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### Course: AP®︎ Calculus AB (2017 edition)>Unit 11

Lesson 8: Volume: shell method (optional)

# Shell method for rotating around horizontal line

Find the volume of a solid of revolution by rotating around the x-axis using the shell method. Created by Sal Khan.

## Want to join the conversation?

• What does "solid of revolution" mean?
• A solid of revolution is a solid figure formed by rotating a two-dimensional figure around an axis. For example, you can create a sphere by rotating a half-circle around the line formed by its endpoints, or a cone by rotating a slanted line segment around a vertical line dropped from its higher endpoint to a point level with the line segment's lower endpoint. Many other figures can be rotated to produce other types of shapes. Donuts, anyone?
• when would you have to solve for y?
• Since in this video you are affecting the y, you have to solve for x. If you were affecting the x, you would have to solve for y.
• Wouldn't this problem just be simpler using disks in terms of x, from 0 to 8?
• In this case, you really could have done it easily either way. However, in some cases using the disk method is not always easy. For example, if we were rotating part of the graph y=(x-3)^2*(x-1) around the y-axis (Sal actually does this in the video titled Shell method for rotating around vertical line), it would require writing x as a function of y, which is not very easy to do in this case. Using the shell method allows us to use the function as it is in terms of x, making it a lot easier to find the volume.
• Why did you use y for the radius instead of y^3?
• The radius of the shape rotated around the y-axis is equal to the y, or the y value of the function. The y-value and radius both change going up or down the shape. Is this what you meant?
(1 vote)
• I don't understand why the height is equal to the function when revolving about the y-axis, but it's equal to some x-value minus the function when revolving about the x-axis.
• It only seems that way because of the examples used so far, which were bounded by the x axis, or, y=0.
What you are doing is making sure you are only integrating within the correct region.
Here is a similar solid as in the video, but rotated about the y axis, whose region of integration is bounded by y=8 above and y=x^3 below - which means the height is given by h = 8 - x^3. http://bajasound.com/khan/khan0010.jpg

Identifying the limits of integration is a very important skill to develop.
• Sal, when you evaluated the integral from 0-2 you only found the volume of the shape above the x-axis. To get the volume of the entire shape you should have multiplied that by two or took the integral from -2 - 2. So the actual volume should be 192pi/5. Right?
• nope the idea of shells is that you integrate circumferences*heights which covers both sides (since you get a full circle) from integrating just 1 side (you can see it from the picture that as you integrate outwards you integrate the top and bottom areas at the same time)
• Do we get the same answer if we solved with Disk method?
• Yes, either method you use to solve the volume problems will give you the same answer as long as you set up the integral correctly.
• Again randomly inserting a y-value, is it the small radius? I see no x=y function anywhere in this problem at all.
• You change the y equation to x=y by solving for x.
(1 vote)
• I didn't get how did he sum up all the cylinders inside. By doing so, you surpass the volume of the the intended shell.