# 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=6 and x=7-(y-1)^2 rotated about the x-axis?

Aug 15, 2015

$\frac{27 \pi}{2}$

#### Explanation:

Since we are revolving the around the x axis using the shell method we will be integrating with respect to y.

To proceed we need to find where the functions intersect

Solve $x + y = 6$ for $x$

$x = 6 - y$

Now set the two functions equal to each other

$6 - y = 7 - {\left(y - 1\right)}^{2}$

$6 - y = 7 - \left({y}^{2} - 2 y + 1\right)$

$6 - y = 7 - {y}^{2} + 2 y - 1$

$6 - y = 6 - {y}^{2} + 2 y$

$- y = - {y}^{2} + 2 y$

${y}^{2} - y - 2 y = 0$

${y}^{2} - 3 y = 0$

$y \left(y - 3\right) = 0$

$y = 0$ and $y = 3$

Our radius will be some value $y$ over the interval
$0 \le y \le 3$

Our cylinder height is
$7 - {\left(y - 1\right)}^{2} - \left(6 - y\right)$

The integral for the volume is

$2 \pi {\int}_{0}^{3} y \left[7 - {\left(y - 1\right)}^{2} - \left(6 - y\right)\right] \mathrm{dy}$

2piint_0^3y[7-(y^2-2y+1)-(6-y)dy

$2 \pi {\int}_{0}^{3} y \left[7 - {y}^{2} + 2 y - 1 - 6 + y\right] \mathrm{dy}$

$2 \pi {\int}_{0}^{3} y \left[- {y}^{2} + 3 y\right] \mathrm{dy}$

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

Integrating we have

$2 \pi \left[- {y}^{4} / 4 + {y}^{3}\right]$

Now evaluating we get

$2 \pi \left[- {\left(3\right)}^{4} / 4 + {3}^{3}\right] = 2 \pi \left[- \frac{81}{4} + 27\right]$

$2 \pi \left[- \frac{81}{4} + \frac{108}{4}\right] = 2 \pi \left[\frac{27}{4}\right] = \frac{27 \pi}{2}$