Chapter A6, Problem 70E

To determine

To find: the area of the surface

Answer to Problem 70E

The area of the surface is $S\approx 248.889$

Explanation of Solution

Given:

  $y=4\cosh \left(\frac{x}{4}\right),$

  $16\le x\le 1n81,$

  

Calculation:

The given figure is shown below,

For,

  $\begin{array}{l}y=4\cosh \frac{x}{4}\\ 1+{\left(\frac{dy}{dx}\right)}^2=1+{\sinh }^2\left(\frac{x}{4}\right)\\ ={\cosh }^2\left(\frac{x}{4}\right)\end{array}$

Therefore the surface area S is,

  $\begin{array}{l}S={\displaystyle \underset{-1n\text{ 16}}{\overset{1n81}{\int }}2\pi y\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}\cdot dx}\\ =8\pi {\displaystyle \underset{-1n\text{ 16}}{\overset{1n81}{\int }}{\cosh }^2\left(\frac{x}{4}\right)\cdot dx}\\ 4\pi {\displaystyle \underset{-1n\text{ 16}}{\overset{1n81}{\int }}1+\cosh \left(\frac{x}{2}\right)\cdot dx}\\ 4\pi {\left[x+2\sinh \left(\frac{x}{2}\right)\right]}_{1n\text{ 16}}^{1n\text{ 81}}\end{array}$

Continuation of the above step,

  $\begin{array}{l}=4\pi \left[1n\left(.16\right)+2\sinh \left(1n9\right)+2\sinh \left(1n4\right)\right]\\ =4\pi \left[2\text{ 1n 36+}\left(9-\frac{1}{9}\right)+\left(4-\frac{1}{4}\right)\right]\\ =4\pi \left(4\text{ 1n 6+}\frac{320+135}{36}\right)\\ =16\pi \text{ 1n 6+}\frac{455\pi }{9}\\ \boxed{S\approx 248.889}\end{array}$

Conclusion:

The area of the surface is $S\approx 248.889$