# 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= sqrt(5x), x=5 rotated about the y axis?

May 26, 2016

Please see the explanation section, below.

#### Explanation:

$y = \sqrt{5 x}$ and $x = 5$ (and, presumably, the $x$ axis)

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.

From the given information, we conclude that $x$ varies from $0$ to $5$.

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: $\sqrt{5 x}$

The volume of the representative shell is: $2 \pi x \sqrt{5 x} \mathrm{dx}$

The volume of the resulting solid is:

$V = {\int}_{0}^{5} 2 \pi x \sqrt{5 x} \mathrm{dx} \mathrm{dx}$

$= 2 \pi \sqrt{5} {\int}_{0}^{5} {x}^{\frac{3}{2}} \mathrm{dx}$

$= 2 \pi \sqrt{5} {\left[\frac{2 {x}^{\frac{5}{2}}}{5}\right]}_{0}^{5} = 100 \pi$