2.2 Determining Volumes by Slicing (part 2)

P. 150: 58, 59, 74 – 80 all, 98 – 102 all
Find the volume of the solid obtained by rotating the region bounded by the curves y = x^2, y = 12-x, x = 0 and x ≥ 0 about (a) the x–axis; (b) the line y = -2; (c) the line y = 15; (d) the y-axis; (e) the line x = -5; (f) the line x = 7.

Webwork: Shells and Washers due 5/17

Lecture notes: https://www.dropbox.com/s/lugsw916xjk5j3l/Note%20May%2011%2C%202020.pdf?dl=0

Back to volumes by slicing

Last week we saw that the area of a 2-dimensional region can be calculated by integrating lengths of 1-dimensional segments: $A = \int_a^b f(x) dx$.

We also saw that the volume of a 3-dimensional shape can be calculated by integrating areas of 2-dimensional cross sections: $V = \int_a^b A(x)dx$.

Back to the disk method

We applied this volume formula to the special case of a solid of revolution, which given by rotating the graph of a function $y=f(x)$ around the $x$-axis. In this case, the shapes of the 2-dimensional cross sections are all disks, and the radii are given by $f(x)$. So $A(x)$ becomes $\pi (f(x))^2$ and the volume formula above becomes $V = \int_a^b \pi (f(x))^2dx$.

Remember how we figured out how the cross section was a disk: we fixed a position “$x$” and looked at the vertical segment connecting the $x$ axis to the graph $y=f(x)$. This segment is perpendicular to the axis of rotation, which we said was the $x$-axis. When you spin this segment around the $x$-axis in 3-dimensional space, it sweeps out a disk.

Washer method

The next special case of the volume formula is another solid of revolution; this is called the washer method but it’s really just the disk method in disguise. A washer is a disk with a concentric disk inside it removed. (This is like the washers that go with screws, not the washers that go with driers.)

When would you use the washer method instead of the disk method? When the cross-section is a washer instead of a disk. Fix a position “$x$” again and look at the vertical segment connecting the $x$ axis to the graph $y=f(x)$. If one end of that segment isn’t the axis of rotation, then rotating that segment around the axis of rotation will sweep out a washer instead of a disk.

Scroll down to Figure 2.2 here.

But if the cross section is a disk with a disk removed, we can use the disk method for both those disks.

For example, assume you are rotating the region between the graphs $y=f(x)$ and $y=g(x)$, between $x=a$ and $x=b$ around the $x$-axis. Assume for now that $0 \leq f(x) \leq g(x)$. Then the 3-dimensional shape you get is the one you would get if you just rotated $y=g(x)$ around the $x$-axis, minus the one you would get if you rotated $y=f(x)$ around the $x$-axis.

So applying the disk method once to $y=g(x)$ and one to $y=f(x)$ would give us the volume formula:

$V = \int_a^b \pi (g(x))^2 dx – \int_a^b \pi (f(x))^2 dx$

but we can simplify the right hand side to get the formula that’s usually referred to as the washer method:

$V = \int_a^b \pi \left( (g(x))^2 – (f(x))^2 \right) dx$.

  • This video (8 minutes) has an explanation of the washer method along with an example where the axis of rotation is the $x$ axis.
  • This video (7 minutes) shows an example where the axis of rotation is a horizontal line that’s not the $x$-axis.
  • This video (7 minutes) shows an example where the axis of rotation is the $y$-axis.