The washer method of calculating volumes of revolution

This method is a logical extension of the disk method for finding volumes of revolution. You might want to review that section before going in with this one.

These 3D objects are obtained by revolving the white outline by 360˚ about the $y$-axis.

Area of a "washer"

By "washer," we're referring to a flat ring like the washers used to keep bolts from pulling through their holes.

Finding the area of a washer is simple: Just find the area of the outer circle and subtract the area of the inner circle.

All we need is the two radii ($r$ for the inner circle, $R$ for the outer), which in turn must be expressed as a function of the coordinate axis along which we'll be integrating. Getting those functional relationships will be the crux of getting volumes of revolution right.

Example 1 : Volume of a torus

Now the area of the washer as a function of $y$ is:

$$ \begin{align} A_{washer} &= \pi \left( [h + (r^2 - y^2)^{1/2}]^2 - [h - (r^2 - y^2)^{1/2}]^2 \right) \\[5pt] &= \pi [h^2 + (r^2 - y^2) + 2h (r^2 - y^2)^{1/2} - h^2 + 2h (r^2 - y^2)^{1/2} - (r^2 + y^2)] \\[5pt] &= 4\pi h(r^2 - y^2)^{1/2} \end{align}$$

Then we can set up the integral. Here we'll take advantage of the symmetry of the circle to introduce zero as the lower limit. In other words, given that one half of a circle has the same area as the other, integrating between $-r$ and $r$ is the same as twice the integral from $0$ to $r$, where $r$ is the radius of the circle.

$$V = 4\pi h \int_{-r}^r (r^2 - y^2)^{1/2} \, dy \; = \; 8\pi h\int_0^r (r^2 - y^2)^{1/2} \, dy$$

Evaluation of this integral is a bit tricky. It involves evaluation of the limit of the inverse tangent function as the argument approaches zero (i.e. the denominator of the argument approaches zero). That limit is $\pi/2$

$$ \begin{align} &4 \pi h \left[ y \sqrt{r^2 - y^2} + y^2 \tan^{-1} \left( \frac{y}{\sqrt{r^2 - y^2}} \right) \right]_0^r \\[5pt] &= 4 \pi h \lim_{y\to r} \, y^2 \tan^{-1} \left( \frac{y}{\sqrt{r^2 - y^2}} \right) \\[5pt] &= 4\pi h \cdot \frac{r^2\pi}{2} \text{ units}^3 \end{align}$$

Example 2

Revolution of an area between two 2D curves to form a solid

In this example we'll revolve the region trapped between the curves $f(x) = x^2$ and $g(x)=x^3$ around the $y$-axis to obtain a bowl-like 3D figure.

The washer method is perfect for calculating the volume of this figure. Washers perpendicular to the $y$-axis have the radii shown (→). They were simply found by solving for $x$ from the functions.

The area function can then be found from the radii, $R_1$ and $R_2$:

$$ \begin{align} A &= \pi [y^{1/3})^2 - (y^{1/2})^2] \\[5pt] &= \pi (y^{2/3} - y) \end{align}$$

Then we can set up and solve the relatively simple integral:

$$V = \pi \int_0^1 (y^{2/3} - y) \, dy = \pi \left[ \frac{3 y^{5/3}}{5} - \frac{y^2}{2} \right]_0^1 = \pi \left[ \frac{3}{5} - \frac{1}{2} \right] = \frac{\pi}{10} \text{ units}^3$$

Example 3

Revolution about an off-axis line

Problem: Find the volume of the figure generated by revolving the area trapped between $f(x) = (x + 2)^{1/2}$ and $g(x) = e^x$ around the horizontal line $y = 2$.

Notice that this figure is not revolved around the $x$- or $y$-axis, but a different line, the horizontal line $y = 2$. We have to account for this in our integral or we won't get the right volume. One option is to just translate the figure downward by two units so that we actually revolve around the y-axis, but the distance is the same as in the original problem. Let's do it that way by making $f(x) = (x+2)^{1/2}- 2$ and $g(x) = e^x- 2$.

