# 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=6x+7 and y=x^2 rotated about the line y=49?

Sep 23, 2015

Given a choice, I would use washers, but. . .

#### Explanation:

Here is a (non-interactive) graph using Desmos (desmos.com)
I have included the line $y = 49$ in red. Here is the region using the Socratic grapher. It isn't quite accurate, but you can zoom in and out and drag the graph around

graph{(y-x^2)(y-6x-7) (sqrt(x+1))/(sqrt(x+1))<= 0 [-8.29, 20.21, -6, 8.26]}

$y = 6 x + 7$ if and only if $x = \frac{1}{6} y - \frac{7}{6}$

$y = {x}^{2}$ if and only if $x = \pm \sqrt{y}$

To use shells we take our representative slices horizontally. So the bounds become

For $y = 0$ to $y = 1$, $x$ goes from $- \sqrt{y}$ to $\sqrt{y}$.

From $y - 1$ to $y = 49$, $x$ goes from $\frac{1}{6} y - \frac{7}{6}$ to $\sqrt{y}$.

Throughout the problem, the radius of the cylindrical shell will be $49 - y$

So we need to evaluate two integrals:

$2 \pi {\int}_{0}^{1} \left(49 - y\right) \left(\sqrt{y} - \left(- \sqrt{y}\right)\right) \mathrm{dy} = 4 \pi {\int}_{0}^{1} \left(49 {y}^{\frac{1}{2}} - {y}^{\frac{3}{2}}\right) \mathrm{dy}$

$= \frac{1936}{15} \pi$

And

$2 \pi {\int}_{1}^{49} \left(49 - y\right) \left(\sqrt{y} - \left(\frac{1}{6} y - \frac{7}{6}\right)\right) \mathrm{dy} = \frac{25296}{5} \pi$

The volume is the sum of the two integrals.

Washers

$\pi {\int}_{-} {1}^{7} \left[{\left(49 - {x}^{2}\right)}^{2} - {\left(49 - \left(6 x + 7\right)\right)}^{2}\right] \mathrm{dx} = \frac{77824}{15} \pi$