Recall in supplemental notes 26 that I stated that there are two basic methods of calculating the volume of a solid. The first one was the disk method, and the second one is the method of cylindrical shells.

When you rotate a curve about the y-axis, you generate a solid. To estimate the volume of this solid, we can slice the base into many partitions. Each partition can be made into a rectangle with height f (x). If you rotate this rectangular partition about the y-axis, it will form a right circular cylinder or a cylindrical shell.

The volume of the shell swept out by the rectangle is D V = 2p * radius * shell height * thickness = 2p x f (x)dx. If I partition this interval up into an infinite number of partitions,

the result will be the volume of the solid. Again, this can become very tedious, so from the concept of a Riemann sum, we get the following for formula for finding the volume of a solid.

There are three steps in calculating the volume of the solid, and they are the following.

1.

Draw the region and identify the height of the shell.

2.

Set up the integral.

3.

Integrate to find the volume.

EXAMPLE 1:

Find the volume of the solid generated by revolving the region bounded by the curves

about the y-axis.

The bounds for integration are x = 0 and x = 4. Now, let us set up the integral for the volume.

To find the volume of this solid using the washer method, we must solve for x in terms of y.

Now, let us again look at the graph.

The outer radius, R (y), is x = 4, and the inner radius, r (y), is x = y ². Now let us solve for the limits of integration. 4 = y ² ® y = 2. Recall that one of the bounds was given in the original problem. Therefore, the limits of integration are y = 0 and y = 2. The integral for this volume is the following.

So which method works the best? Well, let us compare the two solutions. For the washer method, we had to solve for x in terms of y. In this case; it was not that hard, but there are cares where it is not worth it. If it is hard to solve for x in terms of y, then use the cylindrical shell method.

EXAMPLE 2:

Find the volume of the solid generated by revolving the region bounded by the curves

about the y-axis.

Let us look at the graph of this region.

To use cylindrical shell method to solve this problem, we will have to divide the region into two pieces. Region 1 will be the region over the interval 0 £ x £ 1/2 and shell height of y = 3. Region 2 will be the region over the interval 1/2 £ x £ 2. To find the shell height for this region, we must solve the equation for the curve for y in terms of x.

The total volume of this solid will be the sum of the volume of the solid generated by region 1 and the volume of the solid generated by region 2. Here is the integral for this volume.

Which method would be easier? When we did this problem in supplemental notes 26 (ex 4), we only had to use one region. The shell method required us to use two regions to calculate the volume. For one of the regions, we had to solve one of the equations for y in terms of x. It was not that hard, but it was one more step that we had to do to set up the integrals.

Let us do some examples where the region is revolved about the x-axis. The shell formula for revolution about the x-axis is the following.

EXAMPLE 3:

Find the volume of the solid generated by revolving the region bounded by the curves x = 2y - y ² and x = y about the x-axis.

The shell height for this problem is the outer curve - inner curve = 2y - y ² - y = y - y ².

We have to find the limits of integration for this region.

2y - y ² = y ® y - y ² = 0 ® y (1 - y) = 0 ® y = 0 or y = 1.

If I had wanted to find the volume of this solid with the washer method, then I would have had to solve x = 2y - y ² for y in terms of x. To do this, we would have had to complete the square. The final equation would have looked like the following.

Which one of the roots would we have to use? If we had chosen incorrectly, then our answer would be wrong. Therefore, the shell method was the best way to calculate this volume.

In deciding what method you need to use, you must look at two things. The first one is whether or not the equation can be easily solved for the other variable in terms of the first. The second is whether or not the region will have to be divided up into two or more regions. A graph of the region will help you in this process.

EXAMPLE 4:

Find the volume of the solid generated by revolving the region bounded by the curves x = 12(y ² - y ³) and x = 0 about (a) the x-axis, and (b) about the line y = 1.

I will use the shell method to solve both of these problems.

The shell height is x = 12(y ² - y ³), and the limits for integration are y = 0 and y = 1.

The shell height is x = 12(y ² - y ³), and the limits for integration are y = 0 and y = 1. The shell's radius is 1 - y because we are revolving about the line y = 1. The distance from the axis of rotation and y is 1 - y.