Now we find the area of a washer (thick black line in the figure above), making sure that we use $f(x)$ (black curve) as the outer radius, and $g(x)$ as the inner radius. This will ensure that we calculate a positive area & volume.

$$ \begin{align} A(x) = \pi [(g(x))^2 - (f(x))^2] &\pi[((x + 2)^{1/2})^2 - (e^x)^2] \\[5pt] &= \pi(x + 2 - e^{2x}) \end{align}$$

Then we have a fairly simple integral, only with pesky non-integer limits:

$$ \begin{align} V = \pi \int_{-1.981}^{0.448} (x + 2 - e^{2x}) \, dx &= \pi \left[ \frac{x^2}{2} + 2x - \frac{e^{2x}}{2} \right]_{-1.981}^{0.448} \\[5pt] &= 1.78 \pi = 5.59 \text{ units}^3 \end{align}$$

Rather than take a formulaic approach to off-axis revolution problems, I encourage you to simply apply the appropriate horizontal or vertical translation to an off axis problem to force it to be an on-axis one. It's a general approach that will help you to solve more complicated problems later.

Practice problems

1. Find the volume of the solid generated by revolving the region bounded by the curves $y=x^2$ and $x=y^2$ about the $x$-axis.

   Solution

   

   Here is the setup. The inner radius corresponds to the $x^2$ function and the outer to the root function. We rotate around the $x$-axis. The intersection point is $(1,1)$, the solution to $x^2 = \sqrt{x}$. The integral is

   The cross section of the area element to be revolved around the $x$-axis is shown below.

   The volume of the washer element is $V = \pi (R-r)^2 dx$, so our integral is

   $$ \begin{align} V &= \int_0^1 \pi(x-x^4) dx \\[5pt] &= \pi \left( \frac{x^2}{2} - \frac{x^5}{5} \right)\bigg|_0^1 \\[5pt] &= \pi \left( \frac{1}{2} - \frac{1}{5} -0 + 0 \right) = \pi \left( \frac{5-2}{10} \right) \\[5pt] &= \frac{3\pi}{10} \text{ units}^3 \\[5pt] &= 0.942 \text{ units}^3 \end{align}$$

   




2. The area bounded by $x = 3\pi/2$, $x = 2\pi$, $y = \cos(x)$ and the $x$-axis is revolved around the $y$-axis. Calculate the volume of the resulting solid.

   Solution

   Here is a graph of the cross section described in this problem:

   

   The volume element (washer) geometry looks like this:

   

   Now the limits along the $x$-axis are $\frac{3\pi}{2}$ to $2\pi$, where the near and far-side radial functions are

   $$ \begin{align} y &= \cos(x) \rightarrow x = \cos^{-1}(y) \\[5pt] R &= 2\pi \text{ (constant)} \end{align}$$

   So the volume of the washer is:

   $$ \begin{align} V_{\text{washer}} &= \pi(R^2-r^2) dy \\[5pt] &= \pi(4\pi^2-[\cos^{-1}(y)]^2) \, dy \end{align}$$

   Now that's going to be a terrible thing to integrate, and it suggests that integration of this volume using the washer method might not be the right choice. It turns out that the shell method is a better choice. I'll put the solution here so you can refer back to it when you learn how to use shells.

   
   

   Here is a schematic of one of our cencentric shells.

   

   Then we have

   $$ \begin{align} h &= \cos(x) \\[5pt] r &= x \\[5pt] \text{limits } &= \left[ \frac{3\pi}{2}, \; 2\pi \right] \end{align}$$

   This time the volume integral is

   $$ \require{cancel} \begin{align} V &= 2\pi \int_{\frac{3\pi}{2}}^{2\pi} x \cos(x) dx \\[5pt] &\color{#005095}{\text{Let } u = x, \phantom{000} dv = \cos(x) dx} \\[5pt] &\color{#005095}{du = dx, \phantom{000} v = \sin(x)} \\[5pt] x\sin(x) &- \int \sin(x) dx \\[5pt] &= x \sin(x) - \cos(x) \\[5pt] &\rightarrow 2\pi \left[ x\sin(x) - \cos(x) \right]_{\frac{3\pi}{2}}^{2\pi} \\[5pt] &= 2\pi \left[ 2\pi \sin(2\pi) - \cos(2\pi) - \frac{3\pi}{2} \sin\left(\frac{3\pi}{2}\right) + \cos \left( \frac{3\pi}{2} \right) \right] \\[5pt] &= 2\pi \left[ -1+\frac{3\pi}{2} \right] \\[5pt] &= \cancel{2}\pi \left[ \frac{3\pi-2}{\cancel{2}} \right] \\[5pt] &= 3\pi^2 - 2\pi = 23.32 \text{ units}^3 \end{align}$$




3. Find the volume of the solid formed when the region trapped between $y = (x-5)^2-2$ and $y = -(x-5)^2 + 2$ is revolved around the $y$-axis.

