Cone
Parametarizing the Volume
We can do this using cylindrical coordinates
\[\b{r}(s, \phi, t) = s \; \u{s} + z \; \u{z} \\[10pt] V = \{ \b{r}(s, \phi, z) \ : \ 0 \leq s \leq R \quad 0 \leq \phi < 2\pi \quad \tfrac{H}{R}s \leq z \leq H \}\]This is the first object where the parameters bounds depend on the another parameters.
\[\frac{\partial \b{r}}{\partial s} = \u{s} \qquad\qquad \frac{\partial \b{r}}{\partial \phi} = s \; \u{\phi} \qquad\qquad \frac{\partial \b{r}}{\partial z} = \u{z}\]Therefore
\[dV = \left \lvert \begin{array}{ccc} ds & 0 & 0 \\ 0 & s \; d\phi & 0 \\ 0 & 0 & dz \end{array} \right \rvert = s \; ds \; d\phi \; dz\]Mass
\[\begin{align} M &= \int \; dm \\[10pt] &= \rho \int \; dV \\[10pt] &= \rho \int_{0}^{R} \int_{0}^{2\pi} \int_{\frac{H}{R}s}^{H} s \; ds \; d\phi \; dz \\[10pt] &= \rho \left ( \int_{0}^{2\pi} \; d\phi \right ) \left ( \int_{0}^{R} \int_{\frac{H}{R}s}^{H} s \; ds \; dz \right ) \\[10pt] &= \rho \left ( 2\pi \right ) \left ( \int_{0}^{R} s(H - \tfrac{H}{R}s) \; ds \right ) \\[10pt] &= \rho \cdot 2\pi \left ( \int_{0}^{R} (Hs - \tfrac{H}{R}s^2) \; ds \right ) \\[10pt] &= \rho \cdot 2\pi \left ( \tfrac{1}{2}HR^2 - \tfrac{1}{3}\tfrac{H}{R}R^3 \right ) \\[10pt] &= \rho \cdot 2\pi \left ( \tfrac{1}{6}R^2H \right ) \\[10pt] &= \rho \cdot \tfrac{1}{3} \pi R^2 H \end{align}\]Moment of Inertia About Central Axis
\[\begin{align} M &= \int r_{axis}^2 \; dm \\[10pt] &= \rho \int r_{axis}^2 \; dV \\[10pt] &= \rho \int_{0}^{R} \int_{0}^{2\pi} \int_{\frac{H}{R}s}^{H} s^2 \cdot s \; ds \; d\phi \; dz \\[10pt] &= \rho \left ( \int_{0}^{2\pi} \; d\phi \right ) \left ( \int_{0}^{R} \int_{\frac{H}{R}s}^{H} s^3 \; ds \; dz \right ) \\[10pt] &= \rho \left ( 2\pi \right ) \left ( \int_{0}^{R} s^3(H - \tfrac{H}{R}s) \; ds \right ) \\[10pt] &= \rho \cdot 2\pi \left ( \int_{0}^{R} (Hs^3 - \tfrac{H}{R}s^4) \; ds \right ) \\[10pt] &= \rho \cdot 2\pi \left ( \tfrac{1}{4}HR^4 - \tfrac{1}{5}\tfrac{H}{R}R^5 \right ) \\[10pt] &= \rho \cdot 2\pi \left ( \tfrac{1}{20}R^4H \right ) \\[10pt] &= \rho \cdot \tfrac{1}{10} \pi R^4 H \\[10pt] &= \tfrac{3}{10} M R^2 \end{align}\]Moment of Inertia About Central Diameter
