91Ó°ÊÓ

The function $f\left(x\right)=x^2$on the interval$[-1,1]$.

Remember that using a definite integral, you may derive the surface area of a solid of rotation by rotating the graph of a function around the $x-$axis from a to b.

$S=2\mathrm{π}{∫}_a^{b}f\left(x\right)\sqrt{1+{\left(f^{\text{'}}\left(x\right)\right)}^2}dx……\text{(1)}$

Keep in mind that the function $f\left(x\right)=x^2$has a continuous derivative in the range $[-1,1]$and is differentiable. In order to obtain the function's derivative and integral on the right side of equation $\left(1\right)$, we must first differentiate the function.

localid="1660474461534">S=2π∫-11x21+(2x)2dx=2π∫-11x21+4x2dx=π4∫-11x·8x1+4x2dx

Since the above integral is not in the conventional form, it must first be evaluated using the integration by parts method.

localid="1661329262326" $\begin{array}{rcl}S& =& \frac{\mathrm{π}}{4}\left[{\left.x·\frac{2}{3}{\left(1+4x^2\right)}^{3/2}\right|}_{-1}^1-\frac{2}{3}{∫}_{-1}^1{\left(1+4x^2\right)}^{3/2}dx\right]\\ & =& \frac{\mathrm{π}}{6}\left[5\sqrt{5}+5\sqrt{5}-{∫}_{-1}^1{\left(1+4x^2\right)}^{3/2}dx\right]\\ & =& \frac{5\sqrt{5}\mathrm{π}}{3}-\frac{\mathrm{π}}{6}{∫}_{-1}^1{\left(1+4x^2\right)}^{3/2}dx\end{array}$

Assuming $I=∫{\left(1+4x^2\right)}^{3/2}dx$, integrate using the substitution approach. Place the integration limits after that, and then assess the definite integral. Take$2x=tanu;x=\frac{1}{2}{\sec }^{2}udu$in the integral and integrate as a result.

localid="1661329269472" $\begin{array}{rcl}I& =& ∫{\left(1+{\tan }^{2}u\right)}^{3/2}·\frac{1}{2}{\sec }^{2}udu\\ & =& \frac{1}{2}∫{\sec }^{5}udu\end{array}$

Recall the reduction formula for integration of ${\sec }^{n}x(n>2)$given by

$∫{\sec }^{n}xdx=\frac{{\sec }^{n-2}x\tan x}{n-1}+\frac{n-2}{n-1}∫{\sec }^{n-2}xdx+C$

Use above integral to integrate I as

localid="1661329309848" role="math" $\begin{array}{rcl}I& =& \frac{1}{2}\left[\frac{{\sec }^{3}u\tan u}{4}+\frac{3}{4}∫{\sec }^{3}udu\right]\\ & =& \frac{1}{2}\left[\frac{{\sec }^{3}u\tan u}{4}+\frac{3}{4}\left\{\frac{\sec u\tan u}{2}+\frac{1}{2}∫\sec udu\right\}\right]\\ & =& \frac{1}{16}\left[2{\sec }^{3}u\tan u+3\{\sec u\tan u+\ln |\sec u+\tan u\left|\right\}\right]\\ & =& \frac{1}{16}\left[2{\sec }^{3}u\tan u+3\sec u\tan u+3\ln |\sec u+\tan u|\right]\end{array}$

Apply the trigonometric identity ${\sec }^{2}x=1+{\tan }^{2}x$to the result shown above, then rewrite the result in the variable $x$.

$$\begin{gathered} I=\frac{1}{16}\left[2{\left(1+4x^2\right)}^{3/2}·2x+3{\left(1+4x^2\right)}^{1/2}·2x+3\ln \left|2x+\sqrt{1+4x^2}\right|\right] \\ =\frac{1}{4}x{\left(1+4x^2\right)}^{3/2}+\frac{3}{8}x\sqrt{1+4x^2}+3\ln \left|2x+\sqrt{1+4x^2}\right| \end{gathered}$$

Assess the value of this integral now using the integration limitations.

$\begin{array}{rcl}[I{]}_{-1}^1& =& \frac{1}{4}[5\sqrt{5}+5\sqrt{5}]+\frac{3}{8}[\sqrt{5}+\sqrt{5}]+3\ln |2+\sqrt{5}|-3\ln |-2+\sqrt{5}|\\ & =& \frac{13}{4}\sqrt{5}+3\ln \frac{\sqrt{5}+2}{\sqrt{5}-2}\left(âˆ\mu \ln m-\ln n=\ln \frac{m}{n}\right)\\ & =& \frac{13\sqrt{5}}{4}+6\ln (\sqrt{5}+2)\left(âˆ\mu \frac{\sqrt{5}+2}{\sqrt{5}-2}×\frac{\sqrt{5}+2}{\sqrt{5}+2}=(\sqrt{5}+2{)}^2\right)\end{array}$

In the formula for the surface area, substitute this number to obtain.

$\begin{array}{rcl}S& =& \frac{5\sqrt{5}\mathrm{Ï€}}{3}-\frac{\mathrm{Ï€}}{6}\left[\frac{13\sqrt{5}}{4}+6\ln (\sqrt{5}+2)\right]\\ & =& \frac{27\sqrt{5}}{24}\mathrm{Ï€}-\mathrm{Ï€}\ln (\sqrt{5}+2)\end{array}$

This means that the surface area produced by rotating the graph of $f\left(x\right)=x^2$around the localid="1660483502909" $x-$axis on the range $[-1,1]$is $\frac{27\sqrt{5}}{24}\mathrm{Ï€}-\mathrm{Ï€}\ln (\sqrt{5}+2)$.