920. Find the area of the part of the surface:
1) of the cone $z^{2}=2 x y$, located in the first octant between the planes $x=2, y=4$;
2) of the sphere $x^{2}+y^{2}+z^{2}=R^{2}$, located inside the cylinder $x^{2}+y^{2}=R x$
3) of the cylinder $x^{2}+y^{2}=R x$, located inside the sphere $x^{2}+$ $+y^{2}+z^{2}=R^{2}$
Solution. 1) We apply formula (1). Using the equation of the cone, we transform the surface integral into a double integral with variables $x$ and $y$:

$$
\begin{gathered}
S=\iint_{\boldsymbol{\sigma}} d s=\iint_{\sigma_{x y}} \sqrt{1+\left(z_{x}^{\prime}\right)^{2}+\left(z_{y}^{\prime}\right)^{2}} d x d y=\frac{1}{\sqrt{2}} \iint_{\sigma_{x y}}\left(\sqrt{\frac{x}{y}}+ \\
+\sqrt{\frac{y}{x}}\right) d x d y
\end{gathered}
$$

where $\sigma_{x y}$ is the projection of $\sigma$ onto the $x O y$ plane - a rectangle $O A B C$ (Fig. 194).

Evaluating the double integral, we get

$$
\begin{gathered}
S=\frac{1}{\sqrt{2}} \int_{0}^{2} d x \int_{0}^{4}\left(\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{x}}\right) d y= \\
=\left.\frac{1}{\sqrt{2}} \int_{0}^{2}\left(2 \sqrt{x y}+\frac{2}{3} \sqrt{\frac{y^{3}}{x}}\right)\right|_{y=0} ^{y=4} d x= \\
=2 \sqrt{2} \int_{0}^{2}\left(\sqrt{x}+\frac{4}{3 \sqrt{x}}\right) d x=\left.2 \sqrt{2}\left(\frac{2}{3} \sqrt{x^{3}}+\frac{8}{3} \sqrt{x}\right)\right|_{0} ^{2}=16
\end{gathered}
$$

2) The given surface * is symmetric with respect to the planes $x O y$ and $x O z$; in the first octant, one-fourth of it ( $\sigma$ ) is located (Fig. 195), for which the ordinate $z=\sqrt{R^{2}-x^{2}-y^{2}}$. Therefore, according to formula (1), the desired area is

$$
S=4 \iint_{\sigma} d s=4 \iint_{\sigma_{1}} \sqrt{1+\left(z_{x}^{\prime}\right)^{2}+\left(z_{y}^{\prime}\right)^{2}} d x d y=4 R \iint_{\sigma_{1}} \frac{d x d y}{\sqrt{R^{2}-x^{2}-y^{2}}},
$$

where $\sigma_{1}$ is a semicircle bounded by the circle $x^{2}+y^{2}=R x$ and the $O x$ axis $\left(x^{2}+y^{2} \leqslant R x, y \geqslant 0\right)$.[^35]

Switching to polar coordinates and integrating, we have: $0 \leqslant \rho \leqslant R \cos \varphi, \quad 0 \leqslant \rho \leqslant \frac{\pi}{2} ;$

$$
S=4 R \iint_{\sigma_{1}} \frac{\rho d t h}{\sqrt{ } R^{2}-\frac{\rho^{2}}{}} \frac{d \rho}{-\rho_{\rho}}=-2 R \int_{0}^{\frac{\pi}{3}} d\left(p \int_{0}^{R \cos \varphi}\left(R^{2}-\rho^{2}\right)^{-\frac{1}{2}} d\left(R^{2}-\rho^{2}\right)=\right.
$$

$$
=\left.2 R \int_{\frac{\pi}{2}}^{n}\left[2\left(R^{2}-\rho^{2}\right)^{\frac{1}{2}}\right]\right|_{0} ^{R \cos \varphi} d \varphi=4 R^{2} \int_{0}^{\frac{\pi}{2}}(1-\sin \varphi) d \varphi=2 R^{2}(\pi-2)
$$

$\left(\rho=R \cos \varphi\right.$ is the polar equation of the circle $\left.x^{2}+y^{2}=R x\right)$.

![](https://cdn.mathpix.com/cropped/2024_05_22_8a069bcd8684ea7ea3dbg-323.jpg?height=315&width=450&top_left_y=635&top_left_x=92)

Fig. 194

![](https://cdn.mathpix.com/cropped/2024_05_22_8a069bcd8684ea7ea3dbg-323.jpg?height=390&width=530&top_left_y=571&top_left_x=633)

Fig. 195

3) The given surface * (Fig. 195) is also symmetric with respect to the planes $x O y$ and $x O z$; in the first octant, one-fourth of it $(T)$ is located, for which the ordinate $y=\sqrt{R x-x^{2}}$. Therefore, transforming the surface integral of formula (1) into a double integral with variables $x$ and $z$, we get

$$
S=4 \iint_{T_{x 2}} \sqrt{1+\left(y_{x}^{\prime}\right)^{2}+\left(y_{z}^{\prime}\right)^{2}} d x d z=2 R \iint_{T_{x 3}} \frac{d x d z}{\sqrt{R x-x^{2}}},
$$

where $T_{x z}$ is a planar region bounded by the $O x$ and $O z$ axes and the parabola $l$, whose equation is $z^{2}=R^{2}-R x$, obtained by eliminating $y$ from the given equations. (This parabola is the projection onto the $x O z$ plane of the line of intersection of the cylinder and the sphere.) Integrating, we find

$$
S=2 R \int_{0}^{R} \frac{d x}{\sqrt{R x-x^{2}}} \int_{0}^{\sqrt{R^{2}}-\overline{R x}} d z=2 R \sqrt{R} \int_{0}^{R} \frac{d x}{\sqrt{ }} \bar{x}=\left.4 R \sqrt{R x}\right|_{0} ^{R}=4 R^{2}
$$[^36]