find the surface area of the solid generated by revolving the region bounded by the graphs of y = x²,y = 0,x…

Explanation:

Step1: Recall surface - area formula

The formula for the surface area $S$ of the solid generated by revolving the curve $y = f(x)$ from $x=a$ to $x = b$ about the $x$-axis is $S=2\pi\int_{a}^{b}y\sqrt{1+(y')^{2}}dx$. Given $y = x^{2}$, then $y'=2x$.

Step2: Substitute into the formula

We have $a = 0$, $b=\sqrt{5}$, and $y=x^{2}$, $y' = 2x$. So $S=2\pi\int_{0}^{\sqrt{5}}x^{2}\sqrt{1+(2x)^{2}}dx=2\pi\int_{0}^{\sqrt{5}}x^{2}\sqrt{1 + 4x^{2}}dx$. Let $x=\frac{1}{2}\tan\theta$, then $dx=\frac{1}{2}\sec^{2}\theta d\theta$. When $x = 0$, $\theta=0$; when $x=\sqrt{5}$, $\tan\theta = 2\sqrt{5}$, so $\theta=\arctan(2\sqrt{5})$. Also, $\sqrt{1 + 4x^{2}}=\sqrt{1+\tan^{2}\theta}=\sec\theta$ and $x^{2}=\frac{1}{4}\tan^{2}\theta$. The integral becomes: [ \begin{align*} S&=2\pi\int_{0}^{\arctan(2\sqrt{5})}\frac{1}{4}\tan^{2}\theta\cdot\sec\theta\cdot\frac{1}{2}\sec^{2}\theta d\theta\ &=\frac{\pi}{4}\int_{0}^{\arctan(2\sqrt{5})}\tan^{2}\theta\sec^{3}\theta d\theta\ &=\frac{\pi}{4}\int_{0}^{\arctan(2\sqrt{5})}(\sec^{2}\theta - 1)\sec^{3}\theta d\theta\ &=\frac{\pi}{4}\int_{0}^{\arctan(2\sqrt{5})}(\sec^{5}\theta-\sec^{3}\theta)d\theta \end{align*} ] The reduction formula for $\int\sec^{n}\theta d\theta=\frac{\sec^{n - 2}\theta\tan\theta}{n - 1}+\frac{n - 2}{n - 1}\int\sec^{n - 2}\theta d\theta$. For $n = 5$: $\int\sec^{5}\theta d\theta=\frac{\sec^{3}\theta\tan\theta}{4}+\frac{3}{4}\int\sec^{3}\theta d\theta$. For $n = 3$: $\int\sec^{3}\theta d\theta=\frac{\sec\theta\tan\theta}{2}+\frac{1}{2}\int\sec\theta d\theta=\frac{\sec\theta\tan\theta}{2}+\frac{1}{2}\ln|\sec\theta+\tan\theta|+C$. After substituting back $\theta=\arctan(2\sqrt{5})$, $\sec\theta=\sqrt{1 + \tan^{2}\theta}=\sqrt{1 + 20}= \sqrt{21}$ and $\tan\theta=2\sqrt{5}$. [ \begin{align*} S&=\frac{\pi}{4}\left[\left(\frac{\sec^{3}\theta\tan\theta}{4}+\frac{3}{4}\left(\frac{\sec\theta\tan\theta}{2}+\frac{1}{2}\ln|\sec\theta+\tan\theta|\right)\right)-\left(\frac{\sec\theta\tan\theta}{2}+\frac{1}{2}\ln|\sec\theta+\tan\theta|\right)\right]_0^{\arctan(2\sqrt{5})}\ &=\frac{\pi}{4}\left[\frac{\sec^{3}\theta\tan\theta}{4}+\frac{3\sec\theta\tan\theta}{8}+\frac{3}{8}\ln|\sec\theta+\tan\theta|-\frac{\sec\theta\tan\theta}{2}-\frac{1}{2}\ln|\sec\theta+\tan\theta|\right]_0^{\arctan(2\sqrt{5})}\ &=\frac{\pi}{4}\left[\frac{\sec^{3}\theta\tan\theta}{4}-\frac{\sec\theta\tan\theta}{8}-\frac{1}{8}\ln|\sec\theta+\tan\theta|\right]_0^{\arctan(2\sqrt{5})}\ &=\frac{\pi}{4}\left[\frac{(\sqrt{21})^{3}\cdot2\sqrt{5}}{4}-\frac{\sqrt{21}\cdot2\sqrt{5}}{8}-\frac{1}{8}\ln|\sqrt{21}+2\sqrt{5}|\right]\ &\approx 53.2565 \end{align*} ]

Answer:

$53.2565$