqrt{R^2+H^2}}\times 2\pi xdx\text{}\\ Also, from similar triangles, \frac{x}{y}=\frac{R}{\sqrt{R^2+H^2}}\\ \end{array}$

$\begin{array}{l}I=\underset{ }{\overset{ }{ }}dm{x}^2 =\frac{M}{\pi R\sqrt{R^2+H^2}}\underset{ }{\overset{ }{ }}2\pi x^{3}dy\\ \text{}I=\frac{2M}{R\sqrt{R^2+H^2}}\times \frac{R^3}{{\left(R^2+H^2\right)}^{\frac{3}{2}}}\times {\left[\frac{y^4}{4}\right]}_0^{\sqrt{R^2+H^2}}=\frac{M{R}^2}{2}\end{array}$