Chapter 7.3, Problem 66E

To determine

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

Answer to Problem 66E

The area of the surface is 3.

Explanation of Solution

Given:

  $x=\sqrt{2y-1},\frac{5}{8}\le y\le 1;y$ -axis.

Concept used:

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

Calculation:

Consider,

  $x=\sqrt{2y-1}$

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

  $\begin{array}{c}\frac{dx}{dy}=\frac{1}{2\sqrt{2y-1}}\cdot 2\\ =\frac{1}{\sqrt{2y-1}}\end{array}$

Now the area of the surface is,

  $\begin{array}{c}S={\displaystyle \underset{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$8$}\right.}{\overset{1}{\int }}2\pi x\sqrt{1+{\left(\frac{dx}{dy}\right)}^2}dy}\\ ={\displaystyle \underset{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$8$}\right.}{\overset{1}{\int }}2\pi \sqrt{2y-1}\sqrt{1+{\left(\frac{1}{\sqrt{2y-1}}\right)}^2}dy}\\ ={\displaystyle \underset{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$8$}\right.}{\overset{1}{\int }}2\pi \sqrt{2y-1}\sqrt{1+\frac{1}{2y-1}}dy}\\ ={\displaystyle \underset{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$8$}\right.}{\overset{1}{\int }}2\pi \sqrt{2y-1}\sqrt{\frac{2y-1+1}{2y-1}}dy}\\ ={\displaystyle \underset{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$8$}\right.}{\overset{1}{\int }}2\pi \sqrt{2y-1}\left(\frac{\sqrt{2y}}{\sqrt{2y-1}}\right)dy}\\ =2\sqrt{2}\pi {\displaystyle \underset{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$8$}\right.}{\overset{1}{\int }}\sqrt{y}dy}\end{array}$

Now on integrating, it follows that

  $\begin{array}{c}S=2\sqrt{2}\pi {\left[\frac{y^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]}_{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$8$}\right.}^1\\ =\frac{4\sqrt{2}\pi }{3}\left[{\left(1\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-{\left(\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$8$}\right.\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]\\ \approx 3\end{array}$