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You CAN Ace Calculus

17calculus > integrals > volume integrals

Topics You Need To Understand For This Page

Calculus Main Topics

Tools

Related Topics and Links

Calculating Volume of Revolution

on this page: ► washer-disc method     ► cylinder-shell method     ► additional things to watch for

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 or the cylinder (shell) method. For other ways to calculate volume, see the links in the related topics panel.

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, 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, which is explained in this first panel.

Describing An Area In The xy-Plane

To describe an area in the xy-plane, the first step is to plot the boundaries and determine the actual region that needs to be described. There are several graphing utilities listed on the tools page. Our preference is to use the free program winplot ( used to plot these graphs; we used gimp to add labels and other graphics ). However, graphing by hand is usually the best and quickest way.

We use the graph to the right to facilitate this discussion. A common way to describe this area is the area bounded by \(f(x)\) (red line), \(g(x)\) (blue line) and \(x=a\) (black line).
[Remember that an equation like \(x=a\) can be interpreted two ways, either the point x whose value is a or the vertical line. You should be able to tell what is meant by the context.]

Okay, so we plotted the boundaries and shaded the area to be described. Now, we need to choose a direction to start, either vertically or horizontally. We will show both ways, starting with vertically, since it is more natural and what you are probably used to seeing. Also, this area is easier to describe vertically than horizontally (you will see why as you read on).

Vertically

Our first step is to draw a vertical arrow on the graph somewhere within the shaded area, like we have done here. Some books draw an example rectangle with the top on the upper graph and the bottom on the lower graph. That is the same idea as we have done with the arrow.

Now we need to think of this arrow as starting at the left boundary and sweeping across to the right boundary of the area. This sweeping action is important since it will sweep out the area. As we think about this sweeping, we need to think about where the arrow enters and leaves the shaded area. Let's look our example graph to demonstrate. Think about the arrow sweeping left to right. Notice that it always enters the area by crossing \(g(x)\), no matter where we draw it. Similarly, the arrow always exits the area by crossing \(f(x)\), no matter where we draw it. Do you see that?

But wait, how far to the right does it go? We are not given that information. What we need to do is find the x-value where the functions \(f(x)\) and \(g(x)\) intersect. You should be able to do that. We will call that point \((b,f(b))\). Also, we will call the left boundary \(x=a\). So now we have everything we need to describe this area. We give the final results below.

vertical arrow

\( g(x) \leq y \leq f(x) \)

arrow leaves through \(f(x)\) and enters through \(g(x)\)

\( a \leq x \leq b \)

arrow sweeps from left (\(x=a\)) to right (\(x=b\))

Horizontally

We can also describe this area horizontally (or using a horizontal arrow). We will assume that we can write the equations of \(f(x)\) and \(g(x)\) in terms of \(y\). ( This is not always possible, in which case we cannot describe the area in this way. ) For the sake of this discussion, we will call the corresponding equations \(f(x) \to F(y)\) and \(g(x) \to G(y)\).

Let's look at the graph. Notice we have drawn a horizontal arrow. Just like we did with the vertical arrow, we need to determine where the arrow enters and leaves the shaded area. In this case, the arrow sweeps from the bottom up. As it sweeps, we can see that it always crosses the vertical line \(x=a\). However, there is something strange going on at the point \((b,f(b))\). Notice that when the arrow is below \(f(b)\), the arrow exits through \(g(x)\) but when the arrow is above \(f(b)\), the arrow exits through \(f(x)\). This is a problem. To overcome this, we need to break the area into two parts at \(f(b)\).

