# 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=2x^2+5, y=x+3, the y-axis, and the line x=3 rotated about the x-axis?

Oct 23, 2015

See the explanation section, below.

#### Explanation:

Here is a graph of part of the region to be rotated about the $x$ axis. In order to use shells, we must take our representative slice parallel to the axis of rotation. In this case, that's the $x$ axis, so the thickness of our shells will be $\mathrm{dy}$.

That means we'll need to rewrite the curves as functions of $y$

$x = \sqrt{\frac{y - 5}{2}}$, $x = y - 3$

Every shell will have $\text{radius} = y$, $\text{thickness} = \mathrm{dy}$ and $\text{height" = x_"right" - x_"left}$

As the graph shows, there are three separate integrals we need to do, because the calculation of ${x}_{\text{right}}$ and ${x}_{\text{left}}$ change as $y$ goes from $3$ (the lower bound) to $23$ (the uppr bound, not on my graph).

(Are we sure we want to use shells for this?)

From $y = 3$ to $y = 5$ , the height of the cylindrical shell will be
$\text{height" = x_"right" - x_"left} = \left(y - 3\right) - 0 = y - 3$

So we need to integrate

${\int}_{3}^{5} 2 \pi y \left(y - 3\right) \mathrm{dy} = 2 \pi {\int}_{3}^{5} \left({y}^{2} - 3 y\right) \mathrm{dy}$ which is left to the student.

From $y = 5$ to $y = 6$, the height of the cylindrical shell will be
$\text{height" = x_"right" - x_"left} = \left(y - 3\right) - \sqrt{\frac{y - 5}{2}}$

So we need to integrate

${\int}_{5}^{6} 2 \pi y \left(y - 3 - \sqrt{\frac{y - 5}{2}}\right) \mathrm{dy} = 2 \pi {\int}_{5}^{6} \left({y}^{2} - 3 y - y \sqrt{\frac{y - 5}{2}}\right) \mathrm{dy}$ the last term of which can be integrated by substitution (or parts).

From $y = 6$ to $y = 23$, the height of the cylindrical shell will be
$\text{height" = x_"right" - x_"left} = 3 - \sqrt{\frac{y - 5}{2}}$

So we need to integrate

${\int}_{6}^{23} 2 \pi y \left(3 - \sqrt{\frac{y - 5}{2}}\right) \mathrm{dy} = 2 \pi {\int}_{6}^{23} \left(3 y - y \sqrt{\frac{y - 5}{2}}\right) \mathrm{dy}$ the last term of which can be integrated by substitution (or parts).

Washers

To use washers take the slices perpendicular to the axis of rotation.

As x varies from $0$ to $3$, the greater radius at $x$ is $R = 2 {x}^{2} + 5$ and the lesser radius is $r = x + 3$,

We need to integrate

${\int}_{0}^{3} \pi \left({R}^{2} - {r}^{2}\right) \mathrm{dx} = \pi {\int}_{0}^{3} \left({\left(2 {x}^{2} + 5\right)}^{2} - {\left(x + 3\right)}^{2}\right) \mathrm{dx}$ which is simply the integral of a degree 4 polynomial when we expand it.