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=4(x-2)^2# and #y=x^2-4x+7# rotated about the y axis?

1 Answer
Dec 15, 2015

Please see the explanation section, below.

Explanation:

# y=4(x-2)^2# and #y=x^2-4x+7#

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.

Solving: #4(x-2)^2 = x^2-4x+7#, we find the #x# values at the points of intersection to be #x=1,3#. So #x# varies from #1# to #3#.

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: #(x^2-4x+7)-4(x-2)^2#

The volume of the representative is: #2pix[(x^2-4x+7)-4(x-2)^2]dx#

The volume of the resulting solid is:

#V=int_1^3 2pix[(x^2-4x+7)-4(x-2)^2]dx#

# = 2pi int_1^3 [-3x^3-4x^2+3x]dx#

# = 2pi[8] = 16pi#