 Let's take another look at a volume of a solid of revolution. So let's find the volume of the solid of revolution formed when the region bounded by y equals square root log x and the x-axis over the interval between 1 and 2 is rotated around the x-axis. So first let's graph that region. Now let's try to find the volume using disks. So we'll draw a representative rectangle towards the axis of rotation, which is the x-axis, and then we'll revolve everything around the x-axis. And let's take a look at those disks. The radius of the disk will be our y-value. The height will be a tiny portion of the x-axis, so the volume will be pi y squared dx, which will sum from x equals 1 to x equals 2. And since our differential variable is x, we need to convert this y squared into an expression in x, and we get our integral. Unfortunately, we don't yet know how to find the antiderivative of log x. And while a little later on we'll see how we could do this integral, right now we're not able to. And this is why it's important to have a couple of different methods of approaching the same problem, because if the only tool you have is a hammer, every problem must be treated as a nail. So if the only method we had of finding volumes was using the disk method, then we'd have to evaluate this integral, and the fact that we might not know how to do this would not excuse us from having to do it. Fortunately, we do know a second method of finding volume. So once again we'll graph our region, but this time we'll draw a representative rectangle parallel to the axis of rotation. Now we'll rotate everything around the x-axis, and our representative rectangle becomes a representative wall where the length of the wall is equal to the circumference of a circle with radius y, or 2 pi y. The height of the wall is this length that goes from this x-value, whatever it is, up to this x-value, which is always equal to 2. So the height is going to be 2 minus the x-value, and the thickness of the wall is some tiny portion of the y-axis. So our representative volume will be 2 pi y times 2 minus x dy. And since we're going from x equals 1 to x equals 2, we'll sum these volumes from x equals 1 to x equals 2. Since our differential variable is y, we must express everything in terms of y. So we need to express x in terms of y. So from y equals square root log x, we find that x is equal to... We also need to convert our limits of integration. Our limit x equals 1 is going to be y equals 0, and if x equals 2, then y equals log 2, where we use the positive square roots because we graph the region. Making these substitutions gives us an integrand where the only variable is y. Before trying to evaluate the integral, let's do a little bit of cleanup. This constant factor of 2 pi can come out front, and we can expand the integrand, and we can further split the integral. We can evaluate the antiderivative separately. The antiderivative of 2y is... For the other antiderivative, we'll need to use a u-substitution. So we'll let u equals y squared, making the substitution, putting everything back, and we'll evaluate. y squared from y equals 0 to y equals log 2 is... 1 half e to y squared from y equals 0 to y equals log 2 is... which gives us our volume.