# How do you use the disk method to find the volume of the solid formed by rotating the region bounded by y = 2x and y = x^2 about the y-axis?

May 15, 2015

$y = 2 x$ and $y = {x}^{2}$ intersect at $\left(0 , 0\right)$ and at $\left(2 , 4\right)$

I can't get both curves on one graph, but I'll assume you can graph the line and the parabola.

Rotating around the $y$ axis and using disks, means our independent variable will be $y$.

The representative slice is horizontal with left end (little radius: r) on the line $x = \frac{1}{2} y$ and right end (big radius: R) on the parabola $x = \sqrt{y}$.
The thickness of the disk will be $\mathrm{dy}$, and the limits of integration will be $0$ to $4$.

${\int}_{0}^{4} \pi {R}^{2} - \pi {r}^{2} \mathrm{dy} = \pi {\int}_{0}^{4} \left({\left(\sqrt{y}\right)}^{2} - {\left(\frac{y}{2}\right)}^{2}\right) \mathrm{dy}$

$\pi {\int}_{0}^{4} \left(y - {y}^{2} / 4\right) \mathrm{dy} = \pi {\left[{y}^{2} / 2 - {y}^{3} / 12\right]}_{0}^{4} = \pi \left[8 - \frac{16}{3}\right] = \frac{8 \pi}{3}$