### 2.17. Compute the surface integral of the second kind

$$
\iint_{S}\left(x y^{2}+z^{2}\right) d y d z+\left(y z^{2}+x^{2}\right) d z d x+\left(z x^{2}+y^{2}\right) d x d y
$$

where $S-$ is the outer side of the upper ( $z \geqslant 0$ ) hemisphere $x^{2}+$ $+y^{2}+z^{2}=a^{2}$.
Solution. The projection $S$ onto the $X O Y$ plane is a circle of radius $a$:

$$
x^{2}+y^{2} \leqslant a^{2}
$$

We will convert the given surface integral into a double integral over the variables $x, y$. For this, we will again use formula (2.54). In our case,

$$
\begin{aligned}
& P=x y^{2}+z^{2}, Q=y z^{2}+x^{2}, R=z x^{2}+y^{2}, z=\sqrt{a^{2}-x^{2}-y^{2}}, \\
& d S=\sqrt{1+z_{x}^{2}+z_{y}^{2}} d x d y=\frac{a d x d y}{\sqrt{a^{2}-x^{2}-y^{2}}}, \\
& \dot{n}=\left(-\frac{z_{x}}{\sqrt{1+z_{x}^{2}+z_{y}^{2}}},-\frac{z_{y}}{\sqrt{1+z_{x}^{2}+z_{y}^{2}}}, \frac{1}{\sqrt{1+z_{x}^{2}+z_{y}^{2}}}\right) .
\end{aligned}
$$

(The minus sign corresponds to the outer side of the hemisphere.)

$$
\begin{aligned}
& \vec{n} d S=\bar{d} \vec{S}=\left(-z_{x} d x d y,-z_{y} d x d y, d x d y\right)= \\
& =\left(\frac{x d x d y}{\sqrt{a^{2}-x^{2}-y^{2}}}, \frac{y d x d y}{\sqrt{a^{2}-x^{2}-y^{2}}}, d x d y\right)
\end{aligned}
$$

We will perform the necessary calculations:

$$
\begin{aligned}
& \iint_{S}\left(x y^{2}+z^{2}\right) d y d z+\left(y z^{2}+x^{2}\right) d z d x+\left(z x^{2}+y^{2}\right) d x d y= \\
& =\iint_{x^{2}+y^{2} \leqslant a^{2}}\left[\frac{\left(x y^{2}+a^{2}-x^{2}-y^{2}\right) x}{\sqrt{a^{2}-x^{2}-y^{2}}}+\frac{\left(x y^{2}+y\left(a^{2}-x^{2}-y^{2}\right)\right) y}{\sqrt{a^{2}-x^{2}-y^{2}}}+\right. \\
& \left.+\left(\sqrt{a^{2}-x^{2}-y^{2}} x^{2}+y^{2}\right)\right] d x d y
\end{aligned}
$$

Notice that due to the symmetry of the integration region with respect to the $O X, O Y$ axes, all integrals of functions containing odd powers of the variables $x, y$ in the numerators will be zero. Due to this observation, only the integral containing even powers of $x, y$ remains, which is further computed by transitioning to polar coordinates:

$$
\begin{gathered}
\iint_{x^{2}+y^{2} \leqslant a^{2}}\left(\frac{x^{2} y^{2}}{\sqrt{a^{2}-x^{2}-y^{2}}}+\left(x^{2}+y^{2}\right) \sqrt{a^{2}-x^{2}-y^{2}}+y^{2}\right) d x d y= \\
=\int_{0}^{2 \pi} d \varphi \int_{0}^{a}\left(\frac{r^{5} \cos ^{2} \varphi \sin ^{2} \varphi}{\sqrt{a^{2}-r^{2}}}+r^{3} \sqrt{a^{2}-r^{2}}+r^{3} \sin ^{2} \varphi\right) d r= \\
=\frac{1}{8} \int_{0}^{2 \pi} \sin ^{2} 2 \varphi d \varphi \int_{0}^{a} \frac{r^{4} d r^{2}}{\sqrt{a^{2}-r^{2}}}+\pi \int_{0}^{a} r^{2} \sqrt{a^{2}-r^{2}} d r^{2}+ \\
+\int_{0}^{2 \pi} \frac{1-\cos 2 \varphi}{2} d \varphi \int_{0}^{a} r^{3} d r=\frac{\pi}{8} \int_{0}^{a^{2}} \frac{t^{2} d t}{\sqrt{a^{2}-t}}+\pi \int_{0}^{a^{2}} t \sqrt{a^{2}-t} d t+\frac{\pi a^{4}}{4}= \\
=\frac{\pi}{8}\left(-\left.2 \sqrt{a^{2}-t} \cdot t^{2}\right|_{0} ^{a^{2}}-\left(\left.\frac{4}{3} \sqrt{\left(a^{2}-t\right)^{3}} \cdot 2 t\right|_{0} ^{a^{2}}\right)+\left(-\left.\frac{8}{15} \sqrt{\left(a^{2}-t\right)^{5}} \cdot 2\right|_{0} ^{a^{2}}\right)\right)+ \\
+\pi\left(-\left.\frac{2}{3} \sqrt{\left(a^{2}-t\right)^{3}} \cdot t\right|_{0} ^{a^{2}}-\left(\left.\frac{4}{15} \sqrt{\left(a^{2}-t\right)^{5}}\right|_{0} ^{a^{2}}\right)\right)+ \\
+\frac{\pi a^{4}}{4}=\frac{2 \pi a^{5}}{15}+\frac{4 \pi a^{5}}{15}+\frac{\pi a^{4}}{4}=\frac{2 \pi a^{5}}{5}+\frac{\pi a^{4}}{4}=\frac{\pi a^{4}}{20}(8 a+5) .
\end{gathered}
$$

Problems related to the Gauss-Ostrogradsky formula.