Find the moment of inertia of a homogeneous solid cone
Find the moment of inertia of a homogeneous solid cone with base radius $R$ and height $H$ with respect to its axis.
Explanation of Method
Slice the cone into disks of radius $r(z)=Rz/H$ at height $z$, use the standard disk formula $dI=\frac12\,dm\cdot r^2$ for each thin disk, and integrate over $z\in[0,H]$.
Step 1: Cross-sectional radius
By similar triangles, at height $z$ (apex at $z=0$): $r(z)=\dfrac Rz H$.
Step 2: Set up the triple integral
\[I_z=\int_0^H\int_0^{2\pi}\int_0^{Rz/H}\rho^3\,d\rho\,d\theta\,dz = 2\pi\int_0^H\frac{R^4z^4}{4H^4}\,dz\]
Step 3: Evaluate
\[I_z=\frac{\pi R^4}{2H^4}\cdot\frac{H^5}5=\frac{\pi R^4H}{10}\]
\[\boxed{I_z=\frac{\pi R^4H}{10}\cdot\delta}\qquad(\delta=\text{density})\]
Step 4: In terms of mass
With $m=\delta\cdot\frac13\pi R^2H$: \[I_z=\frac{\pi R^4H}{10}\cdot\delta = m\cdot\frac{3R^2}{10}\]
\[\boxed{I_z=\frac{3mR^2}{10}}\]
Step 5: Comparison and radius of gyration
A solid cylinder of the same radius has $I=\frac12mR^2$; the cone's moment is $\frac35$ of that, reflecting mass concentrated toward the (smaller-radius) apex. Radius of gyration: $k=R\sqrt{3/10}\approx0.548R$.
Result
Final answer:
\[\boxed{I_z = \dfrac{\pi R^4H}{10}\delta = \dfrac{3mR^2}{10}}\]