# 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?

Jul 27, 2015

This is what you will be revolving.

$y = 2 {x}^{2} - {x}^{3}$:
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.

$V \left(x\right) = {\sum}_{x = a}^{b} 2 \pi x f \left(x\right) \Delta x$

$= {\int}_{a}^{b} 2 \pi x f \left(x\right) \mathrm{dx}$

where:
$x$ is the radius of the shell
$f \left(x\right)$ is the height of the shell
$\mathrm{dx}$ is the thickness of the shell
$2 \pi$ 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.

$0 = 2 {x}^{2} - {x}^{3}$

$0 = {x}^{2} \left(2 - x\right)$

$x = 2 , 0$

So we have:

$= 2 \pi {\int}_{0}^{2} x \left(2 {x}^{2} - {x}^{3}\right) \mathrm{dx}$

$= 2 \pi {\int}_{0}^{2} 2 {x}^{3} - {x}^{4} \mathrm{dx}$

$= 2 \pi \left[{x}^{4} / 2 - {x}^{5} / 5\right] {|}_{0}^{2}$

$= 2 \pi \left\{\left[{\left(2\right)}^{4} / 2 - {\left(2\right)}^{5} / 5\right] - 0\right\}$

$= 2 \pi \left[\frac{16}{2} - \frac{32}{5}\right]$

$= 2 \pi \left[\frac{40}{5} - \frac{32}{5}\right]$

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