# 13. Volume

## d1. Volume using Thin Washers

Next we consider the problem of computing the volume of a solid of revolution using thin washers.

The region between the functions \(y=f(x)\) and \(y=g(x)\) between \(x=a\) and \(x=b\) is rotated about the \(x\)-axis. Find the volume of the solid swept out.

This is an \(x\) integral. So we draw a vertical rectangle and rotate it about the \(x\)-axis obtaining a thin washer.

The situation is exactly like the case for a disk except that we need to subtract off the volume of the hole. So the volume is: \[ V=\int_a^b (\pi f(x)^2-\pi g(x)^2)\,dx \]

Now turn your head sideways:

The region between the functions \(x=f(y)\) and \(x=g(y)\) between \(y=a\) and \(y=b\) is rotated about the \(y\)-axis. Find the volume of the solid swept out.

This is a \(y\) integral. So we draw a horizontal rectangle and rotate it about the \(y\)-axis obtaining a thin washer.

Again we subtract off the volume of the hole. So the volume is: \[ V=\int_a^b (\pi f(y)^2-\pi g(y)^2)\,dy \]

If the region below the function \(y=f(x)\) above the function \(y=g(x)\) between \(x=a\) and \(x=b\) is rotated about the \(x\)-axis, the volume swept out is: \[ V=\int_a^b \pi(f(x)^2-g(x)^2)\,dx \] If the region to the left of the function \(x=f(y)\) and to the right of the function \(x=g(y)\) between \(y=a\) and \(y=b\) is rotated about the \(y\)-axis, the volume swept out is: \[ V=\int_a^b \pi(f(y)^2-g(y)^2)\,dy \] Both formulas can be summarized as \[ V=\int_a^b \pi(\text{outer radius})^2-\pi(\text{inner radius})^2\,dv \] where \(dv\) is either \(dx\) or \(dy\).

Notice that finding the volume of a solid of revolution by washers is really a special case of finding a volume by slicing \[ V=\int_a^b A(v)\,dv \] in which each cross-sectional area is the area between two circles, \(A=\pi(\text{outer radius})^2-\pi(\text{inner radius})^2\).

The area between the curves \(y=x\) and \(y=\sqrt{x}\) can be computed as either an \(x\)-integral or as a \(y\)-integral. It can also be rotated about the \(x\)-axis or about the \(y\)-axis. We will look at all four possibilities on this and the next page.

The region between the curves \(x=y\) and \(x=y^2\) is rotated about the \(y\)-axis. Find the volume swept out. (Do the integral as a \(y\) integral.)

Since the curves are given with \(x\) as functions of \(y\), we will compute using a \(y\)-integral. We cut up the \(y\)-axis and see that the rectangles are horizontal and rotate into thin washers. So the volume is: \[\begin{aligned} V&=\pi\int_a^b (\text{right}^2-\text{left}^2)\,dy =\pi\int_0^1 ((y)^2-(y^2)^2)\,dy \\ &=\pi\int_0^1 (y^2-y^4)\,dy =\pi\left[\dfrac{y^3}{3}-\dfrac{y^5}{5}\right]_0^1 =\dfrac{2}{15}\pi \end{aligned}\]

Now it's your turn:

The region between the curves \(y=x\) and \(y=\sqrt{x}\) is rotated about the \(x\)-axis. Find the volume swept out. (Do the integral as an \(x\) integral.)

Notice this is the same \(2\)-dimensional area as in example 1, but it is rotated about a different axis. So the volume is totally different.

The volume is to be computed as an \(x\)-integral. So cut up the \(x\)-axis.

\(\displaystyle V=\dfrac{\pi}{6}\)

We cut up the \(x\)-axis and see that the rectangles are vertical and they rotate into thin washers. So the volume is: \[\begin{aligned} V&=\pi\int_a^b (\text{upper}^2-\text{lower}^2)\,dx \\ &=\pi\int_0^1 \left((\sqrt{x})^2-(x)^2\right)\,dx \\ &=\pi\int_0^1 (x-x^2)\,dx =\pi\left[\dfrac{x^2}{2}-\dfrac{x^3}{3}\right]_0^1=\dfrac{\pi}{6} \end{aligned}\]

Here's another example:

Example |

You can practice computing Volumes of Revolution by Disks and Washers by using the following Maplet (requires Maple on the computer where this is executed):

Volume of a Solid of Revolution Rate It

For now, when you click on Show in Step 1, only do those problems which are Disks or Washers.

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