## 17Calculus Integrals - Volume of Revolution - Washer-Disc Method

This page covers single volume integrals when an area is rotated about a vertical or horizontal line. The area is defined by equations in the form $$y=f(x)$$ or $$x=f(y)$$ and we use the washer (disc) method. For other ways to calculate volume, see the links in the related topics panel.
If you want a full video lecture on this topic, we recommend this video and this instructor.

### Prof Leonard - Volume of Solids By Disks and Washers Method [2hr-47mins-48secs]

video by Prof Leonard

Overview

When calculating the volume of rotation, there are 3 factors that determine how to set up the integral.
1. method (washer-disc or cylinder-shell)
2. axis of rotation
3. function (graph and form of the equations)

On this page, we discuss the washer-disc method where the axis of rotation will always be either an axis or a straight line that is parallel to one of the axes. However, before we discuss the rotation of an area, we need to know how to describe an area in the plane. Go to the main volume integrals page to review that material.

washer-disc method

x-axis rotation y-axis rotation Now we will discuss each of these plots separately and explain each part of the plots.

Getting Started

Here are some key things that you need to do and know to get started.
1. Draw a rough plot of the area that is being rotated. This is usually best done by hand since you will need to label it.
2. Decide what method you will use, washer-disc or cylinder-shell.
3. On the rough plot from point 1, label the axis of rotation and draw a representative rectangle somewhere in the area.
4. Label R and r.

Once those steps are done, you are ready to set up your integral.

Alternate Names

Disc Method

Disk Method

Ring Method

What Is A Washer?

If you are not familiar with a washer (other than to wash clothes), this wiki page has pictures and explains what a washer is. In short, it is a disc with a circular hole in it whose center is the same as the full disc.

We choose to use the term washer-disc method to refer to this technique. We think it covers the two most common names. This, of course, is a personal preference for this site and you need to check with your instructor to see what they require.

The volume integral using the washer-disc method is based on the volume of a washer or disc. Let's think a bit about the volume of a washer-disc. If we start with a full disc (no hole in the middle), the volume is the surface area times the thickness. Since the disc is a circle, the area of a circle is $$\pi R^2$$ where $$R$$ is the radius of the circle. The volume is $$\pi R^2 t$$ where $$t$$ is the thickness. We choose to use a capital R here as the radius of the disc.

Now, with a washer, we take the disc we just discussed and put a circular hole in it with it's center the same as the full disc. (Think of a CD or DVD disc.) Now the volume is reduced by what we have taken out of the center. This empty space has volume $$\pi r^2 t$$, where $$r$$ is the radius of the small hole. The thickness, $$t$$, is the same as the full disc.

So now we have what we need to put together an equation for the washer-disc with a hole in the middle. We take the volume of the full disc and subtract the volume of the hole to get $$V = \pi R^2 t - \pi r^2 t = \pi t(R^2-r^2)$$. [Notice the special case when there is no hole in the middle, can be thought of as $$r=0$$ giving the volume of the disk as just $$V=\pi R^2 t$$.]

summary of the washer-disc method

the representative rectangle is perpendicular to axis of revolution

$$R$$ is the distance from the axis of rotation to the far end of the representative rectangle

$$r$$ is the distance from the axis of rotation to the closest end of the representative rectangle

x-axis rotation equation

$$\displaystyle{ V = \pi \int_{a}^{b}{R^2-r^2~dx} }$$

y-axis rotation equation

$$\displaystyle{ V = \pi \int_{c}^{d}{R^2-r^2~dy} }$$

Note - Notice that $$R$$ and $$r$$ are distances, so they are always positive (although since we square them, the sign doesn't make any difference in the equations). washer-disc method with x-axis rotation

If you feel like you need further explanation of this, here is a video that tries to explain this method by drawing in three dimensions. In this video, notice that the axis of rotation runs along one side of the figure and, consequently, $$r=0$$.

