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 y-axis?

Answer 1

`12pi`

Explanation:

With a problem like this, it is always a good idea to start with a graph.

We are going to rotate this shape around the `y`-axis, which will result in a cylinder with a parabolic indentation in the top. In order to evaluate this using shells, we will be integrating a bunch of ring surfaces between the `x`-axis and the function, thus we will get an area function, `A(x)` for surfaces at each `x` and add them all together to get the volume.

We can find our area function by stacking a bunch of circles of radius `x` until they reach a height of `f(x)`. In other words;

`A(x)=2pi (radius) (height)=2pi x(1+x^2)`

Now that we have an area function, we just need to integrate it from the `y`-axis to `x=2`.

`int_0^2 A(x) dx`

Plug in the area function.

`int_0^2 2pi x(1+x^2) dx`

Move the `2pi` out front and simplify inside the integral.

`2pi int_0^2 x+x^3 dx`

Solve the integral.

`2pi (x^2/2 +x^4/4)_0^2`

`2pi (2^2/2+2^4/4)`

`12pi`