Lower Section - - This section is described by the arrow leaving through \(g(x)\). So the arrow sweeps from \(g(a)\) to \(g(b)\).
Upper Section - - This section is described by the arrow leaving through \(f(x)\). The arrow sweeps from \(f(b)\) to \(f(a)\).
The total area is the combination of these two areas. The results are summarized below.

horizontal arrow

lower section

 

upper section

\( a \leq x \leq G(y) \)

arrow leaves through \(G(y)\) and enters through \(x=a\)

\( a \leq x \leq F(y) \)

arrow leaves through \(F(y)\) and enters through \(x=a\)

\( g(a) \leq y \leq g(b) \)

arrow sweeps from bottom (\(y=g(a)\)) to top (\(y=g(b)\))

\( f(b) \leq y \leq f(a) \)

arrow sweeps from bottom (\(y=f(b)\)) to top (\(y=f(a)\))

Type 1 and Type 2 Regions

Some instructors may describe regions in the plane as either Type 1 or Type 2 (you may see II instead of 2). As you know from the above discussion, some regions are better described vertically or horizontally. Type 1 regions are regions that are better described vertically, while Type 2 regions are better described horizontally. The example above was a Type 1 region.

Here is a quick video clip going into more detail on Type 1 and Type 2 regions.

Krista King Math - type I and type 2 regions

like? 19

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 two techniques we discuss on this page are the washer-disc method and the cylinder-shell 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.

washer-disc method

x-axis rotation

y-axis rotation

cylinder-shell method

x-axis rotation

y-axis rotation

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 or p and h (depending on your method; details below).

Once those steps are done, you are ready to set up your integral. Let's look at each technique separately.

Washer-Disc Method

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 of 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

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

x-axis rotation - washer-disc method

Practice 1

area: \(y=\sqrt{x-1}, y=0, x=5\)
axis: x-axis
method: washer-disc

answer

solution

Practice 2

area: \(y=x^2, x=y^2\)
axis: x-axis
method: washer-disc

answer

solution

Practice 3

area: \(y=x^3, y=x, x \geq 0\)
axis: x-axis
method: washer-disc

answer

solution

Practice 4

area: \(y=\sqrt{x}, y \geq 0, x=4\)
axis: x-axis
method: washer-disc

answer

solution

Practice 5

area: \(y=\sqrt{x}, y=x^2\)
axis: x-axis
method: washer-disc

solution

Practice 6

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

answer

solution

Practice 7

area: \(y=\sqrt{x}, y\geq0, x=1\)
axis: x-axis
method: washer-disc

solution

Practice 8

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

solution

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 - washer-disc method

Practice 9

area: \(y=x^2, x=y^2\)
axis: \(y=1\)
method: washer-disc

answer

solution

Practice 10

area: \(y=x^3, y=x, x\geq0\)
axis: \(y=-2\)
method: washer-disc

answer

solution

Practice 11

area: \(y=x^3, y=x, x\geq0\)
axis: \(y=5\)
method: washer-disc

answer

solution

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 - washer-disc method

Practice 12

area: \(x=2\sqrt{y}, x=0, y=9\)
axis: y-axis
method: washer-disc

answer

solution

Practice 13

area: \(y=1-x^2, y=0\)
axis: y-axis
method: washer-disc

answer

solution

Practice 14

area: \(y=x^2, y=0, x=1\)
axis: y-axis
method: washer-disc

solution

parallel to y-axis rotation - washer-disc method

Practice 15

area: \(y=x^3, y=0, x=1\)
axis: \(x=2\)
method: washer-disc

answer

solution

Practice 16

area: \(y=2\sqrt{x}, y=0, x=4\)
axis: \(x=5\)
method: washer-disc

answer

solution

Cylinder-Shell Method

We choose to use the term cylinder-shell method since it seems to bring to mind a cylinder more readily than the term shell, which in general usage could be almost any shape, and it covers the two most common references to this technique. This, of course, is a personal preference for this site and you need to check with your instructor to see what they require.

summary of the cylinder-shell method

the representative rectangle is parallel to axis of revolution

\(p\) is the distance from the axis of rotation to the representative rectangle

\(h\) is the height of the representative rectangle

x-axis rotation equation

\(\displaystyle{ V = 2\pi \int_{c}^{d}{ph~dy} }\)

y-axis rotation equation

\(\displaystyle{ V = 2\pi \int_{a}^{b}{ph~dx} }\)

Notes
1. Notice that \(p\) and \(h\) are distances, so they are always positive.
2. The distance \(p\) is from the axis of rotation to the representative rectangle. Where on the rectangle doesn't matter since the rectangle is infinitely thin. If it helps, you can think of the distance to the center of the rectangle.

cylinder-shell 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. He also works a simple example using this technique.