### Khan Academy - Disc Method [10min-4secs]

Okay, now let's work some problems using the washer-disc method revolving an area about the x-axis.

 x-axis rotation practice
 area: $$y=\sqrt{x-1}, y=0, x=5$$ axis of rotation: x-axis method: washer-disc

Problem Statement

 area: $$y=\sqrt{x-1}, y=0, x=5$$ axis of rotation: x-axis method: washer-disc

$$V=8\pi$$

Problem Statement

 area: $$y=\sqrt{x-1}, y=0, x=5$$ axis of rotation: x-axis method: washer-disc

Solution

### 324 video

video by Krista King Math

$$V=8\pi$$

 area: $$y=x^2, x=y^2$$ axis of rotation: x-axis method: washer-disc

Problem Statement

 area: $$y=x^2, x=y^2$$ axis of rotation: x-axis method: washer-disc

$$\displaystyle{V=\frac{3\pi}{10}}$$

Problem Statement

 area: $$y=x^2, x=y^2$$ axis of rotation: x-axis method: washer-disc

Solution

### 877 video

video by Krista King Math

$$\displaystyle{V=\frac{3\pi}{10}}$$

 area: $$y=x^3, y=x, x \geq 0$$ axis of rotation: x-axis method: washer-disc

Problem Statement

 area: $$y=x^3, y=x, x \geq 0$$ axis of rotation: x-axis method: washer-disc

$$V=4\pi/21$$

Problem Statement

 area: $$y=x^3, y=x, x \geq 0$$ axis of rotation: x-axis method: washer-disc

Solution

### 1174 video

video by PatrickJMT

$$V=4\pi/21$$

 area: $$y=\sqrt{x}, y \geq 0, x=4$$ axis of rotation: x-axis method: washer-disc

Problem Statement

 area: $$y=\sqrt{x}, y \geq 0, x=4$$ axis of rotation: x-axis method: washer-disc

$$8\pi$$

Problem Statement

 area: $$y=\sqrt{x}, y \geq 0, x=4$$ axis of rotation: x-axis method: washer-disc

Solution

In this problem, he doesn't finish the integration, so here are the details.
$$\displaystyle{ \pi \left[ \frac{x^2}{2} \right]_0^4 = \pi [(4^2)/2 - (0^2)/2] = 8\pi}$$

### 1359 video

video by PatrickJMT

$$8\pi$$

 area: $$f(x)=2-\sin(x)$$, $$x=0$$, $$x=2\pi$$, $$y=0$$ axis of rotation: x-axis method: washer-disc

Problem Statement

 area: $$f(x)=2-\sin(x)$$, $$x=0$$, $$x=2\pi$$, $$y=0$$ axis of rotation: x-axis method: washer-disc

$$V = 9\pi^2$$

Problem Statement

 area: $$f(x)=2-\sin(x)$$, $$x=0$$, $$x=2\pi$$, $$y=0$$ axis of rotation: x-axis method: washer-disc

Solution

### 2003 video

video by Dr Chris Tisdell

$$V = 9\pi^2$$

 area: $$y^2=x-2, x=5$$ in the first quadrant axis of rotation: x-axis method: washer-disc

Problem Statement

 area: $$y^2=x-2, x=5$$ in the first quadrant axis of rotation: x-axis method: washer-disc

$$4.5\pi$$

Problem Statement

 area: $$y^2=x-2, x=5$$ in the first quadrant axis of rotation: x-axis method: washer-disc

Solution

### 2272 video

video by Michel vanBiezen

$$4.5\pi$$

 area: $$y=x^2, y=x$$ axis of rotation: x-axis method: washer-disc

Problem Statement

 area: $$y=x^2, y=x$$ axis of rotation: x-axis method: washer-disc

$$2\pi/15$$

Problem Statement

 area: $$y=x^2, y=x$$ axis of rotation: x-axis method: washer-disc

Solution

### 2273 video

video by Michel vanBiezen

$$2\pi/15$$

 area: $$y=\sqrt{9-x^2}$$ in the first quadrant axis of rotation: x-axis method: washer-disc

Problem Statement

 area: $$y=\sqrt{9-x^2}$$ in the first quadrant axis of rotation: x-axis method: washer-disc

$$V=18\pi$$

Problem Statement

 area: $$y=\sqrt{9-x^2}$$ in the first quadrant axis of rotation: x-axis method: washer-disc

Solution

### 2276 video

video by MIP4U

$$V=18\pi$$

 area: $$y=\sqrt{x}, y=x^2$$ axis of rotation: x-axis method: washer-disc

Problem Statement

 area: $$y=\sqrt{x}, y=x^2$$ axis of rotation: x-axis method: washer-disc

Solution

### 1184 video

 area: $$y=\sqrt{x}, y\geq0, x=1$$ axis of rotation: x-axis method: washer-disc

Problem Statement

 area: $$y=\sqrt{x}, y\geq0, x=1$$ axis of rotation: x-axis method: washer-disc

Solution

### 1182 video

Derive the equation for the volume of a sphere of radius r using the washer-disc method.

