Chapter 8.3, Problem 8E

To determine

To find : the volume of the solid.

Answer to Problem 8E

The volume of the solid is $6\pi$ .

Explanation of Solution

Given information : The shaded region about the y -axis.

  

Calculation :

The shaded regions of the y-axis, the cross section of the solid are circular disks that run from $x=\raisebox{1ex}{$3y$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\text{ to }x=-\raisebox{1ex}{$3y$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ .

The area of these cross sectional disks is given by:

  $A\left(y\right)=\raisebox{1ex}{$9\pi y^2$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$

Hence, the volume of the solid is given by solving the integral.

  $\begin{array}{l}V={\displaystyle \underset{0}{\overset{2}{\int }}\raisebox{1ex}{$9\pi y^2$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\\ V=\frac{9\pi }{4}{\displaystyle \underset{0}{\overset{2}{\int }}{y}^{2}dy}\\ V=\frac{9\pi }{4}{\left[\frac{y^3}{3}\right]}_0^2\\ V=\frac{9\pi }{4\left(3\right)}{\left[y^3\right]}_0^2\\ V=\frac{3\pi }{4}{\left[y^3\right]}_0^2\\ V=\frac{3\pi }{4}{\left[{\left(2\right)}^3-0^3\right]}_0^2\\ V=\frac{3\pi }{4}\left(8\right)\\ V=6\pi \end{array}$