Chapter 7, Problem 51RE

To determine

To find: The area of the surface generated by revolving the curves about the indicated axis.

Answer to Problem 51RE

The area of the surface generated is $S=3.839$ .

Explanation of Solution

Given information:

  $y=\tan x$

Calculation:

The surface area $y=f\left(x\right)$ about the $x$ -axis is given by,

  $S={\displaystyle {\int }_a^{b}2}\pi f(x)\sqrt{1+{\left[f^{\prime }(x)\right]}^2}dx$

Substituting the values,

  $\begin{array}{l}f(x)=\tan x\\ f^{\prime }(x)={\sec }^{2}x\end{array}$

  $S=2\pi {\displaystyle {\int }_0^{\pi /4}\tan }x\sqrt{1+{\sec }^{-4}x}dx$

Now substitute,

  $1+{\sec }^{4}x=u$

On differentiating to get,

  $4{\sec }^{4}x\tan xdx=du$

The limits of the integration vary from,

  $\begin{array}{l}{\displaystyle {\int }_0^{\pi /4}t}o{\displaystyle {\int }_{1+\pi x^4\theta }^{1+{\mathrm{mcc}}^4\pi /4}=}{\displaystyle {\int }_2^5}\\ =\frac{2\pi }{4}{\displaystyle {\int }_2^5\frac{\sqrt{u}}{u-1}}du\end{array}$

Substituting $u=t^2\text{ and }du=2tdt$ to get,

  $\begin{array}{c}S=\frac{\pi }{2}{\displaystyle {\int }_{\sqrt{2}}^{\sqrt{5}}\frac{2t^2}{t^2-1}}dt\\ =\frac{\pi }{2}{\displaystyle {\int }_{\sqrt{2}}^{\sqrt{5}}\frac{2t^2-2}{t^2-1}}+\frac{2}{(t+1)(t-1)}dt\\ =\frac{\pi }{2}{\displaystyle {\int }_{\sqrt{2}}^{\sqrt{5}}2}+\frac{(t+1)-(t-1)}{(t+1)(t-1)}dt\\ =\frac{\pi }{2}{\displaystyle {\int }_{\sqrt{2}}^{\sqrt{5}}2}+\frac{1}{t-1}-\frac{1}{t+1}dt\\ \left.=\frac{\pi }{2}\left[2t+\ln \mid \frac{t-1}{t+1}\right)\right]\sqrt{5}\\ =\pi (\sqrt{5}-\sqrt{2})+\frac{\pi }{2}\ln \left[\frac{(\sqrt{5}-1)(\sqrt{2}+1)}{(\sqrt{5}+1)(\sqrt{2}-1)}\right]\\ \approx 3.839\end{array}$

Therefore, the area of the surface generated by revolving the curves about the indicated axisis $S=3.839$ .