# How do you find the volume of the region enclosed by the curves y=2x, y=x^2 rotated about the x-axis?

Sep 11, 2015

The volume of the solid generated by $y = 2 x$, $y = {x}^{2}$ revolved about the x-axis is $\frac{64 \pi}{15}$.

#### Explanation:

Revolving the area between these two curves about the x-axis, we end up with something that looks sort of like a cone... a hollow cone, with a curved inside. Now, imagine for a second taking a cross section parallel to the y-z plane, cutting the cone down the middle. The cross-section is going to look like a washer, with an inner radius equal to the height of $y = {x}^{2}$ wherever we are in the solid, and the outer radius equal to $y = 2 x$ at that same $x$.

For reference, the area of a washer is equal to $\pi \cdot {\left({r}_{\text{outer")^2 - pi*(r_"inner}}\right)}^{2}$, if ${r}_{\text{outer}}$ is the outer radius and ${r}_{\text{inner}}$ is the inner radius. This fact should be immediately obvious - we're just subtracting the area of a circle from the area of another circle.

Now, to find the volume of the solid, we need to sum (integrate) the area of each of cross-sectional washer in the solid. This procedure is commonly called the method of washers .

So, we'll use the method of washers to find the volume of this solid.

The general formula is

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

where $f \left(x\right)$ is a function giving the outer radius of the washer at any x, and $g \left(x\right)$ is a function giving the inner radius of the washer. Note that the integrands here represent the areas of circles - and together, they give the area of each infinitesimal washer.

In our case, $f \left(x\right)$ is equal to $2 x$ and $g \left(x\right)$ is equal to ${x}^{2}$.

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

Next, we need to worry about the limits of the integrals. Since we're only concerned with rotating the region between ${x}^{2}$ and $2 x$, the limits should involve the locations where $y = {x}^{2}$ and $y = 2 x$ intersect. Intersection occurs at $x = 0$ and $x = 2$.

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

The difficult part - setting up the appropriate integral with correct limits - is done, so I won't walk through evaluating the integral. It evaluates to $\frac{64 \pi}{15}$, but if you don't trust me you can evaluate it yourself as an exercise.