Chapter 7.3, Problem 70E

To determine

To find: the surface area generated by revolving the given curve about the indicated axis.

Answer to Problem 70E

The area of the surface is $\frac{49\pi }{3}$ .

Explanation of Solution

Given:

  $y=\sqrt{x+1},1\le x\le 5;x$ -axis.

Concept used:

Let $y=g\left(x\right)>0$ have a continuous first derivative on $\left[c,d\right]$ , then the area of the surface generated by revolving the curve $y=g\left(x\right)$ about the $y$ -axis is $S={\displaystyle \underset{c}{\overset{d}{\int }}2\pi g\left(x\right)\sqrt{1+{\left(g^{\prime }\left(x\right)\right)}^2}dx}$ .

Calculation:

Consider,

  $y=\sqrt{x+1}$

By differentiating with respect to $x$ , it follows that

  $\frac{dy}{dx}=\frac{1}{2\sqrt{x+1}}$

Now the area of the surface is,

  $\begin{array}{c}S={\displaystyle \underset{1}{\overset{5}{\int }}2\pi y\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}dx}\\ ={\displaystyle \underset{1}{\overset{5}{\int }}2\pi \sqrt{x+1}\sqrt{1+{\left(\frac{1}{2\sqrt{x+1}}\right)}^2}dx}\\ ={\displaystyle \underset{1}{\overset{5}{\int }}2\pi \sqrt{x+1}\sqrt{1+\frac{1}{4\left(x+1\right)}}dx}\\ ={\displaystyle \underset{1}{\overset{5}{\int }}2\pi \sqrt{x+1}\sqrt{\frac{4\left(x+1\right)+1}{4\left(x+1\right)}}dx}\\ ={\displaystyle \underset{1}{\overset{5}{\int }}2\pi \sqrt{x+1}\left(\frac{\sqrt{4x+5}}{2\sqrt{x+1}}\right)dx}\\ =\pi {\displaystyle \underset{1}{\overset{5}{\int }}\sqrt{4x+5}dx}\end{array}$

Now by integrating, it follows that

  $\begin{array}{c}S=\pi {\left[\frac{{\left(4x+5\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\cdot \frac{1}{4}\right]}_1^5\\ =\frac{\pi }{6}\left[{\left(4\left(5\right)+5\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-{\left(4\left(1\right)+5\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]\\ =\frac{\pi }{6}\left[{\left(25\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-{\left(9\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]\\ =\frac{\pi }{6}\left[125-27\right]\\ =\frac{49\pi }{3}\end{array}$