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Expert-verified Found in: Page 1120 ### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861 # 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.

Ans: The required surface area is ${\int }_{0}^{2\pi } {\int }_{0}^{3} \left(\sqrt{{r}^{2}\left({e}^{{r}^{2}\mathrm{sin}2\theta }+{e}^{-{r}^{2}\mathrm{sin}2\theta }+2\right)+1}\right)rdrd\theta$

See the step by step solution

## Step 1. Given information.

given,

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

## Step 2. Consider the following surface:

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 $x=f\left(y,z\right)$ : If a surface S is given by $x=f\left(y,z\right)\text{for}f\left(y,z\right)\in D\subseteq {\mathrm{ℝ}}^{2}$, then the surface area of the smooth surface is,

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

## Step 3. Here, the surface S is the portion of the following surface;

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

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,

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

## Step 4. Now,

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,

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

In this case,

$y=r\mathrm{cos}\theta ,z=r\mathrm{sin}\theta ,{y}^{2}+{z}^{2}={r}^{2},and\phantom{\rule{0ex}{0ex}}dA=rdrd\theta .$

## Step 5. Then,

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" $\begin{array}{r}{\int }_{S} dS={\iint }_{D} \left(\sqrt{{\left(\frac{\mathrm{\partial }x}{\mathrm{\partial }y}\right)}^{2}+{\left(\frac{\mathrm{\partial }x}{\mathrm{\partial }z}\right)}^{2}+1}\right)dA\\ \phantom{\rule{1em}{0ex}}={\iint }_{D} \left(\sqrt{{\left(z\left({e}^{yz}+{e}^{-yz}\right)\right)}^{2}+{\left(y\left({e}^{yz}+{e}^{-yz}\right)\right)}^{2}+1}\right)dA\\ ={\iint }_{D} \left(\sqrt{{z}^{2}{\left({e}^{yz}+{e}^{-yz}\right)}^{2}+{y}^{2}{\left({e}^{yz}+{e}^{-yz}\right)}^{2}+1}\right)dA\\ ={\iint }_{D} \left(\sqrt{\left({y}^{2}+{z}^{2}\right){\left({e}^{yz}+{e}^{-yz}\right)}^{2}+1}\right)dA\\ ={\int }_{0}^{2\pi } {\int }_{0}^{3} \left(\sqrt{{r}^{2}{\left({e}^{r\mathrm{cos}\theta \mathrm{sin}\theta }+{e}^{-r\mathrm{cos}\theta r\mathrm{sin}\theta }\right)}^{2}+1}\right)rdrd\theta \\ ={\int }_{0}^{2z} {\int }_{0}^{3} \left(\sqrt{{r}^{2}{\left({e}^{\frac{1}{2}{r}^{2}\mathrm{sin}2\theta }+{e}^{-\frac{1}{2}{r}^{2}\mathrm{sin}2\theta }\right)}^{2}+1}\right)rdrd\theta \\ ={\int }_{0}^{2\pi } {\int }_{0}^{3} \left(\sqrt{{r}^{2}\left({e}^{{r}^{2}\mathrm{sin}2\theta }+{e}^{-{r}^{2}\mathrm{sin}2\theta }+2\right)+1}\right)rdrd\theta \end{array}$

Therefore, the required surface area is the above integral. ### Want to see more solutions like these? 