2.12. Compute surface integrals of the first kind:
a) $\iint_{\sigma}|x| d S, \quad \sigma: x^{2}+y^{2}+z^{2}=1, \quad z \geqslant 0$

b) $\iint_{\sigma}\left(x^{2}+y^{2}\right) d S, \quad \sigma: x^{2}+y^{2}=2 z, \quad z=1$;

c) $\iint_{\sigma}\left(x^{2}+y^{2}+z^{2}\right) d S$, where $\sigma$ is the part of the cone $z^{2}-x^{2}-y^{2}=0$, $z \geqslant 0$, cut off by the cylinder $x^{2}+y^{2}-2 x=0$.
Solution.

a) Let's represent the hemisphere $x^{2}+y^{2}+z^{2}=1, z \geqslant 0$ with parametric equations:

$x=\sin \theta \cos \varphi, \quad y=\sin \theta \sin \varphi, \quad z=\cos 0, \quad \varphi \in[0,2 \pi], \quad \theta \in\left[0, \frac{\pi}{2}\right]$.

The element of area in this representation is

$$
d S=\sin \theta d \varphi d \theta
$$

After all substitutions, the surface integral of the first kind reduces to a double integral over the region $D$:

$$
\begin{gathered}
\varphi \in[0,2 \pi], \quad \theta \in\left[0, \frac{\pi}{2}\right]: \\
\iiint_{\sigma}|x| d S=\iint_{D} \sin ^{2} \theta|\cos \varphi| d \theta d \varphi=\int_{0}^{2 \pi}|\cos \varphi| d \varphi \int_{0}^{\frac{\pi}{2}} \frac{1-\cos 2 \theta}{2} d \theta= \\
=\left.4 \int_{0}^{\frac{\pi}{2}} \cos \varphi d \varphi \cdot\left(\frac{\theta}{2}-\frac{\sin 2 \theta}{4}\right)\right|_{0} ^{\frac{\pi}{2}}=\left.\pi(+\sin \varphi)\right|_{0} ^{\frac{\pi}{2}}=\pi
\end{gathered}
$$

b) The element of area of the paraboloid $2 z=x^{2}+y^{2}$ is calculated in the usual way:

$$
d S=\sqrt{1+z_{x}^{2}+z_{y}^{2}} d x d y=\sqrt{1+x^{2}+y^{2}} d x d y
$$

The surface integral reduces to a double integral over the circle $x^{2}+y^{2} \leqslant 2$ on the $XOY$ plane:

$$
\iint_{\sigma}\left(x^{2}+y^{2}\right) d S=\iint_{x^{2}+y^{2} \leqslant 2}\left(x^{2}+y^{2}\right) \cdot \sqrt{1+x^{2}+y^{2}} d x d y
$$

To compute this integral, we switch to polar coordinates:

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

c) The element of area of the cone $z=\sqrt{x^{2}+y^{2}}$ is

$$
d S=\sqrt{1+z_{x}^{2}+z_{y}^{2}} d x d y=\sqrt{2} d x d y
$$

The surface integral reduces to a double integral over the circle $(x-1)^{2}+y^{2} \leqslant 1$ on the $XOY$ plane (considering that $\left.z^{2}=x^{2}+y^{2}\right)$

$$
\iint_{\sigma}\left(x^{2}+y^{2}+z^{2}\right) d S=\sqrt{2} \iiint_{(x-1)^{2}+y^{2} \leqslant 1} 2\left(x^{2}+y^{2}\right) d x d y
$$

In this case, we switch to polar coordinates $x=r \cos \varphi, y=r \sin \varphi$, with the integration region $D$:

$$
\begin{gathered}
0 \leqslant r \leqslant 2 \cos \varphi, \varphi \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]: \\
\sqrt{2} \iiint_{(x-1)^{2}+y^{2} \leqslant 1} 2\left(x^{2}+y^{2}\right) d x d y= \\
=2 \sqrt{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} d \varphi \int_{0}^{2 \cos \varphi} r^{3} d r=8 \sqrt{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos ^{4} \varphi d \varphi=3 \sqrt{2} \pi
\end{gathered}
$$