906. Compute surface integrals of the first kind (over the area of the surface):
1) $I=\iint_{\sigma}(6 x+4 y+3 z) d s$, where $\sigma$ is the part of the plane $x+2 y+$ $+3 z=6$ located in the first octant.
2) $K=\iint_{W}\left(y+z+V \sqrt{a^{2}-x^{2}}\right) d s$, where $W-$ is the surface of the cylinder $x^{2}+y^{2}=a^{2}$, bounded by the planes $z=0$ and $z=h$.
Solution. 1) The surface of integration $\sigma$ is the triangle $ABC$ (Fig. 190). Using its equation and formula (1), we transform the given surface integral into a double integral with variables $x$ and $y$:

$$
\begin{aligned}
& z=\frac{1}{3}(6-x-2 y), d s=\sqrt{1+\left(z_{x}^{\prime}\right)^{2}+\left(z_{4}^{\prime}\right)^{2}} d x d y=\frac{\sqrt[V]{\overline{1}}}{3} d x d y \\
& l=\frac{V}{3} \iint_{\sigma_{x j}} \int_{0}(5 x+2 y+6) d x d y
\end{aligned}
$$

where $\sigma_{x y}$ is the triangle $ABO$, which is the projection of $\sigma$ onto the plane $x O y$.[^34]

The obtained double integral is computed by repeated integration:

$$
\begin{gathered}
I=\frac{\sqrt[V]{14}}{3} \int_{y_{0}=0}^{y_{B}=3} d y \int_{x_{p}=0}^{x_{Q}=\beta-2 y}(5 x+2 y+6) d x= \\
=\cdots \frac{\sqrt{14}}{3} \int_{0}^{3}\left(\frac{5}{2} x^{2}+2 x y+\left.6 x\right|_{x=0} ^{x=-2 y}\right) d y=2 \sqrt{14} \int_{\overline{3}}^{3}\left(y^{2}-10 y+21\right) d i y== \\
=\left.2 \sqrt{14}\left(\frac{y^{3}}{3}-5 y^{2}+21 y\right)\right|_{0} ^{3}=54 \sqrt{14}
\end{gathered}
$$

2) For the entire surface $W$, it is impossible to express one of the coordinates as a single-valued function of the other two coordinates. The parts of the cylindrical surface located on different sides of the vertical coordinate planes (Fig. 191) have different explicit equations: the part $\mathbb{W}_{1}$, located to the left of the plane $x O z$, has the equation $y=-\sqrt{a^{2}-x^{2}}$, and the part $\mathbb{W}_{2}^{\prime}$, located to the right of this plane, has the equation $y=\sqrt{a^{2}-x^{2}}$. Therefore, we compute the given integral $K$ over the surface $\mathbb{W}$ as the sum of integrals $K_{1}$ and $K_{2}$ over the constituent parts $\mathbb{W}_{1}$ and $\mathbb{W}_{2}$.

Transforming the surface integrals $K_{1}$ and $K_{2}$ into double integrals with variables $x$ and $z$, we obtain:

$$
\begin{aligned}
& d s=\sqrt{1+\left(y_{1}\right)^{2}+\left(I_{2}\right)^{2}} d x d z=\frac{a d x d z}{\sqrt{-a^{2}-x^{2}}}, \\
& K_{1}=a \int_{\left(w_{1}\right)_{*}} \frac{z d x d z}{\sqrt{a^{2}-x^{2}}}, \quad K_{2}=a \int_{\left(x_{x^{2}}\right.} \int_{n^{2}}\left(2+\frac{z}{\sqrt{a^{2}-x^{2}}}\right) d x d z .
\end{aligned}
$$

Therefore,

$$
K=K_{1}+K_{2}=2 a \iint_{A I C C D}\left(1+\frac{z}{V}\right) d x d z,
$$

since the rectangle $ABCD$ is the common projection of the surfaces $\mathbb{W}_{1}$ and $\mathbb{W}_{2}$ onto the plane $x O z$.

Computing the obtained double integral, we get:

$$
\begin{gathered}
K=2 a \int_{0}^{h} d z \int_{-a}^{a}\left(1+\frac{2}{\sqrt{a^{2}-x^{2}}}\right) d x=2 a \int_{0}^{h}\left(x+\left.z \arcsin \frac{x}{a}\right|_{x=-a} ^{x=a}\right) d z= \\
=2 a \int_{0}^{h}(2 a+\pi z) d z=\left.2 a\left(2 a z+\frac{\pi z^{2}}{2}\right)\right|_{0} ^{h}=a h(4 a+\pi h)
\end{gathered}
$$