# How do you use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region y = 1 + x^2, y = 0, x = 0, x = 2 rotated about the line x=4?

Sep 28, 2017

$\frac{76 \pi}{3}$

#### Explanation:

This is a graph of the region that will be revolved around the vertical line $x = 4$ (not pictured).

graph{(y-1-x^2)(y)( sqrt(2-x) )(sqrt(x)) / (sqrt(2-x))/(sqrt(x))<=0 [0, 6, -1.51, 6.39]}

Recall the general form of the volume of a solid of revolution using the shells method when you are revolving about a vertical line:

$V = {\int}_{a}^{b} 2 \pi \cdot r a \mathrm{di} u s \cdot f \left(x\right) \text{ } \mathrm{dx}$

The hardest conceptual part here is the radius. Since the axis of revolution is $x = 4$, the form of the radius is the expression $4 - x$.

(Why? Draw a segment from $x = 4$ on the x-axis to $x = 1$ on the x-axis. How long is that segment? It is $4 - 1 = 3$.

Thus:

$V = {\int}_{0}^{2} 2 \pi \cdot \left(4 - x\right) \cdot \left(1 + {x}^{2}\right) \mathrm{dx}$
$= 2 \pi {\int}_{0}^{2} \left(4 + 4 {x}^{2} - x - {x}^{3}\right) \mathrm{dx}$
$= 2 \pi \left(4 x + \frac{4}{3} {x}^{3} - \frac{1}{2} {x}^{2} - \frac{1}{4} {x}^{4}\right) {|}_{0}^{2}$
$= 2 \pi \left(4 \cdot 2 + \frac{4}{3} \cdot {2}^{3} - \frac{1}{2} \cdot {2}^{2} - \frac{1}{4} \cdot {2}^{4}\right)$
$= 2 \pi \left(8 + \frac{32}{3} - 2 - 4\right) = 2 \pi \left(2 + \frac{32}{3}\right) = 2 \pi \left(\frac{38}{3}\right) = \frac{76 \pi}{3}$