Chapter 8, Problem 52RE

To determine

To find : Area of the surface.

Answer to Problem 52RE

Required area $=\pi \times 0.664=2.086$ square unit.

Explanation of Solution

Given information :

  $xy=1$ is the given curve.

  $1\le y\le 2$

The curve is revolving about y-axis.

Formula used :

If the curve is described by the function $x=g\left(y\right)$ , $c\le y\le d$ and rotate about y-axis then the area of the surface of revolution is given by $A=2\pi {\displaystyle \underset{c}{\overset{d}{\int }}g\left(y\right)\sqrt{1+{\left[g^{\prime }\left(y\right)\right]}^2}}dy$ .

Calculation :

  

Here, $x=\frac{1}{y}\Rightarrow g\left(y\right)=\frac{1}{y}$

Required area $=A=2\pi {\displaystyle \underset{1}{\overset{2}{\int }}\frac{1}{y}\sqrt{1+{\left(-\frac{1}{y^2}\right)}^2}}dy$

  $\begin{array}{l}A=2\pi {\displaystyle \underset{1}{\overset{2}{\int }}\frac{1}{y}\frac{\sqrt{\left(y^4+1\right)}}{y^2}}dy\\ =2\pi {\displaystyle \underset{1}{\overset{2}{\int }}\frac{\sqrt{\left(y^4+1\right)}}{y^3}}dy\end{array}$

Put, $y^2=u$

  $\begin{array}{l}2ydy=du\\ \Rightarrow dy=\frac{du}{2y}\end{array}$

$u$	$1$	$\sqrt{2}$
$y$	$1$	$2$

  $\begin{array}{c}A=2\pi {\displaystyle \underset{1}{\overset{\sqrt{2}}{\int }}\frac{\sqrt{\left(u^2+1\right)}}{u^2}}\frac{du}{2}\\ =\pi {\displaystyle \underset{1}{\overset{\sqrt{2}}{\int }}\frac{\sqrt{\left(u^2+1\right)}}{u^2}}du\end{array}$

Now,

  $\begin{array}{l}\pi {\displaystyle \underset{1}{\overset{\sqrt{2}}{\int }}\frac{\sqrt{\left(u^2+1\right)}}{u^2}}du\\ ={\frac{\sqrt{\left(u^2+1\right)}}{-u}]}_1^{\sqrt{2}}-{\displaystyle \underset{1}{\overset{\sqrt{2}}{\int }}\left(-\frac{1}{4}\right)}\times \frac{1\times 2u}{2\sqrt{\left(u^2+1\right)}}du\\ ={\frac{\sqrt{\left(u^2+1\right)}}{-u}]}_1^{\sqrt{2}}+{\displaystyle \underset{1}{\overset{\sqrt{2}}{\int }}\frac{du}{\sqrt{\left(u^2+1\right)}}}\\ =\frac{\sqrt{2+1}}{-\sqrt{2}}+\frac{\sqrt{2}}{1}+{\left[\ln \left(\sqrt{4^2+1}+4\right)\right]}_1^{\sqrt{2}}\\ =\frac{\sqrt{3}}{-\sqrt{2}}+\sqrt{2}+\ln \left(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{2}+1}\right)\\ \approx 0.664\end{array}$

Therefore required area $=\pi \times 0.664=2.086$ square unit.