Question

How do you find the volume of the solid generated by revolving the region bounded by the curves `y=2x^2 -x^3` and y = 0 rotated about the y-axis?

Answer 1

This is what you will be revolving.

`y = 2x^2 - x^3`:
graph{2x^2 - x^3 [-0.1, 10, -0.3, 5]}

What we could do here is use the shell method, which is far more convenient in this case than the traditional revolution method.

`V(x) = sum_(x = a)^b 2pixf(x)Deltax`

`= int_a^b 2pixf(x)dx`

where:
`x` is the radius of the shell
`f(x)` is the height of the shell
`dx` is the thickness of the shell
`2pi` indicates the radian measure of the "circumference" of the shell (not a true circular circumference, hence the quotes)

Now where is the x-intercept? We can see it's `2`, so let's show that.

`0 = 2x^2 - x^3`

`0 = x^2(2 - x)`

`x = 2, 0`

So we have:

`= 2piint_0^2 x(2x^2 - x^3)dx`

`= 2piint_0^2 2x^3 - x^4dx`

`= 2pi [x^4/2 - x^5/5]|_(0)^(2)`

`= 2pi {[(2)^4/2 - (2)^5/5] - 0}`

`= 2pi [16/2 - 32/5]`

`= 2pi [40/5 - 32/5]`

`= color(blue)((16pi)/5 "rad") ~~ 10.0531 "rad"`