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.
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 (washerdisc or cylindershell)
2. axis of rotation
3. function (graph and form of the equations)
On this page, we discuss the washerdisc 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.
Okay, so now that you know how to describe an area in the plane, we will use that knowledge to calculate a volume of revolution. The technique we discuss on this page is the washerdisc method. We will use these figures extensively in the discussion of these techniques. Click here to download a one page pdf document of these pictures, with space to write notes. Feel free to copy this page to use while studying, working practice problems, in exams (if allowed by your instructor) and to give to fellow students. If you are an instructor and you think it will help your students, you may make as many copies as you need to use in your classes. For everyone, we ask that you keep the information that you got it from this website at the bottom of the sheet.
washerdisc method  

xaxis rotation 
yaxis 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, washerdisc or cylindershell.
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 washerdisc 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 washerdisc method is based on the volume of a washer or disc. Let's think a bit about the volume of a washerdisc. 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 washerdisc with a hole in the middle. We take the volume of the full disc and subtract the volume of the hole to get \(V = \pi R^2 t  \pi r^2 t = \pi t(R^2r^2)\). [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 washerdisc 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  
xaxis rotation equation  \(\displaystyle{ V = \pi \int_{a}^{b}{R^2r^2~dx} }\) 
yaxis rotation equation  \(\displaystyle{ V = \pi \int_{c}^{d}{R^2r^2~dy} }\) 
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).
washerdisc method with xaxis 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, \(r=0\).
video by Khan Academy 

Okay, now let's work some problems using the washerdisc method revolving an area about the xaxis.
xaxis rotation practice 
area: \(y=\sqrt{x1}, y=0, x=5\) 
axis of rotation: xaxis 
method: washerdisc 
Problem Statement 

area: \(y=\sqrt{x1}, y=0, x=5\) 
axis of rotation: xaxis 
method: washerdisc 
Final Answer 

Problem Statement 

area: \(y=\sqrt{x1}, y=0, x=5\) 
axis of rotation: xaxis 
method: washerdisc 
Solution 

video by Krista King Math 

Final Answer 

\(V=8\pi\) 
close solution

area: \(y=x^2, x=y^2\) 
axis of rotation: xaxis 
method: washerdisc 
Problem Statement 

area: \(y=x^2, x=y^2\) 
axis of rotation: xaxis 
method: washerdisc 
Final Answer 

Problem Statement 

area: \(y=x^2, x=y^2\) 
axis of rotation: xaxis 
method: washerdisc 
Solution 

video by Krista King Math 

Final Answer 

\(\displaystyle{V=\frac{3\pi}{10}}\) 
close solution

area: \(y=x^3, y=x, x \geq 0\) 
axis of rotation: xaxis 
method: washerdisc 
Problem Statement 

area: \(y=x^3, y=x, x \geq 0\) 
axis of rotation: xaxis 
method: washerdisc 
Final Answer 

Problem Statement 

area: \(y=x^3, y=x, x \geq 0\) 
axis of rotation: xaxis 
method: washerdisc 
Solution 

video by PatrickJMT 

Final Answer 

\(V=4\pi/21\) 
close solution

area: \(y=\sqrt{x}, y \geq 0, x=4\) 
axis of rotation: xaxis 
method: washerdisc 
Problem Statement 

area: \(y=\sqrt{x}, y \geq 0, x=4\) 
axis of rotation: xaxis 
method: washerdisc 
Final Answer 

Problem Statement 

area: \(y=\sqrt{x}, y \geq 0, x=4\) 
axis of rotation: xaxis 
method: washerdisc 
Solution 

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

Final Answer 

\(8\pi\) 
close solution

area: \(f(x)=2\sin(x)\), \(x=0\), \(x=2\pi\), \(y=0\) 
axis of rotation: xaxis 
method: washerdisc 
Problem Statement 

area: \(f(x)=2\sin(x)\), \(x=0\), \(x=2\pi\), \(y=0\) 
axis of rotation: xaxis 
method: washerdisc 
Final Answer 

Problem Statement 

area: \(f(x)=2\sin(x)\), \(x=0\), \(x=2\pi\), \(y=0\) 
axis of rotation: xaxis 
method: washerdisc 
Solution 

video by Dr Chris Tisdell 

Final Answer 

\(V = 9\pi^2\) 
close solution

area: \(y^2=x2, x=5\) in the first quadrant 
axis of rotation: xaxis 
method: washerdisc 
Problem Statement 

area: \(y^2=x2, x=5\) in the first quadrant 
axis of rotation: xaxis 
method: washerdisc 
Final Answer 

