Example 3. Calculate the mass of a part of the cone $36\left(x^{2}+y^{2}\right)=z^{2}$, located inside the cylinder $x^{2}+y^{2}=1(x \geqslant 0, z \geqslant 0)$, if the density is $\rho=\frac{5\left(x^{2}+y^{2}\right)}{6}$.
Solution. The limits of the variable $z: z=0$ and $z=6 \sqrt{x^{2}+y^{2}}$, since a vertical line will intersect the body at points on the base $(z=0)$ and the cone. The region of integration for the double integral is the circle $x^{2}+y^{2} \leqslant 1$, so it is convenient to switch to polar coordinates. We have:

$$
\begin{aligned}
& m=\frac{5}{6} \iint_{W}\left(x^{2}+y^{2}\right) d x d y d z=\frac{5}{6} \iint_{D}\left(x^{2}+y^{2}\right) d x d y \int_{0}^{6 \sqrt{x^{2}+y^{2}}} d z= \\
& =5 \iint_{x^{2}+y^{2} \leqslant 1}\left(x^{2}+y^{2}\right)^{3 / 2} d x d y=5 \int_{0}^{2 \pi} d \varphi \int_{0}^{1} r^{4} d r=5 \cdot 2 \pi \cdot \frac{1}{5}=2 \pi
\end{aligned}
$$

Answer. $2 \pi$.