Problem Statement

Derive the equation for the volume of a sphere of radius r using the washer-disc method.

Solution

### 1183 video

Looking back at the plots, you will notice that the axis of revolution is always a coordinate axis, either the x-axis or the y-axis. A twist you will see is when the axis of revolution is another line. On this site, we will discuss only axes that are parallel to one of the coordinate axes.
In this case, the equations that will change are the ones that describe the distance from the axes of rotation. We suggest that you set up a sum from the parallel coordinate axis to the axis of rotation and then solve for whatever variable you need. This concept, especially, requires you to think over in your mind several times and look at examples.

 parallel to x-axis rotation practice
 area: $$y=x^2, x=y^2$$ axis of rotation: $$y=1$$ method: washer-disc

Problem Statement

 area: $$y=x^2, x=y^2$$ axis of rotation: $$y=1$$ method: washer-disc

$$\displaystyle{V=11\pi/30}$$

Problem Statement

 area: $$y=x^2, x=y^2$$ axis of rotation: $$y=1$$ method: washer-disc

Solution

### 1172 video

video by Krista King Math

$$\displaystyle{V=11\pi/30}$$

 area: $$y=x^3, y=x, x\geq0$$ axis of rotation: $$y=-2$$ method: washer-disc

Problem Statement

 area: $$y=x^3, y=x, x\geq0$$ axis of rotation: $$y=-2$$ method: washer-disc

$$V=25\pi/21$$

Problem Statement

 area: $$y=x^3, y=x, x\geq0$$ axis of rotation: $$y=-2$$ method: washer-disc

Solution

In the video, he set up the integral but did not evaluate it. His integral was $$\displaystyle{ \int_{0}^{1}{\pi[(2+x)^2 - \pi(2+x^3)^2]dx} }$$. This evaluates to $$\displaystyle{ \pi \left[ x^2 + \frac{x^3}{3} - x^4 - \frac{x^7}{7} \right]_{0}^{1} }$$

### 1175 video

video by PatrickJMT

$$V=25\pi/21$$

 area: $$y=x^3, y=x, x\geq0$$ axis of rotation: $$y=5$$ method: washer-disc

Problem Statement

 area: $$y=x^3, y=x, x\geq0$$ axis of rotation: $$y=5$$ method: washer-disc

$$V=97\pi/42$$

Problem Statement

 area: $$y=x^3, y=x, x\geq0$$ axis of rotation: $$y=5$$ method: washer-disc

Solution

### 1176 video

video by PatrickJMT

$$V=97\pi/42$$

 area: $$y=x^2, y=x$$ axis of rotation: $$y=5$$ method: washer-disc

Problem Statement

 area: $$y=x^2, y=x$$ axis of rotation: $$y=5$$ method: washer-disc

$$23\pi/15$$

Problem Statement

 area: $$y=x^2, y=x$$ axis of rotation: $$y=5$$ method: washer-disc

Solution

### 2275 video

video by Michel vanBiezen

$$23\pi/15$$

 area: $$f(x)=x, g(x)=x^2-3x$$ axis of rotation: $$y=5$$ method: washer-disc

Problem Statement

 area: $$f(x)=x, g(x)=x^2-3x$$ axis of rotation: $$y=5$$ method: washer-disc

Hint Problem Statement

 area: $$f(x)=x, g(x)=x^2-3x$$ axis of rotation: $$y=5$$ method: washer-disc

$$\displaystyle{ V = \frac{1472\pi}{15} }$$

Problem Statement

 area: $$f(x)=x, g(x)=x^2-3x$$ axis of rotation: $$y=5$$ method: washer-disc

Hint Solution

1. Draw the plot and the example rectangle and label r and R. See the hint or the animation. We drew the example rectangle perpendicular to the axis of revolution since we were told to use the washer-disc method. The distance r is the distance from the axis of revolution to closest end of the rectangle. The distance R is from the axis of revolution to the farthest end of the example rectangle.

2. Choose The Integral - The volume integral we need is $$V = \pi\int_a^b{R^2-r^2~dx}$$. We chose this integral because we are told to use the washer-disc method, so we need an integral with r and R. We integrate with respect to x since the example rectangle is vertical and consequently it moves horizontally, sweeping in the x-direction.