Problem Statement 

area: \(y^2=x2, x=5\) in the first quadrant 
axis of rotation: xaxis 
method: washerdisc 
Solution 

video by Michel vanBiezen 

Final Answer 

\( 4.5\pi \) 
close solution

area: \( y=x^2, y=x \) 
axis of rotation: xaxis 
method: washerdisc 
Problem Statement 

area: \( y=x^2, y=x \) 
axis of rotation: xaxis 
method: washerdisc 
Final Answer 

Problem Statement 

area: \( y=x^2, y=x \) 
axis of rotation: xaxis 
method: washerdisc 
Solution 

video by Michel vanBiezen 

Final Answer 

\( 2\pi/15 \) 
close solution

area: \(y=\sqrt{9x^2}\) in the first quadrant 
axis of rotation: xaxis 
method: washerdisc 
Problem Statement 

area: \(y=\sqrt{9x^2}\) in the first quadrant 
axis of rotation: xaxis 
method: washerdisc 
Final Answer 

Problem Statement 

area: \(y=\sqrt{9x^2}\) in the first quadrant 
axis of rotation: xaxis 
method: washerdisc 
Solution 

video by MIP4U 

Final Answer 

\( V=18\pi \) 
close solution

area: \(y=\sqrt{x}, y=x^2\) 
axis of rotation: xaxis 
method: washerdisc 
Problem Statement 

area: \(y=\sqrt{x}, y=x^2\) 
axis of rotation: xaxis 
method: washerdisc 
Solution 

video by Khan Academy 

close solution

area: \(y=\sqrt{x}, y\geq0, x=1\) 
axis of rotation: xaxis 
method: washerdisc 
Problem Statement 

area: \(y=\sqrt{x}, y\geq0, x=1\) 
axis of rotation: xaxis 
method: washerdisc 
Solution 

video by Khan Academy 

close solution

Problem Statement 

Derive the equation for the volume of a sphere of radius r using the washerdisc method.
Solution 

video by Khan Academy 

close solution

Looking back at the plots, you will notice that the axis of revolution is always a coordinate axis, either the xaxis or the yaxis. 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 xaxis rotation practice 
area: \(y=x^2, x=y^2\) 
axis of rotation: \(y=1\) 
method: washerdisc 
Problem Statement 

area: \(y=x^2, x=y^2\) 
axis of rotation: \(y=1\) 
method: washerdisc 
Final Answer 

Problem Statement 

area: \(y=x^2, x=y^2\) 
axis of rotation: \(y=1\) 
method: washerdisc 
Solution 

video by Krista King Math 

Final Answer 

\(\displaystyle{V=11\pi/30}\) 
close solution

area: \( y=x^3, y=x, x\geq0 \) 
axis of rotation: \(y=2\) 
method: washerdisc 
Problem Statement 

area: \( y=x^3, y=x, x\geq0 \) 
axis of rotation: \(y=2\) 
method: washerdisc 
Final Answer 

Problem Statement 

area: \( y=x^3, y=x, x\geq0 \) 
axis of rotation: \(y=2\) 
method: washerdisc 
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 \(\displaystyle{ \pi \left[ x^2 + \frac{x^3}{3}  x^4  \frac{x^7}{7} \right]_{0}^{1} }\)
video by PatrickJMT 

Final Answer 

\(V=25\pi/21\) 
close solution

area: \( y=x^3, y=x, x\geq0 \) 
axis of rotation: \(y=5\) 
method: washerdisc 
Problem Statement 

area: \( y=x^3, y=x, x\geq0 \) 
axis of rotation: \(y=5\) 
method: washerdisc 
Final Answer 

Problem Statement 

area: \( y=x^3, y=x, x\geq0 \) 
axis of rotation: \(y=5\) 
method: washerdisc 
Solution 

video by PatrickJMT 

Final Answer 

\(V=97\pi/42\) 
close solution

area: \( y=x^2, y=x \) 
axis of rotation: \(y=5\) 
method: washerdisc 
Problem Statement 

area: \( y=x^2, y=x \) 
axis of rotation: \(y=5\) 
method: washerdisc 
Final Answer 

