Chapter 6.3, Problem 12E

To determine

Find the volume enclosed by the given curves.

Answer to Problem 12E

The volume of the solid is $\frac{27}{2}\pi$

Explanation of Solution

Given information: Given curves $x+y=3,x=4-{\left(y-1\right)}^2$

Calculation:

The region is shown in the picture:

  

When rotate a thin horizontal strip as shown in figure, about the $x-axis$ .

Get a cylindrical shell with radius $y$ .

The height of the cylindrical shell is $\left(4-{\left(y-1\right)}^2\right)-\left(3-y\right)=3y-y^2$

  $\begin{array}{l}V=2\pi {\displaystyle \underset{0}{\overset{3}{\int }}\left(\text{Radius}\right)}\left(\text{Height}\right)dy\\ =2\pi {\displaystyle \underset{0}{\overset{3}{\int }}y\left(3y-y^2\right)}dy\\ =2\pi {\displaystyle \underset{0}{\overset{3}{\int }}\left(3y^2-y^3\right)}dy\\ =2\pi {\left[y^3-\frac{y^4}{4}\right]}_0^3\\ =2\pi \left[3^3-\frac{3^4}{4}\right]\\ =\frac{27}{2}\pi \end{array}$

Thus, the volume of the solid is $\frac{27}{2}\pi$