3. Determine r and R - From the plot, let's start on the axis of revolution and move down to the x-axis of revolution. This distance is 5 units. Moving back up to the end of the rectangle that lands on $$y=x$$, we move y units. What we are left with is r, so $$r=5-y$$. However, we need to replace y with expression for y in terms of x. Since $$y=x$$ is the line that we are working with, we have $$r=5-x$$.
Determining the expression for R is a bit trickier. Starting on the axis of revolution, we move down to the x-axis which is 5 units. However, when $$x < 3$$ we need to go a little further in the same direction to get the full distance R. Let's put that aside for a minute and think about the part of the graph for $$x > 3$$. In this case, we need to go back up y units, so $$R=5-y$$. This looks the same as r but in this case, we are landing on the plot $$y=x^2-3x$$. So $$R=5-(x^2-3x) = -x^2+3x+5$$. Let's plug in a few values and compare the values to graph to see if they match.

 $$x=3$$ $$R=-3^2+3(3)+5=5$$ ✔ $$x=4$$ $$R=-4^2+3(4)+5=1$$ ✔

So far, so good. Let's plug in a few values $$x < 3$$ and see what we get.

 $$x=0$$ $$R=-0^2+3(0)+5=5$$ ✔ $$x=2$$ $$R=-2^2+3(2)+5=7$$ ✔

So it looks like we have the correct equation for R. We can do the same with r to check our equation. This does not guarantee that we have the right equations but it may give an indication if they are incorrect. I usually check both endpoints and at least one other point, two other points is even better.

4. Set up and evaluate the integral - If we look at the animation above, we can tell that the rectangle sweeps across the area from $$x=0$$ to $$x=4$$. So our integral is $$\displaystyle{ V = \pi \int_0^4{ (-x^2+3x+5)^2 - } }$$ $$\displaystyle{ (5-x)^2 ~ dx }$$. Let's evaluate it.

 $$\displaystyle{ V = \pi\int_0^4{ (-x^2+3x+5)^2 - (5-x)^2 ~ dx } }$$ $$\displaystyle{ V = \pi\int_0^4{ (x^4-3x^3-5x^2-3x^3+ } }$$ $$9x^2+15x-5x^2+15x+25) -(25-10x+x^2) ~dx$$ $$\displaystyle{ V = \pi\int_0^4{ x^4-6x^3-2x^2+40x ~dx } }$$ $$\displaystyle{ V = \pi\left[ \frac{x^5}{5} - \frac{6x^4}{4} -\frac{2x^3}{3}+\frac{40x^2}{2} \right]_0^4 }$$ $$\displaystyle{ V = \pi\left[ \frac{4^5}{5} - \frac{6(4)^4}{4} -\frac{2(4)^3}{3}+\frac{40(4)^2}{2} \right] - 0 }$$ $$\displaystyle{ V = \pi\left[ \frac{1024}{5} - \frac{1536}{4} -\frac{128}{3}+\frac{640}{2} \right] }$$ $$\displaystyle{ V = \pi \frac{1472}{15} }$$

$$\displaystyle{ V = \frac{1472\pi}{15} }$$

Now, let's work some problems with the y-axis as the axis of rotation. Here is the plot that contains all the information you need to work these problems. washer-disc method with y-axis rotation

 y-axis rotation practice
 area: $$x=2\sqrt{y}, x=0, y=9$$ axis of rotation: y-axis method: washer-disc

