Find ${\int }_S 1dS$, where S is the portion of the surface with equation $x=e^{yz}-e^{-yz}$that lies on the positive side of the circle of radius 3 and centered at the origin in the yz-plane.

given,

$x=e^{yz}-e^{-yz}$

The surface S is the portion of the surface determined by$yx=e^{yz}-e^{-yz}$ 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${\int }_S 1dS$.
The formula for Finding the Area of a Surface$x=f(y,z)$ :
If a surface S is given by $x=f(y,z)\text{for}f(y,z)\in D\subseteq {\mathrm{ℝ}}^2$, then the surface area of the smooth surface is,

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

$x=e^{yz}-e^{-yz}$

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

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

and,

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

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,

$D=\left\{\right(r,\theta )\mid 0\le r\le 3,0\le \theta \le 2\pi \}.$

In this case,

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

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

localid="1650476425232" $\begin{array}{r}{\int }_S dS={\iint }_D \left(\sqrt{{\left(\frac{\mathrm{\partial }x}{\mathrm{\partial }y}\right)}^2+{\left(\frac{\mathrm{\partial }x}{\mathrm{\partial }z}\right)}^2+1}\right)dA\\ ={\iint }_D \left(\sqrt{{\left(z\left(e^{yz}+e^{-yz}\right)\right)}^2+{\left(y\left(e^{yz}+e^{-yz}\right)\right)}^2+1}\right)dA\\ ={\iint }_D \left(\sqrt{z^2{\left(e^{yz}+e^{-yz}\right)}^2+y^2{\left(e^{yz}+e^{-yz}\right)}^2+1}\right)dA\\ ={\iint }_D \left(\sqrt{\left(y^2+z^2\right){\left(e^{yz}+e^{-yz}\right)}^2+1}\right)dA\\ ={\int }_0^{2\pi } {\int }_0^3 \left(\sqrt{r^2{\left(e^{r\cos \theta \sin \theta }+e^{-r\cos \theta r\sin \theta }\right)}^2+1}\right)rdrd\theta \\ ={\int }_0^{2z} {\int }_0^3 \left(\sqrt{r^2{\left(e^{\frac{1}{2}{r}^2\sin 2\theta }+e^{-\frac{1}{2}{r}^2\sin 2\theta }\right)}^2+1}\right)rdrd\theta \\ ={\int }_0^{2\pi } {\int }_0^3 \left(\sqrt{r^2\left(e^{r^2\sin 2\theta }+e^{-r^2\sin 2\theta }+2\right)+1}\right)rdrd\theta \end{array}$

Therefore, the required surface area is the above integral.