marinacabello82g3

2023-02-18

How to find the volume of the region enclosed by the curves $y=2x$, $y={x}^{2}$ rotated about the x-axis?

Quincy Doyle

When we rotate the area between these two curves about the x-axis, we get something that looks like a cone... a hollow cone with a curved inside. Consider taking a cross section parallel to the y-z plane and 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 $y=2x$ at that same x.
For reference, the area of a washer is equal to $\pi \cdot {\left({r}_{\text{outer}}\right)}^{2}-\pi \cdot {\left({r}_{\text{inner}}\right)}^{2}$, if ${r}_{\text{outer}}$ is the outer radius and ${r}_{\text{inner}}$ is the inner radius. This fact should be immediately obvious - we're simply subtracting the area of one circle from the area of another.
To calculate the volume of the solid, add (integrate) the areas of each cross-sectional washer in the solid. This method is also known as the washers method.
Thus, we'll use the method of washers to find the volume of this solid.
The general formula is
$V={\int }_{a}^{b}\pi {\left(f\left(x\right)\right)}^{2}dx-{\int }_{a}^{b}\pi {\left(g\left(x\right)\right)}^{2}dx$
where f(x) is a function giving the outer radius of the washer at any x, and g(x) 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(x) is equal to 2x and g(x) is equal to ${x}^{2}$.
$V={\int }_{a}^{b}\pi {\left(2x\right)}^{2}dx-{\int }_{a}^{b}\pi {\left({x}^{2}\right)}^{2}dx$
Next, we need to worry about the limits of the integrals. Since we're only concerned with rotating the region between ${x}^{2}$ and 2x, the limits should involve the locations where $y={x}^{2}$ and $y=2x$ intersect. Intersection occurs at $x=0$ and $x=2$.
$V={\int }_{0}^{2}\pi {\left(2x\right)}^{2}dx-{\int }_{0}^{2}\pi {\left({x}^{2}\right)}^{2}dx$
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.

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