Khan Academy - Shell Method

x-axis rotation - cylinder-shell method

Additional Things To Watch For

It would be nice if, after learning these two techniques, that was all there was to it. However, as usual in mathematics, that is not the case. Here are a couple twists you will probably see in your homework and maybe even on an exam. We will just touch on them and how to work with them and allow you to fill in the details.

Twist 1. Multiple Integrals

Looking back at the plots used in the discussions, you will notice that no matter where we draw the representative rectangle, each end will always be on the same curve. That will not always be the case. Here is how you handle that situation.
Look back at the discussion on describing area in the xy-plane. When one end of the rectangle 'jumps' from one graph to another, you need to break the integral at that point. Then you will have more than one integral and you just add the result of each together.

Twist 2. Axis of Revolution Not A Coordinate Axis

We touched on this previously but let's look at a quick example that may help you understand this more deeply. We will just set up part of this problem and you can take it from there. The complete solution is found in the practice problems (coming soon).

Problem Statement - - Calculate the volume of revolution of the area defined by \(y=x^2-3x\) and \(y=x\) rotated about the line \(y=5\).
We have set up the plot shown on the right with all the pertinent data. Everything should look familiar to you except that we added an additional line from the x-axis to the rectangle and the new variable y, both in red. The trick here is to find the equation for p (assuming we use the cylinder-shell method, which is not the best way here). We know that the distance from the x-axis to the axis of rotation is always 5. So we build an equation using that information, \(y+p=5\) and solve for p, \(p=5-y\).
We deliberatedly chose y for that distance since the distance is parallel to the y-axis. This allows us to use the equation straight-away. If the distance had been parallel to the x-axis, we would have used x instead.
Take some time to think about this and then look as some examples and practice problems. You will be able to comprehend it soon.
If you decide to work this problem, notice that setting this up as we have shown will produce an integral that is difficult to solve. Try using the washer-disc method instead.

parallel to x-axis rotation - cylinder-shell method

Practice 17

area: \(y=x^3, y=0, x=1\)
axis: \(y=1\)
method: cylinder-shell

answer

solution

Practice 18

area: \(y=\sqrt{x}, y=x\)
axis: \(y=-3\)
method: cylinder-shell

solution

Practice 19 [ exam link ]

[20 points] For the region R bounded by the graphs of \(y=2x\), the x-axis and \(x=1\), consider the solid generated by revolving the region about the line \(y=3\). Sketch a different plot of region R for each method, clearly showing the region by shading, the axes of revolution, the representative rectangle, R, r, p and h. a. Set up the integral to calculate the volume using the disc-washer method. b. Set up the integral to calculate the volume using the shell-cylinder method. c. Calculate the volume of the solid by evaluating one of the integrals you set up in parts a and b.

answer

solution

Okay, now let's look at some problems with y-axis rotation. Here is the plot with all the information you need to work these problems.

cylinder-shell method with y-axis rotation

y-axis rotation - cylinder-shell method

Practice 20

area: \(y=\sin(x^2), y=0, x:[0,\sqrt{\pi}] \)
axis: y-axis
method: cylinder-shell

answer

solution

Practice 21

area: \(y=-x^2+x, y=0\)
axis: y-axis
method: cylinder-shell

solution

parallel to y-axis rotation - cylinder-shell method

Practice 22

area: \(y=4x-x^2, y=3\)
axis: \(x=1\)
method: cylinder-shell

answer

solution

Practice 23

area: \(y=\sqrt{x}, y=x\)
axis: \(x=-4\)
method: cylinder-shell

solution

Practice 24

area: \(y=-x^2+x, y=0\)
axis: \(x=-3\)
method: cylinder-shell

solution

Practice 25

area: \(y=x^2, x=1, x=2, y=4\)
axis: \(x=-2\)
method: cylinder-shell

solution

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