# 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=4(x-2)^2 and y=x^2-4x+7 rotated about the y axis?

Dec 15, 2015

Please see the explanation section, below.

#### Explanation:

$y = 4 {\left(x - 2\right)}^{2}$ and $y = {x}^{2} - 4 x + 7$

The region bounded by the functions is shown below in blue.
We are using the shell method, so we take a representative slice parallel to the axis of revolution. In this case, that is the $y$-axis. So the thickness of the slice is $\mathrm{dx}$.
The dashed line shows the radius of revolution for the representative.

We will be integrating with respect to $x$, so we need the minimum and maximum $x$ values.

Solving: $4 {\left(x - 2\right)}^{2} = {x}^{2} - 4 x + 7$, we find the $x$ values at the points of intersection to be $x = 1 , 3$. So $x$ varies from $1$ to $3$.

The volume of a cylindrical shell is the surface area of the cylinder times the thickness:

$2 \pi r h \times \text{thickness}$

In the picture, the radius is shown by the dashed line, it is $x$.

The height of the slice is the upper $y$ value minus the lower: $\left({x}^{2} - 4 x + 7\right) - 4 {\left(x - 2\right)}^{2}$

The volume of the representative is: $2 \pi x \left[\left({x}^{2} - 4 x + 7\right) - 4 {\left(x - 2\right)}^{2}\right] \mathrm{dx}$

The volume of the resulting solid is:

$V = {\int}_{1}^{3} 2 \pi x \left[\left({x}^{2} - 4 x + 7\right) - 4 {\left(x - 2\right)}^{2}\right] \mathrm{dx}$

$= 2 \pi {\int}_{1}^{3} \left[- 3 {x}^{3} - 4 {x}^{2} + 3 x\right] \mathrm{dx}$

$= 2 \pi \left[8\right] = 16 \pi$