91影视

given,

$x=e^{yz}鈭�e^{鈭�yz}$

The surface S is the portion of the surface determined by$yx=e^{yz}鈭�e^{鈭�yz}$ that lies on the positive side of the circle of radius 3 and is centered at the origin in the yz-plane.
The objective is to find the value of${鈭�}_S鈥�1dS$.
The formula for Finding the Area of a Surface$x=f(y,z)$ :
If a surface S is given by $x=f(y,z)\text{for}f(y,z)鈭�D鈯�{\mathrm{鈩�}}^2$, then the surface area of the smooth surface is,

role="math" localid="1650475395157" $\begin{array}{r}{鈭�}_S鈥�1dS={鈭�}_S鈥�\sqrt{{\left(\frac{\mathrm{鈭�}x}{\mathrm{鈭�}y}\right)}^2+{\left(\frac{\mathrm{鈭�}x}{\mathrm{鈭�}z}\right)}^2+1}dA\\ ={鈭�}_D鈥�\sqrt{{\left(\frac{\mathrm{鈭�}x}{\mathrm{鈭�}y}\right)}^2+{\left(\frac{\mathrm{鈭�}x}{\mathrm{鈭�}z}\right)}^2+1dA}...\left(1\right)\end{array}$

$x=e^{yz}鈭�e^{鈭�yz}$

Now, first, find $\frac{\mathrm{鈭�}x}{\mathrm{鈭�}y}\text{and}\frac{\mathrm{鈭�}x}{\mathrm{鈭�}z}$. The partial derivative of x is,

$\begin{array}{r}\frac{\mathrm{鈭�}x}{\mathrm{鈭�}y}=\frac{\mathrm{鈭�}}{\mathrm{鈭�}y}\left(e^{yz}鈭�e^{鈭�yz}\right)\\ =z\left(e^{yz}+e^{鈭�yz}\right)\end{array}$

and,

$\begin{array}{r}\frac{\mathrm{鈭�}x}{\mathrm{鈭�}z}=\frac{\mathrm{鈭�}}{\mathrm{鈭�}z}\left(e^{yz}鈭�e^{鈭�yz}\right)\\ =y\left(e^{yz}+e^{鈭�yz}\right)\end{array}$

The surface S lies on the positive side of the circle of radius 3 and is centered at the origin in the -plane. The region of integration D is the disk$y^2+z^2=9$ is shown in the following figure.
In polar coordinates, the region of integration is described as follows,

$D=\left\{\right(r,胃)鈭�0鈮�r鈮�3,0鈮�胃鈮�2蟺\}.$

In this case,

$$\begin{gathered} y=r\cos 鈦�胃,z=r\sin 鈦�胃,y^2+z^2=r^2,and \\ dA=rdrd胃. \end{gathered}$$

Substitute $\frac{\mathrm{鈭�}x}{\mathrm{鈭�}y}=z\left(e^{yz}+e^{鈭�yz}\right)\text{and}\frac{\mathrm{鈭�}x}{\mathrm{鈭�}z}=y\left(e^{jz}+e^{鈭�yz}\right)$into formula (1), then the area of the surface is,

localid="1650476425232" $\begin{array}{r}{鈭�}_S鈥�dS={鈭�}_D鈥�\left(\sqrt{{\left(\frac{\mathrm{鈭�}x}{\mathrm{鈭�}y}\right)}^2+{\left(\frac{\mathrm{鈭�}x}{\mathrm{鈭�}z}\right)}^2+1}\right)dA\\ ={鈭�}_D鈥�\left(\sqrt{{\left(z\left(e^{yz}+e^{鈭�yz}\right)\right)}^2+{\left(y\left(e^{yz}+e^{鈭�yz}\right)\right)}^2+1}\right)dA\\ ={鈭�}_D鈥�\left(\sqrt{z^2{\left(e^{yz}+e^{鈭�yz}\right)}^2+y^2{\left(e^{yz}+e^{鈭�yz}\right)}^2+1}\right)dA\\ ={鈭�}_D鈥�\left(\sqrt{\left(y^2+z^2\right){\left(e^{yz}+e^{鈭�yz}\right)}^2+1}\right)dA\\ ={鈭�}_0^{2蟺}鈥�{鈭�}_0^3鈥�\left(\sqrt{r^2{\left(e^{r\cos 鈦�胃\sin 鈦�胃}+e^{鈭�r\cos 鈦�胃r\sin 鈦�胃}\right)}^2+1}\right)rdrd胃\\ ={鈭�}_0^{2z}鈥�{鈭�}_0^3鈥�\left(\sqrt{r^2{\left(e^{\frac{1}{2}{r}^2\sin 鈦�2胃}+e^{鈭�\frac{1}{2}{r}^2\sin 鈦�2胃}\right)}^2+1}\right)rdrd胃\\ ={鈭�}_0^{2蟺}鈥�{鈭�}_0^3鈥�\left(\sqrt{r^2\left(e^{r^2\sin 鈦�2胃}+e^{鈭�r^2\sin 鈦�2胃}+2\right)+1}\right)rdrd胃\end{array}$

Therefore, the required surface area is the above integral.