\[\begin{align} M &= \int r_{axis}^2 \; dm \\[10pt] &= \rho \int r_{axis}^2 \; dV \\[10pt] &= \rho \int_{0}^{R} \int_{0}^{2\pi} \int_{\frac{H}{R}s}^{H} (s^2 \sin^2 \phi + z^2) \cdot s \; ds \; d\phi \; dz \\[10pt] &= \rho \int_{0}^{R} \int_{0}^{2\pi} \int_{\frac{H}{R}s}^{H} (s^3 \sin^2 \phi + s z^2) \; ds \; d\phi \; dz \\[10pt] &= \rho \int_{0}^{R} \int_{\frac{H}{R}s}^{H} (\pi s^3 + 2\pi s z^2) \; ds \; dz \\[10pt] &= \rho \cdot \pi \int_{0}^{R} (s^3 (H - \tfrac{H}{R}s) + \tfrac{2}{3} s (H - \tfrac{H}{R}s)^3) \; ds \\[10pt] &= \rho \cdot \pi \int_{0}^{R} (\tfrac{H}{R} (Rs^3 - s^4) + \tfrac{2}{3} \tfrac{H^3}{R^3} (R^3 s - 3 R^2 s^2 + 3 R s^3 - s^4)) \; ds \\[10pt] &= \rho \cdot \pi (\tfrac{H}{R} (R \cdot \tfrac{1}{4}R^4 - \tfrac{1}{5}R^5) + \tfrac{2}{3} \tfrac{H^3}{R^3} (R^3 \cdot \tfrac{1}{2}R^2 - 3 R^2 \cdot \tfrac{1}{3}R^3 + 3 R \cdot \tfrac{1}{4}R^4 - \tfrac{1}{5}R^5)) \\[10pt] &= \rho \cdot \pi (\tfrac{H}{R} (\tfrac{1}{20}R^5) + \tfrac{2}{3} \tfrac{H^3}{R^3} (\tfrac{1}{20} R^5)) \\[10pt] &= \rho \cdot (\tfrac{1}{20} \pi R^4 H + \tfrac{1}{30} \pi R^2 H^3 )\\[10pt] &= \tfrac{3}{20} M R^2 + \tfrac{1}{10} M H^2 \\[10pt] \end{align}\]Inertia Tensor
\[\begin{align} \m{I} &= \begin{bmatrix} \tfrac{3}{20} M R^2 + \tfrac{1}{10} M H^2 & 0 & 0 \\ 0 & \tfrac{3}{20} M R^2 + \tfrac{1}{10} M H^2 & 0 \\ 0 & 0 & \tfrac{3}{10} M R^2 \end{bmatrix} \\[10pt] &= \tfrac{1}{20} M \begin{bmatrix} 3 R^2 + 2 H^2 & 0 & 0 \\ 0 & 3 R^2 + 2 H^2 & 0 \\ 0 & 0 & 6 R^2 \end{bmatrix} \end{align}\]Inertia Tensor at Center of Mass
Now, using the parallel axis theorem, we can shift the cone to its center of mass. We are shifting the cone straight down by $\tfrac{3}{4}H$. Therefore, $\b{D} = - \tfrac{3}{4}H \; \u{z}$.
\[\begin{align} \m{I_{\text{cm}}} &= \tfrac{1}{20} M \begin{bmatrix} 3 R^2 + 2 H^2 & 0 & 0 \\ 0 & 3 R^2 + 2 H^2 & 0 \\ 0 & 0 & 6 R^2 \end{bmatrix} - M \begin{bmatrix} \tfrac{9}{16} H^2 & 0 & 0 \\ 0 & \tfrac{9}{16} H^2 & 0 \\ 0 & 0 & 0 \end{bmatrix} \\[10pt] &= \begin{bmatrix} \tfrac{3}{20} M R^2 - \tfrac{37}{80} M H^2 & 0 & 0 \\ 0 & \tfrac{3}{20} M R^2 - \tfrac{37}{80} M H^2 & 0 \\ 0 & 0 & \tfrac{3}{10} M R^2 \end{bmatrix} \\[10pt] &= \tfrac{1}{80} M \begin{bmatrix} 12 R^2 - 37 H^2 & 0 & 0 \\ 0 & 12 R^2 - 37 H^2 & 0 \\ 0 & 0 & 24 R^2 \end{bmatrix} \end{align}\]The center of mass is not necessarily the most elegant want to express the moment of inertia.