921. Find the mass of a hemisphere if the surface density at each of its points is numerically equal to the distance of this point from the radius perpendicular to the base of the hemisphere.
Solution. Place the origin of the rectangular coordinate system at the center of the base of the hemisphere and direct the z-axis perpendicular to this base. Then the equation of the hemisphere will be $z=\sqrt{R^{2}-x^{2}-y^{2}}$, where $R$ is the radius of the hemisphere; the surface density at point $M(x, y, z)$ of the hemisphere will be $\delta(M)=$ $=\sqrt{x^{2}+y^{2}} ; \quad d s=\sqrt{1+\left(z_{x}^{\prime}\right)^{2}+\left(z_{y}^{\prime}\right)^{\prime}} d x d y=\frac{R d x d y}{\sqrt{R^{2}-x^{2}-y^{2}}}$.

Substituting into formula (2), we get:

$$
m=\iint_{\sigma} \delta d s=R \iint_{\sigma_{x y}} \frac{\sqrt{x^{2}+y^{2}} d x d y}{\sqrt{R^{2}-x^{2}-y^{2}}}, \quad \text { where } \sigma_{x y}-\text { circle } x^{2}+y^{2} \leqslant R^{2}
$$

Switching to polar coordinates, we find

$$
m=R \iint_{i \leq R} \frac{\rho^{2} d \varphi d \rho}{\sqrt{R^{2}-\rho^{2}}}=R \int_{0}^{2 \pi} d \varphi \int_{0}^{R} \frac{\rho^{2} d \rho}{\sqrt{R^{2}-\rho^{2}}}=\frac{\pi^{2} R^{3}}{2}
$$

(The inner integral is computed using the substitution $\rho=R \sin t$.)