# 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 x= y^2 x= y+2 rotated about the y-axis?

##### 1 Answer
Sep 28, 2015

See explanation below.

#### Explanation:

To use shells when revoling about a vertical line, we need to take our representative slices vertically, so the thickness is $\mathrm{dx}$. And we need the boundaries written as functions of $x$.

$y = \sqrt{x}$, $y = - \sqrt{x}$, $y = x - 2$

Here is a picture of the region to be revolved about the $y$ axis. I have included two representative slices of the region.

The curves intersect at $\left(1 , - 1\right)$ and at $\left(4 , 2\right)$

The radii of our cylindrical shells will be $x$ from $x = 0$ to $x = 4$.
The thickness is $\mathrm{dx}$ and
the height of each shell is the greater $y$ at $x$ - lesser $y$ at $x$.
(The $y$ above - $y$ below#)

The greater value of $y$ can be found using $y = \sqrt{x}$ for all $x$ from $0$ to $4$.
The lesser value of $y$ changes at $x = 3$. So we'll need to evaluate two integrals.

From $x = 0$ to $x = 1$, the lesser $y$ value is $- \sqrt{x}$, so the height is $\sqrt{x} - \left(- \sqrt{x}\right) = 2 \sqrt{x}$

To get that part of the volume, we need to evaluate:

${\int}_{0}^{1} 2 \pi r h \mathrm{dx} = 2 \pi {\int}_{0}^{1} x \left(2 \sqrt{x}\right) \mathrm{dx} = 4 \pi {\int}_{0}^{1} {x}^{\frac{3}{2}} \mathrm{dx} = \frac{8}{5} \pi$

From $x = 1$ to $x = 4$, the lesser $y$ value is $x - 2$, so the height is $\sqrt{x} - \left(x - 2\right) = \sqrt{x} - x + 2$

To get that part of the volume, we need to evaluate:

${\int}_{1}^{4} 2 \pi r h \mathrm{dx} = 2 \pi {\int}_{1}^{4} x \left(\sqrt{x} - x + 2\right) \mathrm{dx} = 2 \pi {\int}_{1}^{4} \left({x}^{\frac{3}{2}} - {x}^{2} + 2 x\right) \mathrm{dx} = \frac{64}{5} \pi$

The total volume is found by adding these two results, to get:

$\frac{72}{5} \pi$