Problem Statement

 area: $$x=2\sqrt{y}, x=0, y=9$$ axis of rotation: y-axis method: washer-disc

$$V=162\pi$$

Problem Statement

 area: $$x=2\sqrt{y}, x=0, y=9$$ axis of rotation: y-axis method: washer-disc

Solution

### 343 video

video by Krista King Math

$$V=162\pi$$

 area: $$y=1-x^2, y=0$$ axis of rotation: y-axis method: washer-disc

Problem Statement

 area: $$y=1-x^2, y=0$$ axis of rotation: y-axis method: washer-disc

$$\displaystyle{V=\frac{\pi}{2}}$$

Problem Statement

 area: $$y=1-x^2, y=0$$ axis of rotation: y-axis method: washer-disc

Solution

### 878 video

video by Krista King Math

$$\displaystyle{V=\frac{\pi}{2}}$$

 area: $$y=x^2, y=5, x=0$$ axis of rotation: y-axis method: washer-disc

Problem Statement

 area: $$y=x^2, y=5, x=0$$ axis of rotation: y-axis method: washer-disc

$$25\pi/2$$

Problem Statement

 area: $$y=x^2, y=5, x=0$$ axis of rotation: y-axis method: washer-disc

Solution

### 2271 video

video by Michel vanBiezen

$$25\pi/2$$

 area: $$y=x^2, y=x$$ axis of rotation: y-axis method: washer-disc

Problem Statement

 area: $$y=x^2, y=x$$ axis of rotation: y-axis method: washer-disc

$$\pi/6$$

Problem Statement

 area: $$y=x^2, y=x$$ axis of rotation: y-axis method: washer-disc

Solution

### 2274 video

video by Michel vanBiezen

$$\pi/6$$

 area: $$y=\sqrt{x}$$ on $$[0,4]$$ axis of rotation: y-axis method: washer-disc

Problem Statement

 area: $$y=\sqrt{x}$$ on $$[0,4]$$ axis of rotation: y-axis method: washer-disc

$$V=32\pi/5$$

Problem Statement

 area: $$y=\sqrt{x}$$ on $$[0,4]$$ axis of rotation: y-axis method: washer-disc

Solution

### 2277 video

video by MIP4U

$$V=32\pi/5$$

 area: $$x^2+y^2=1, y \geq 1/2, x \geq 0$$ axis of rotation: y-axis method: washer-disc

Problem Statement

 area: $$x^2+y^2=1, y \geq 1/2, x \geq 0$$ axis of rotation: y-axis method: washer-disc

$$V=5\pi/24$$

Problem Statement

 area: $$x^2+y^2=1, y \geq 1/2, x \geq 0$$ axis of rotation: y-axis method: washer-disc

Solution

### 2280 video

video by Michel vanBiezen

$$V=5\pi/24$$

 area: $$y=-3x+6$$, x-axis, y-axis axis of rotation: y-axis method: washer-disc

Problem Statement

 area: $$y=-3x+6$$, x-axis, y-axis axis of rotation: y-axis method: washer-disc

$$V=8\pi$$

Problem Statement

 area: $$y=-3x+6$$, x-axis, y-axis axis of rotation: y-axis method: washer-disc

Solution

### 2281 video

video by Michel vanBiezen

$$V=8\pi$$

 area: $$y=x^2, y=0, x=1$$ axis of rotation: y-axis method: washer-disc

Problem Statement

 area: $$y=x^2, y=0, x=1$$ axis of rotation: y-axis method: washer-disc

Solution

### 1185 video

 parallel to y-axis rotation practice
 area: $$y=x^3, y=0, x=1$$ axis of rotation: $$x=2$$ method: washer-disc

Problem Statement

 area: $$y=x^3, y=0, x=1$$ axis of rotation: $$x=2$$ method: washer-disc

$$V = 3\pi/5$$

Problem Statement

 area: $$y=x^3, y=0, x=1$$ axis of rotation: $$x=2$$ method: washer-disc

Solution

### 1173 video

video by Krista King Math

$$V = 3\pi/5$$

 area: $$y=2\sqrt{x}, y=0, x=4$$ axis of rotation: $$x=5$$ method: washer-disc

Problem Statement

 area: $$y=2\sqrt{x}, y=0, x=4$$ axis of rotation: $$x=5$$ method: washer-disc

$$V=832\pi/15$$

Problem Statement

 area: $$y=2\sqrt{x}, y=0, x=4$$ axis of rotation: $$x=5$$ method: washer-disc

Solution

### 1916 video

video by MIP4U

$$V=832\pi/15$$

 area: $$x^2+y^2=1$$, x-axis, y-axis axis of rotation: $$x=2$$ method: washer-disc

Problem Statement

 area: $$x^2+y^2=1$$, x-axis, y-axis axis of rotation: $$x=2$$ method: washer-disc

$$V=(\pi/12)[3\pi-4]$$

Problem Statement

 area: $$x^2+y^2=1$$, x-axis, y-axis axis of rotation: $$x=2$$ method: washer-disc

Solution

### 2279 video

video by Michel vanBiezen

$$V=(\pi/12)[3\pi-4]$$

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