Example 4. Calculate the integral $I=\iint_{\sigma}(x \cos \alpha+\cos \beta+$ $\left.+x z^{2} \cos \gamma\right) d \sigma$, where $\sigma$ is the part of the sphere $x^{2}+y^{2}+z^{2}=1$ located in the first octant.
Solution. Let $D_{1}, D_{2}, D_{3}$ be the projections of the surface of the unit sphere onto the coordinate planes $O y z, O x z, O x y$ respectively. We will compute each of the component integrals separately:

$$
\begin{aligned}
& I_{\mathrm{I}}=\iint_{\sigma} x \cos \alpha d \sigma=\iint_{D_{1}} x d y d z=\iint_{D_{1}} \sqrt{1-y^{2}-z^{2}} d y d z= \\
& =\int_{0}^{\pi / 2} d \varphi \int_{0}^{1} \sqrt{1-r^{2}} \cdot r d r=\frac{\pi}{2}\left(-\left.\frac{\left(1-r^{2}\right)^{3 / 2}}{3}\right|_{0} ^{1}\right)=\frac{\pi}{6} \\
& I_{2}=\iint_{\sigma} \cos \beta d \sigma=\iint_{D_{2}} d x d z=\frac{\pi}{4} \\
& \begin{aligned}
I_{3}=\iint_{\sigma} x z^{2} \cos \gamma d \sigma=\iint_{D_{3}} x z^{2} d x d y=\iint_{D_{3}} x\left(1-x^{2}-y^{2}\right) d x d y= \\
=\int_{0}^{\pi / 2} \cos \varphi d \varphi \cdot \int_{0}^{1}\left(1-r^{2}\right) r^{2} d r=\left.\left.\sin \varphi\right|_{0} ^{\pi / 2}\left(\frac{r^{3}}{3}-\frac{r^{5}}{5}\right)\right|_{0} ^{1}=\frac{2}{15}
\end{aligned}
\end{aligned}
$$

Thus, $I=I_{1}+I_{2}+I_{3}=\frac{\pi}{6}+\frac{\pi}{4}+\frac{2}{15}=\frac{2}{15}+\frac{5 \pi}{12}$.

Answer. $I=\frac{2}{15}+\frac{5 \pi}{12}$.