 In order to find volumes using the definite integral, we have to consider how these volumes are generated. A solid of revolution is formed by taking a two-dimensional region and revolving it around a fixed line. The boundary of the figure forms the surface of the solid. For example, if a circle is revolved around its diameter, the solid of revolution is a sphere. But if a circle is revolved around an outside line, the solid of revolution is a torus. If a right triangle is revolved around one of its sides, the solid of revolution is a cone. If a parabolic region is revolved around the axis of the parabola, the resulting solid of revolution is a paraboloid. And in general, we can take any region and revolve it around any line to produce a solid of revolution. To find the volume of a solid of revolution, we identify a representative rectangle, revolve it to produce a representative solid, find the volume of the representative solid, and then sum the volumes through integration. We have two primary options. If we draw our representative rectangle perpendicular to, towards the axis of rotation, the representative solid is a short stubby cylinder, with the volume being pi times the radius squared times the height. Summing these volume elements gives us the volume using the short stubby cylinder method. Actually, nobody calls it the short stubby cylinder method, even though that's what it is. They invariably refer to this as the disc method. On the other hand, if we draw our representative rectangle parallel to along the axis of rotation, the representative solid is a wall. To find the volume of this wall, let's unwrap it so that it forms a slab. The length of this slab is the same as the circumference of our original circle, which will be 2 pi times some radius. The height will be some unknown amount, and the thickness is going to be some tiny portion of the radius. Summing these elements of volume gives us the volume using the wall method. And again, no one actually calls us the wall method. They call this the method of shells. So let's take an example. Suppose we want to find the volume of the solid of revolution formed by revolving the region between y equals 4 and y equals x squared around the y axis. We'll use the disc method to compute the volume and then verify using the shell method. So first of all, let's actually graph the region. Since we're revolving this around the y axis, the disc method will use representative rectangles drawn perpendicular to the y axis. Now we'll revolve our region and our representative rectangle around the y axis. And now let's take a look at our representative volume. It's a short stubby cylinder, a disc with radius equal to the x value, and height a tiny portion of the y axis. So the volume will be pi radius squared times thickness, pi x squared dy. And we'll want to sum these, and we might go from our lowest value has y equals 0 to our highest value, y equals 4. Because our differential variable is y, we need to make sure that everything is expressed in terms of y, which means I'll have to do something with this x squared. So replacing x squared with y, we now have an integral in terms of y only. We'll find the antiderivative, evaluate, and so the volume of this paraboloid will be 8 pi. To find the volume using walls, I mean to use shells, we'll draw a representative rectangle parallel to the axis of rotation. We're still rotating everything around the y axis, and that produces a wall, I mean shell, where the volume is the length times the height times the thickness. The length of the wall will be equal to the circumference of a circle with radius x, so that's 2 pi x. The height is going to be the distance from this top line to the curve itself for minus y. And the thickness of the wall is going to be this tiny portion of the x axis, which we designate as dx. So that gives us the volume of the shell, circumference times height times thickness, and we'll sum these from y equals 0 to y equals 4, and that allows us to express our volume as a definite integral. Since our differential is dx, the only permitted variable is x. Since y equals x squared, we can replace y with x squared. Our limits of integration also need to be changed, because they are currently in y. Our lower limit is y equals 0, and since y equals x squared, we can replace y with x squared and solve for x. 0. And our upper limit, y equals 4, tells us x equals plus or minus 2, but which one do we use? Because we graphed it, we know we should be using x equals 2. And this gives us a definite integral where everything is in terms of x. So let's integrate and evaluate, which gives us the volume using shells, 8 pi. And we see that this is the same volume that we got before, which is something of a relief because we don't want to get two different volumes for the same object. Now you might wonder, since both of these methods give us the volume, why do we actually need two methods? And the reason is this. If the only tool you have is a hammer, then every problem must be treated as a nail. And that's great if you're hacking a picture. Not so great if you're opening a jar of pickles. Some volume problems will be easier using the disk method, and some volume problems will be easier using the shell method. Having two methods allows us to choose between two different integrals, and we can do either to get the correct answer.