## Thomas' Calculus 13th Edition

$$\frac{\pi}{4}$$
We orient the cone with its vertex at the origin and axis along the $z$ -axis $\Rightarrow \phi=\frac{\pi}{4}.$ We use the $x$ -axis which is through the vertex and parallel to the base of the cone: \begin{align*} I_{x}&=\int_{0}^{2 \pi} \int_{0}^{1}\left(r^{2} \sin ^{2} \theta+z^{2}\right) d z r d r d \theta\\ &=\int_{0}^{2 \pi} \int_{0}^{1}\left(r^{3} \sin ^{2} \theta-r^{4} \sin ^{2} \theta+\frac{r}{3}-\frac{r^{4}}{3}\right) d r d \theta\\ &=\int_{0}^{2 \pi}\left(\frac{\sin ^{2} \theta}{20}+\frac{1}{10}\right) d \theta\\ &= \int_{0}^{2 \pi}\left(\frac{1}{40}-\frac{\cos 2 \theta}{40}+\frac{1}{10}\right) d \theta \\ &=\left[\frac{\theta}{40}-\frac{\sin 2 \theta}{80}+\frac{\theta}{10}\right]_{0}^{2 \pi}\\ &=\frac{\pi}{20}+\frac{\pi}{5}=\frac{\pi}{4} \end{align*}