SUNY Geneseo Department of Mathematics

Volumes of Revolution by the Disk Method

Friday, December 11

Math 221 02
Fall 2020
Prof. Doug Baldwin

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Volumes of Revolution

From “Solids of Revolution” and “The Disk Method” in section 6.2 of the textbook, and this discussion of volumes of revolution.

Key Idea

Integrate f(x)² from the left end of the volume to the right end

(This is really volume by slices, with circular cross-sections produced by rotating a curve around an axis.)

Example: Trumpet

Imagine taking the graph of y = 1/x between x = 1 and x = 2, and rotating it around the x axis to create a shape that looks vaguely like a trumpet. What is the volume of this trumpet?

To see where the “key idea” above comes from, start with the area of a circular cross-section:

Graph of 1 over x squared rotated around x to make a trumpet shape with circular cross-sections

Now use that area in the volume by slices integral, i.e., the integral of area over the length of the solid:

Integrating pi over x squared from 1 to 2 yields pi over 2

Axis of Rotation

What if you take the graph of y = x² + 1 between x = -1 and x = 1 and rotate it around the x axis? What is the resulting volume?

Plug the function y = x² + 1 directly into the volume of revolution idea:

Rotate x squared plus 1 around x to form a pinched tube; integrate pi times x squared plus 1 squared to get volume,

Notice that we evaluated this integral in careful detail, in particular breaking it up into 3 integrals according to the sum rule for integrals. This is fine, although as you get more familiar with integration you might find that you just use the power rule on each term of the sum without explicitly referencing the sum rule:

Separating integral of 3 terms into 3 integrals equivalent to integrating sum all at once

Nothing says you can only rotate around the x axis. What if you take the graph of y = x² + 1 but rotate it around the y axis? What’s the volume of that volume of revolution, between y = 1 and y = 2?

This time we’ll be integrating over a range of y coordinates, and the radius of each circular cross-section will be the corresponding x value. So the first thing to do is solve for x in terms of y:

y equals x squared plus 1 means x equals square root of y minus 1

Now use the volume of revolution integral, integrating along the y axis and remembering to plug the formula for x into the πr² formula for the area of a circle:

Integrate pi times square root of y minus 1 squared from 1 to 2 yields pi over 2

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Use the rest of the semester to finish grading problem sets or any “redos” you want to do.

The last day of grading is Monday, December 21.

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