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?

1 Answer
May 26, 2016

Please see the explanation section, below.

Explanation:

y=sqrt(5x) 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 dx.
The dashed line shows the radius of revolution for the representative.

enter image source here

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:

2pirh xx "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(5x)

The volume of the representative shell is: 2pixsqrt(5x)dx

The volume of the resulting solid is:

V=int_0^5 2pixsqrt(5x)dxdx

= 2pisqrt5 int_0^5 x^(3/2)dx

= 2pisqrt5[(2x^(5/2))/5]_0^5= 100pi