Chapter 7.3, Problem 17E

To determine

To find: The volume of the solid.

Answer to Problem 17E

The volume of the given solid $V=\frac{117}{5}\pi$ .

Explanation of Solution

Given information:

The solid generated by revolving the region bounded by the lines and the curves:

  $\begin{array}{c}y=x^2+1,\\ y=x+3\end{array}$

Calculation:

Limits of integration are $y=x^2+1$ and

  $y=x+3$ intersect.

So, the intersections are:

  $\begin{array}{c}{x}^2+1=x+3\\ \Rightarrow x^2-x-2=0\\ \Rightarrow (x+1)(x-2)=0\\ x=-1\text{ ,}\\ x=2\end{array}$

Area of each cross section is:

  $\begin{array}{c}\pi {(}^{\text{ outer radius }}-\pi {(}^{\text{ inner radius }}\\ =\pi {(}^x-\pi {\left(x^2+1-0\right)}^2\\ =\pi (x+3)(x+3)-\pi \left(x^2+1\right)\left(x^2+1\right)\\ =\pi \left(x^2+6x+9\right)-\pi \left(x^4+2x^2+1\right)\end{array}$

The volume of the given solid is:

  

  $\begin{array}{c}=\pi {\left[-\frac{1}{5}{x}^5-\frac{1}{3}{x}^3+3x^2+8x\right]}_{-1}^2\\ =\pi \left[\left(-\frac{32}{5}-\frac{8}{3}+12+16\right)-\left(+\frac{1}{5}+\frac{1}{3}+3-8\right)\right]\\ =\pi \left[-\frac{32}{5}-\frac{1}{5}-\frac{8}{3}-\frac{1}{3}+28+5\right]\\ =\pi \left[-\frac{33}{5}-3+33\right]\\ =\pi \left[-\frac{33}{5}+\frac{150}{5}\right]\\ =\frac{117}{5}\pi \end{array}$

The graph is shown below:

  

Therefore, the required volume of the given solid $V=\frac{117}{5}\pi$ .