   Solution

   
   

   To find the points of intersection of these curves, set them equal to each other and solve:

   $$ \begin{align} (x-5)^2-2 &= -(x-5)^2+2 \\[5pt] 2(x-5)^2 &= 4 \\[5pt] (x-5)^2 &= 4 \\[5pt] x-5 &= \pm \sqrt{2} \\[5pt] x &= 5\pm \sqrt{2} \end{align}$$

   The volume of our washer is derived from this figure:

   

   The volume of a washer element is then

   $$ \begin{align} V_{\text{washer}} &= \pi(R^2-r^2) dy \\[5pt] &= \pi(((x-5)^2 + 2)^2 - ((x-5)^2 -2)^2) dy \end{align}$$

   
   

   Here is a graph of the area element that gets revolved around the axis:

   

   The volume integral is

   $$ \require{cancel} \begin{align} V &= 2\pi \int_0^2 \cancel{(x-5)^4}-\cancel{4(x-5)^2}+4-(\cancel{(x-5)^4}-\cancel{4(x-5)^2}-4) dy \\[5pt] &= 2\pi \int_0^2 8 \, dy \\[5pt] &= 16\pi y \bigg|_0^2 = 32\pi \text{ units}^2 \end{align}$$

   The $2$ in the $2\pi$ of our integral just recognizes the symmetry of the problem. The area element above is the same above and below the $x$-axis.




4. The area bounded by $f(x) = \ln(x)$, $x = 1$ and $x = 4$ is revolved about the $y$-axis. Calculate the volume of the resulting solid.

   Solution

   
   

   Here is the setup for this problem. We rotate around the y-axis and integrate along x:

   

   The volume of a washer element is

   $$ \begin{align} V_{\text{washer}} &= \pi (R^2 - r^2) dy \\[5pt] &= \pi\left(4^2-(e^y)^2 \right) dy \end{align}$$

   
   

   So the integral is

   $$ \begin{align} V &= \pi \int_0^{ln(4)} 16-e^{2y} \, dy \\[5pt] &= \pi \left[ 16y - \frac{1}{2} e^{2y} \right]_0^{ln(4)} \\[5pt] &= \pi [ 16 \ln(4)] \\[5pt] V &= 16\pi \ln(4) \text{ units}^3 \\[5pt] &\approx 69.7 \text{ units}^3 \end{align}$$




5. The area bounded by $f(x)=e^x$, $x = 0$ and $x = 4$ is revolved about the $y$-axis. Calculate the volume of the resulting solid.

   Solution

   
   

   The setup for this problem is graphed below:

   

   The volumen element is

   $$ \begin{align} V_{\text{washer}} &= \pi (R^2-r^2) dy \\[5pt] R = 4, \; r: \; e^x &= y \rightarrow x = \ln(y) \\[5pt] V &= \pi \int_0^{e^4} 16 - (\ln(y))^2 dy \end{align}$$

   
   

   Well, once a gain, that's a very unattractive integral to contemplate. Without sketching the diagram, here's the setup and solution if we were to do this integral by the method of concentric shells:

   $$ \begin{align} V_{\text{shell}} &= 2\pi rh = 2\pi x e^x dx \\[5pt] V &= 2\pi \int_0^4 x e^x dx \\[5pt] \text{Let } u &= x, \; dv = e^x dx \\[5pt] \text{then } dv &= dx, \text{ and } v = e^x \\[5pt] V &= 2\pi \left[x e^x - \int e^x dx \right] \\[5pt] &= 2\pi \left[x e^x - e^x \right]_0^4 \\[5pt] &= 2\pi e^x (x-1)\bigg|_0^4 \\[5pt] &= 2\pi e^4 (4-1) - 2\pi e^0 (0-1) \\[5pt] &= 2\pi(3 e^4 + 1) = 1035 \text{ units}^3 \end{align}$$

Video examples

1. Volume of revolution | Washer method ($x$-axis)

Minutes of your life: 3:18