Question

The area centered on the line $x=-1$between the graph of $f\left(x\right)=x^2+2$and the x-axis on $[1,3].$

Short answer

The volume of the solid of revolution is $\frac{248}{3}\pi .$

Step-by-step solution

Given information

Consider the given function,

$f\left(x\right)=x^2+2$$f\left(x\right)=x^2+2$

Calculation.

The region is enclosed by the expression $f\left(x\right)=x^2+2$and the x-axis on $[-1,3].$

Use the shell method with shells on the x-axis having radius $x-(-1)=x+1$, where x ranges from localid="1661340316266" $-1\mathrm{to}3$, and height $f\left(x\right)=x^2+2$, as the region is revolved around the line $x=-1.$

Thus, the volume is provided by

localid="1661340332870" $\begin{array}{rcl}\text{Volume}& =& 2\pi {\int }_{-1}^3(x+1)\left(x^2+2\right)dx\\ & =& 2\pi {\int }_{-1}^3\left(x^3+x^2+2x+1\right)dx\\ & =& 2\pi {\left[\frac{x^4}{4}+\frac{x^3}{3}+x^2+x\right]}_{-1}^3\\ & =& 2\pi \left[\left(\frac{81}{4}+9+9+3\right)-\left(\frac{1}{4}-\frac{1}{3}+1-1\right)\right]\end{array}$

Therefore the volume of the solid of revolution is localid="1661340343180" $\frac{248}{3}\pi$.