Chapter 6, Problem 11RE

To determine

To find : The volume of the solid obtained by obtained by rotating the region bounded by the given curves about the specified axis.

Answer to Problem 11RE

Required volume is $\frac{936\pi }{5}$ .

Explanation of Solution

Given information :

Given curves are : $x=0$ ; $x=9-y^2$

The solid is obtained by rotating about $x=-1$

Formula used :

Volume of washer $V_w=\pi {\displaystyle \underset{a}{\overset{b}{\int }}\left[{\left(r_{out}\right)}^2-{\left(r_{in}\right)}^2\right]}dy$

Here, the axis of rotation is $x=-1$ and $a,b$ are lower and upper bounds of y.

Calculation :

  

Here, $r_{\text{out}}=-1+9-y^2=8-y^2$

  $r_{in}=-1-0=-1$

  $x=0$ Intersect the curve $x=9-y^2$ at $y=\pm 3$

Then the solid can be divided into two equal parts about x-axis. We first take the solid in 1^st quadrant.

So, required volume $=2\times \text{ volume of each solid washer}$

  $\begin{array}{l}=2\times \pi {\displaystyle \underset{0}{\overset{3}{\int }}\left[{\left(8-y^2\right)}^2-{\left(-1\right)}^2\right]}dy\\ =2\pi {\displaystyle \underset{0}{\overset{3}{\int }}\left(64-16y^2+y^4-1\right)dy}\\ =2\pi {\displaystyle \underset{0}{\overset{3}{\int }}\left(63-16y^2+y^4\right)dy}\\ =2\pi {{\displaystyle \left[63y-\frac{16}{3}{y}^3+\frac{y^5}{5}\right]}}_0^3\\ =2\pi \left(189-\frac{16}{3}\times 27+\frac{3^5}{5}\right)\\ =\frac{936\pi }{5}\end{array}$