Q.68 - Calculus

Question

On the interval $[-1,1]$, the graph of the function $f\left(x\right)=e^{-x}$revolves around the $x$-axis. Set up and evaluate the definite integral to ascertain the precise area of the surface of rotation.

Step-by-Step Solution

Answer

The graph of the function $f\left(x\right)={e^{-}}^x$ is revolved around the $x$-axis on an interval $[1,-1]$ is

$\pi \left[1+\left(e-\frac{1}{e^2}\right)\sqrt{e^2+1}+\ln \left|\frac{e+\sqrt{e^2+1}}{1+\sqrt{e^2+1}}\right|\right]$.

Step 1: Given Information

The function $f\left(x\right)={e^{-}}^x$ on an interval $[1,-1]$.

Step 2 : Calculation

Remember that the surface area of a solid of revolution is given by rotating a function's graph around the x-axis from point a to point b using a definite integral.

$S=2\pi {\int }_a^{b}f\left(x\right)\sqrt{1+{\left(f^{\text{'}}\left(x\right)\right)}^2}dx$

Keep in mind that the function f(x) has a continuous derivative in the range $[-1,1]$ and is differentiable.

Calculate the resulting integral by differentiating the function and applying the derivative to the integral on the right side of the equation $\left(1\right)$.

$$\begin{gathered} S=2\pi {\int }_{-1}^1{e}^{-x}\sqrt{1+{\left(-e^{-x}\right)}^2}dx \\ =2\pi {\int }_{-1}^1{e}^{-x}\sqrt{1+e^{-2x}}dx \end{gathered}$$

Since the above integral is not in the conventional form, it can be reduced to one using the substitution approach. 

Take  ${e^{-}}^x=u,{e^{-}}^{x}dx=du$ and adjust the integration limits to obtain.

Step 3 : Further Calculation.

Find the integration and solve it.

$\begin{array}{rcl}S& =& 2\pi {\int }_e^{1/e}\sqrt{1+u^2}(-du)\\ & =& -2\pi {\int }_e^{1/e}\sqrt{1+u^2}du\\ & =& 2\pi {\int }_{1/e}^e\sqrt{1+u^2}du \left(\because {\int }_a^{b}f\left(x\right)dx=-{\int }_b^{a}f\left(x\right)dx\right)\\ & =& 2\pi {\left[\frac{u}{2}\sqrt{1+u^2}+\frac{1}{2}\ln \left|u+\sqrt{1+u^2}\right|\right]}_{1/e}^e\end{array}$

Further, reduce the aforementioned statement to obtain.

$\begin{array}{rcl}S& =& \pi \left[e\sqrt{e^2+1}+\ln \left|e+\sqrt{e^2+1}\right|-\frac{1}{e}\sqrt{\frac{1}{e^2}+1}-\ln \left|\frac{1}{e}+\sqrt{\frac{1}{e^2}+1}\right|\right]\\ & =& \pi \left[e\sqrt{e^2+1}+\ln \left|e+\sqrt{e^2+1}\right|-\frac{1}{e^2}\sqrt{e^2+1}-\ln \frac{1+\sqrt{e^2+1}}{e}\right]\\ & =& \pi \left[\left(e-\frac{1}{e^2}\right)\sqrt{e^2+1}+\ln \left|e+\sqrt{e^2+1}\right|-\ln \left|1+\sqrt{e^2+1}\right|+\ln e\right]\\ & =& \pi \left[1+\left(e-\frac{1}{e^2}\right)\sqrt{e^2+1}+\ln \left|\frac{e+\sqrt{e^2+1}}{1+\sqrt{e^2+1}}\right|\right] \left(\because \ln e=1;\ln m-\ln n=\ln \frac{m}{n}\right)\end{array}$

As a result, the surface area obtained by rotating the graph  $f\left(x\right)={e^{-}}^x$ around the $x$-axis on the range$[1,-1]$is defined as,

$S=\pi \left[1+\left(e-\frac{1}{e^2}\right)\sqrt{e^2+1}+\ln \left|\frac{e+\sqrt{e^2+1}}{1+\sqrt{e^2+1}}\right|\right].$