1.19. Find the area of the part of the surface of the cylinder $x^{2}+$ $+y^{2}=2 a x$, cut out by the cylinder $z^{2}=2 a(2 a-x)$.
Solution. As is known (see formula (1.83)), the area of a part of the surface $z=f(x, y)$ (the projection onto the $XOY$ plane is $\Omega_{\text{XOY}}$) is calculated by the formula

$$
\mu(S)=\iint_{\Omega_{X O Y}} \sqrt{1+z_{x}^{2}+z_{y}^{2}} d x d y
$$

In our case, the equation of the surface is $y= \pm \sqrt{2 a x-x^{2}}$. Due to symmetry, we can limit ourselves to the equation $y=\sqrt{2 a x-x^{2}}$ and double the result in the calculations. This surface projects onto the region $\Omega_{x o z}$ of the $XOZ$ plane, where

$$
\Omega_{X O z}=\left\{(x, z):|z| \leqslant 2 a, a>0, \quad 0 \leqslant x \leqslant 2 a-\frac{z^{2}}{2 a}\right\}
$$

The desired area is

$$
\begin{aligned}
\mu(S)= & 2 \iint_{\Omega_{x o z}} \sqrt{1+y_{x}^{2}} d x d z=2 \int_{0}^{2 a} \sqrt{1+y_{x}^{2}} d x \int_{-\sqrt{2 a(2 a-x)}}^{\sqrt{2 a(2 a-x)}} d z= \\
& =2 \int_{0}^{2 a} \sqrt{1+\left(\frac{a-x}{\sqrt{2 a x-x^{2}}}\right)^{2}} 2 \sqrt{2 a(2 a-x)} d x= \\
= & 4 \int_{0}^{2 a} \sqrt{2 a x-x^{2}+a^{2}-2 a x+x^{2}} \sqrt{\frac{2 a(2 a-x)}{x(2 a-x)}} d x= \\
& =4 a \int_{0}^{2 a} \sqrt{\frac{2 a}{x}} d x=\left.4 a \sqrt{2 a} 2 \sqrt{x}\right|_{0} ^{2 a}=16 a^{2}
\end{aligned}
$$