Problem Statement 

area: \( y=x^2, y=x \) 
axis of rotation: \(y=5\) 
method: washerdisc 
Solution 

video by Michel vanBiezen 

Final Answer 

\( 23\pi/15 \) 
close solution

area: \( f(x)=x, g(x)=x^23x \) 
axis of rotation: \(y=5\) 
method: washerdisc 
Problem Statement 

area: \( f(x)=x, g(x)=x^23x \) 
axis of rotation: \(y=5\) 
method: washerdisc 
Hint 

Problem Statement 

area: \( f(x)=x, g(x)=x^23x \) 
axis of rotation: \(y=5\) 
method: washerdisc 
Final Answer 

Problem Statement 

area: \( f(x)=x, g(x)=x^23x \) 
axis of rotation: \(y=5\) 
method: washerdisc 
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 washerdisc 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 \( V = \pi\int_a^b{R^2r^2~dx} \). We chose this integral because we are told to use the washerdisc 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 xdirection.
3. Determine r and R  From the plot, let's start on the axis of revolution and move down to the xaxis 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 \(r=5y\). 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 \(r=5x\).
Determining the expression for R is a bit trickier. Starting on the axis of revolution, we move down to the xaxis 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 \(R=5y\). This looks the same as r but in this case, we are landing on the plot \(y=x^23x\). So \(R=5(x^23x) = x^2+3x+5\). 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
\(\displaystyle{ V = \pi \int_0^4{ (x^2+3x+5)^2  } }\) \(\displaystyle{ (5x)^2 ~ dx }\). Let's evaluate it.
\(\displaystyle{ V = \pi\int_0^4{ (x^2+3x+5)^2  (5x)^2 ~ dx } }\) 
\(\displaystyle{ V = \pi\int_0^4{ (x^43x^35x^23x^3+ } }\) \(9x^2+15x5x^2+15x+25) (2510x+x^2) ~dx \) 
\(\displaystyle{ V = \pi\int_0^4{ x^46x^32x^2+40x ~dx } }\) 
\(\displaystyle{ V = \pi\left[ \frac{x^5}{5}  \frac{6x^4}{4} \frac{2x^3}{3}+\frac{40x^2}{2} \right]_0^4 }\) 
\(\displaystyle{ V = \pi\left[ \frac{4^5}{5}  \frac{6(4)^4}{4} \frac{2(4)^3}{3}+\frac{40(4)^2}{2} \right]  0 }\) 
\(\displaystyle{ V = \pi\left[ \frac{1024}{5}  \frac{1536}{4} \frac{128}{3}+\frac{640}{2} \right] }\) 
\(\displaystyle{ V = \pi \frac{1472}{15} }\) 
Final Answer 

\(\displaystyle{ V = \frac{1472\pi}{15} }\) 
close solution

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

yaxis rotation practice 
area: \(x=2\sqrt{y}, x=0, y=9\) 
axis of rotation: yaxis 
method: washerdisc 
Problem Statement 

area: \(x=2\sqrt{y}, x=0, y=9\) 
axis of rotation: yaxis 
method: washerdisc 
Final Answer 

Problem Statement 

area: \(x=2\sqrt{y}, x=0, y=9\) 
axis of rotation: yaxis 
method: washerdisc 
Solution 

video by Krista King Math 

Final Answer 

\(V=162\pi\) 
close solution

area: \(y=1x^2, y=0\) 
axis of rotation: yaxis 
method: washerdisc 
Problem Statement 

area: \(y=1x^2, y=0\) 
axis of rotation: yaxis 
method: washerdisc 
Final Answer 

Problem Statement 

area: \(y=1x^2, y=0\) 
axis of rotation: yaxis 
method: washerdisc 
Solution 

video by Krista King Math 

Final Answer 

\(\displaystyle{V=\frac{\pi}{2}}\) 
close solution

area: \(y=x^2, y=5, x=0\) 
axis of rotation: yaxis 
method: washerdisc 
Problem Statement 

area: \(y=x^2, y=5, x=0\) 
axis of rotation: yaxis 
method: washerdisc 
Final Answer 

Problem Statement 

area: \(y=x^2, y=5, x=0\) 
axis of rotation: yaxis 
method: washerdisc 
Solution 

video by Michel vanBiezen 

Final Answer 

\( 25\pi/2 \) 
close solution

area: \( y=x^2, y=x \) 
axis of rotation: yaxis 
method: washerdisc 
Problem Statement 

area: \( y=x^2, y=x \) 
axis of rotation: yaxis 
method: washerdisc 
Final Answer 

