Chapter 6.3, Problem 11E

To determine

Find the volume enclosed by the given curves.

Answer to Problem 11E

The volume of the solid is $\frac{16}{3}\pi$

Explanation of Solution

Given information: Given curves $x=1+{\left(y-2\right)}^2,x=2$

Calculation:

The region is shown in the picture:

  

  $x=1+{\left(y-2\right)}^2,x=2\text{ about the }x-axis$

Rotating with the cylinder method about the $x-axis$ means it must be an $dy$ integral.

The given parabola equation is already solved for $x$ . The intersections:

  $\begin{array}{l}2=1+{\left(y-2\right)}^2\\ 0=-1+\left(y^2-4y+4\right)\\ 0=y^2-4y+3\\ 0=\left(y-1\right)\left(y-3\right)\end{array}$

To integrate from $y=1\text{ to }3$

The radius of each cylinder will be $x=y$ and the height will be;

  $\begin{array}{l}h=2-\left(1+{\left(y-2\right)}^2\right)\\ =2-1-\left(y^2-4y+4\right)\\ =1-y^2+4y-4\\ =-y^2+4y-3\end{array}$

  $\begin{array}{l}V={\displaystyle \underset{a}{\overset{b}{\int }}\left(2\pi rh\right)}dy\\ =2\pi {\displaystyle \underset{1}{\overset{3}{\int }}y\left(-y^2+4y-3\right)dy}\\ =2\pi {\displaystyle \underset{1}{\overset{3}{\int }}\left(-y^3+4y^2-3y\right)dy}\\ =2\pi {\left[-\frac{1}{4}{y}^4+\frac{4}{3}{y}^3-\frac{3}{2}{y}^2\right]}_1^3\\ =2\pi \left[-\frac{1}{4}{\left(3\right)}^4+\frac{4}{3}{\left(3\right)}^3-\frac{3}{2}{\left(3\right)}^2-\left(-\frac{1}{4}+\frac{4}{3}-\frac{3}{2}\right)\right]\\ =2\pi \left[-\frac{81}{4}+36-\frac{27}{2}+\frac{1}{4}-\frac{4}{3}+\frac{3}{2}\right]\\ =2\pi \left[-\frac{80}{4}+36-\frac{24}{2}-\frac{4}{3}\right]\\ =2\pi \left[-20+36-12-\frac{4}{3}\right]\\ =2\pi \left[\frac{12-4}{3}\right]\\ =2\pi \left[\frac{8}{3}\right]\\ =\frac{16}{3}\pi \end{array}$

Thus, the volume of the solid is $\frac{16}{3}\pi$