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#