# 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, y=x^2, and x=0 rotated about the line y=2?

Sep 22, 2015

$\frac{28 \pi}{15}$ cubic units

#### Explanation:

Since we are revolving around a horizontal line using the method of shells we will integrate with respect to $y$.

We are bounded by the $y$ axis, the horizontal line $y = 1$, and the function $y = {x}^{2}$

Solve $y = {x}^{2}$ for $x$

$x = \sqrt{y}$

We are in quadrant $I$ so we do not have to worry about the negative square root.

Our representative cylinder height is our function $\sqrt{y}$

Our representative radius is $2 - y$ over the interval $0 \le y \le 1$

The integral for the volume is

$2 \pi {\int}_{0}^{1} \left(2 - y\right) \left({y}^{\frac{1}{2}}\right) \mathrm{dy}$

$2 \pi {\int}_{0}^{1} 2 {y}^{\frac{1}{2}} - {y}^{\frac{3}{2}} \mathrm{dy}$

Integrating we get

$2 \pi \left[\frac{4}{3} {y}^{\frac{3}{2}} - \frac{2}{5} {y}^{\frac{5}{2}}\right]$

Evaluating we get

$2 \pi \left[\frac{4}{3} - \frac{2}{5} - 0\right]$

$2 \pi \left[\frac{20}{15} - \frac{6}{15}\right] = 2 \pi \left[\frac{14}{15}\right] = \frac{28 \pi}{15}$ cubic units