Problem Statement 

area: \( y=x^2, y=x \) 
axis of rotation: yaxis 
method: washerdisc 
Solution 

video by Michel vanBiezen 

Final Answer 

\( \pi/6\) 
close solution

area: \( y=\sqrt{x} \) on \([0,4]\) 
axis of rotation: yaxis 
method: washerdisc 
Problem Statement 

area: \( y=\sqrt{x} \) on \([0,4]\) 
axis of rotation: yaxis 
method: washerdisc 
Final Answer 

Problem Statement 

area: \( y=\sqrt{x} \) on \([0,4]\) 
axis of rotation: yaxis 
method: washerdisc 
Solution 

video by MIP4U 

Final Answer 

\( V=32\pi/5 \) 
close solution

area: \( x^2+y^2=1, y \geq 1/2, x \geq 0 \) 
axis of rotation: yaxis 
method: washerdisc 
Problem Statement 

area: \( x^2+y^2=1, y \geq 1/2, x \geq 0 \) 
axis of rotation: yaxis 
method: washerdisc 
Final Answer 

Problem Statement 

area: \( x^2+y^2=1, y \geq 1/2, x \geq 0 \) 
axis of rotation: yaxis 
method: washerdisc 
Solution 

video by Michel vanBiezen 

Final Answer 

\( V=5\pi/24 \) 
close solution

area: \( y=3x+6 \), xaxis, yaxis 
axis of rotation: yaxis 
method: washerdisc 
Problem Statement 

area: \( y=3x+6 \), xaxis, yaxis 
axis of rotation: yaxis 
method: washerdisc 
Final Answer 

Problem Statement 

area: \( y=3x+6 \), xaxis, yaxis 
axis of rotation: yaxis 
method: washerdisc 
Solution 

video by Michel vanBiezen 

Final Answer 

\( V=8\pi \) 
close solution

area: \(y=x^2, y=0, x=1\) 
axis of rotation: yaxis 
method: washerdisc 
Problem Statement 

area: \(y=x^2, y=0, x=1\) 
axis of rotation: yaxis 
method: washerdisc 
Solution 

video by Khan Academy 

close solution

parallel to yaxis rotation practice 
area: \(y=x^3, y=0, x=1\) 
axis of rotation: \(x=2\) 
method: washerdisc 
Problem Statement 

area: \(y=x^3, y=0, x=1\) 
axis of rotation: \(x=2\) 
method: washerdisc 
Final Answer 

Problem Statement 

area: \(y=x^3, y=0, x=1\) 
axis of rotation: \(x=2\) 
method: washerdisc 
Solution 

video by Krista King Math 

Final Answer 

\( V = 3\pi/5 \) 
close solution

area: \(y=2\sqrt{x}, y=0, x=4\) 
axis of rotation: \(x=5\) 
method: washerdisc 
Problem Statement 

area: \(y=2\sqrt{x}, y=0, x=4\) 
axis of rotation: \(x=5\) 
method: washerdisc 
Final Answer 

Problem Statement 

area: \(y=2\sqrt{x}, y=0, x=4\) 
axis of rotation: \(x=5\) 
method: washerdisc 
Solution 

video by MIP4U 

Final Answer 

\(V=832\pi/15\) 
close solution

area: \( x^2+y^2=1 \), xaxis, yaxis 
axis of rotation: \(x=2\) 
method: washerdisc 
Problem Statement 

area: \( x^2+y^2=1 \), xaxis, yaxis 
axis of rotation: \(x=2\) 
method: washerdisc 
Final Answer 

Problem Statement 

area: \( x^2+y^2=1 \), xaxis, yaxis 
axis of rotation: \(x=2\) 
method: washerdisc 
Solution 

video by Michel vanBiezen 

Final Answer 

\( V=(\pi/12)[3\pi4] \) 
close solution

You CAN Ace Calculus
related topics on other pages 

external links you may find helpful 
Larson Calculus (Google books) Volume  Disk, Washer Methods and Shell, Cylinder Methods pgs 458477 
Interactive Mathematics  Volume of Solid of Revolution by Integration (Disk method) 
Pauls Online Notes  Volumes of Solids of Revolution / Method of Rings 
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