Question

How do you find the surface area of the solid obtained by rotating about the `y`-axis the region bounded by `y=x^2` on the interval `1<=x<=2` ?

Answer 1

First of all, you are missing a bound. We will assume that the other bound is `y=0` or the `x`-axis. The answer is `(15pi)/2`.

The first step is to determine whether you are rotating along an axis that is parallel to the independent axis or the axis of the parameter (`x` in this case). And we are not, so this integration should be done with cylindrical shells.

Always draw a diagram to verify what is the parameter and what is the function.

You should note that `y` is not always a parameter of `x`. For instance, `x=y^2`, `x` is now a parameter of `y`.

The formula for cylindrical shells is:

`V=int_a^b2pirhdr`
`h` is represented by `y`, we have `y=x^2` and `y=0`
`r` is represented by `x`
`V=int_1^2 2 pi x (x^2-0) dx`

Now that the substitutions are done, we can solve:

`V=2 pi int_1^2x^3dx`
`=2pi (x^4)/4|_1^2`
`=2pi([2^4-1^4])/4`
`=(15pi)/2`