2.2 Determining Volumes by Slicing (p. 141 – 149)
  • 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:

  • Volumes (optional) due 5/12
  • Shells and washers (for now, look for volumes that can be calculated by the disk method) due 5/17

 


Motivation

Last class we talked about using integration to find areas of irregular 2-dimensional shapes. Now we want to use integration to find volumes of irregular 3-dimensional shapes. We’ll use a few different techniques for finding volumes and that will take us to the end of the course (finally!).

Background

One important picture from our lesson on Riemann sums is the representation of an irregular shape in the plane as the limit of more and more skinner and skinnier rectangles. I said this picture would be relevant for our volume discussion and now it is.

The picture I like to have in my mind is a bit imprecise, but it helps me keep track of what I’m doing. What’s an infinitely thin rectangle? Well, it’s only got a height and not a base, so it’s just a line segment. So think of an irregular shape in the plane as being made up of infinitely many line segments (probably of different heights) standing next to each other. In class we had vertical line segments standing side by side (integrating with respect to x), but we could have just as easily had horizontal line segments lying on top of each other (integrating with respect to y). The precise math word for this is “foliation,” but that’s outside the scope of this class.

[This would be a nice place for a picture!]

The punchline is that we’re adding up the lengths of a bunch of segments when we set up an integral to find an area. Area is a 2-dimensional measurement; length is a 1-dimensional  measurement. So we’re integrating a bunch of 1-dimensional things to get a 2-dimensional thing. (“Thing” is not a very precise word, but it’ll do for now.)

Volumes by cross-sectional areas

So what does this have to do with finding volumes of 3-dimensional shapes? Well, if you get a 2-dimensional thing (area) by integrating a 1-dimensional thing (length), what do you have to integrate to get a 3-dimensional thing (volume)? Well, that should be a 2-dimensional thing (area)!

I’m running late for something, so I’ll hit post now and add more later if I have time before tomorrow’s class. The previous paragraph is really the only remaining big idea for the course! We’ll just take it and turn it into something usable to calculate volumes of a few different special kinds of shapes.

….

I’m back!

Here’s a video (8.5 minutes) that explains a bit of the theory behind the formula for volume by cross-sectional areas and takes you through the calculations for two examples.

V = \int_a^b A(x)dx

Again, mathispower4u has videos showing a lot more examples. Search for “Determine Volume Of Solids by Slices.”

The disk method

What’s known as the “disk method” is really no different from finding volumes by integrating cross-sectional area, but we apply it to a special type of example so it gets its own name and formula.

The solids whose volumes we’re interested in here are known as “volumes of revolution.” These shapes are created by taking a graph in the x-y plane and rotating it around some line. Think about shawarma or al pastor on a rotating spit. In 3-dimensional space, when you take the cross-section of such a shape, you’ll get a disk (we call it a disk instead of a circle to emphasize that the inside is filled in). What’s the area of a disk? As usual, it’s pi r^2. The thing is, for most shapes, the radius of the cross section will be different as you move along the axis.

So. If you’re rotating the graph of a function y=f(x) on the interval [a, b] around the x-axis, your disks will be perpendicular to the x-axis. So. For every point on the x-axis you have a particular disk. We’ll call it “the disk at position x.” You need to know the radius of that disk to find its area. It helps if you draw some pictures. You’ll see that the radius of that disk at position x is just the height of the function at that point…that means that the radius is just f(x)! So the area A(x) is equal to \pi (f(x))^2.

So. The formula above, applied to this example, becomes

V = \int_a^b pi (f(x))^2 dx.

Of course we have some great videos from our favorite account:

  • This one (10 minutes) shows two examples with lots of pictures.
  • This one (5 minutes) shows another example with a quadratic function rotated about the x-axis.
  • This one (7 minutes) shows an example where the axis of rotation is a horizontal line that is not the x axis.

Next, we’ll talk about something called the “washer method.” This is not really its own method but just the disk method in disguise.