Question

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 = 1 + x^2`, `y = 0`, `x = 0`, `x = 2` rotated about the line `x=4`?

Answer 1

`(76pi)/3`

Explanation:

This is a graph of the region that will be revolved around the vertical line `x =4` (not pictured).

graph{(y-1-x^2)(y)( sqrt(2-x) )(sqrt(x)) / (sqrt(2-x))/(sqrt(x))<=0 [0, 6, -1.51, 6.39]}

Recall the general form of the volume of a solid of revolution using the shells method when you are revolving about a vertical line:

`V = int_a^b 2pi * radius * f(x)" " dx`

The hardest conceptual part here is the radius. Since the axis of revolution is `x = 4`, the form of the radius is the expression `4 - x`.

(Why? Draw a segment from `x = 4` on the x-axis to `x = 1` on the x-axis. How long is that segment? It is `4 - 1 = 3`.

Thus:

`V = int_0^2 2pi * (4-x) * (1 + x^2) dx`
`= 2pi int_0^2 (4 + 4x^2 - x - x^3) dx`
`=2pi (4x + 4/3x^3 - 1/2x^2 - 1/4x^4)|_0^2`
`=2pi(4*2 + 4/3 * 2^3 - 1/2 * 2^2 - 1/4 * 2^4)`
`=2pi(8 + 32/3 - 2 - 4) = 2pi(2 + 32/3) = 2pi(38/3)=(76pi)/3`