 Historically, the very first problem solved using calculus was to compute volumes. We'll proceed as follows. Given an n-dimensional object, we can cut it and describe an n-1-dimensional cross-section. For example, if you take a 1-dimensional line, you can cut it and its cross-sections will be 0-dimensional points. A 2-dimensional figure can be cut and its cross-sections will be 1-dimensional lines. If we cut a 3-dimensional object, we will get 2-dimensional cross-sections, a 4-dimensional object can be cut and its cross-sections will be 3-dimensional objects, and so on. And so these objects can be viewed as a sum of slightly thick cross-sections. For example, let's use a definite integral to find the volume of a cone with a base of 3 feet and a height of 6 feet, and since this is a new process, let's go ahead and verify that our answer is correct using a geometric formula. So let's take a look at our cone, and what we want to do is we want to take a slice through the cone. The cone is a 3-dimensional object, so its cross-sections will be 2-dimensional figures. What figures? We see that they are disks whose radius changes as we move from bottom to top. Now, if we imagine our disks as slightly thick so they are short cylinders, then the volume of the cone will be the sum of the volumes of the cylinders. So the volume of a cylinder will be pi times the radius squared times the height, where, for now at least, we'll leave our radius as r, and our height a tiny portion of the height, dh. So the volume of these cylinders will be pi r squared dh, and we'll sum these from the lowest height, h equals 0, to the highest height, h equals 6. Since our differential is dh, the only permitted variable in our integrand is going to be h. So let's check the individual factors. Pi is a constant, so we don't have to worry about that, but r, the radius, does change. For the cross-sections here at the bottom, the radius is large, but for cross-sections here at the top, the radius is smaller. So r is a variable, and we can't leave it in the integrand. We have to express it in terms of h. So let's change our viewpoint slightly. So if we throw down an x and y axis running through the base and the axis of the cone, we see that the side of the cone is actually on a straight line, and the radius of each of these cross-sections will correspond to the x value of a point on that line. So what's the equation of the line? What we see down here at the bottom, we have the radius 3, because that's the radius of the base when the height is 0. Then up here at the very top of the cone, the radius is 0 when the height is 6. And so that means this line passes through the points 0, 6, and 3, 0. So its equation will be h equals 6 minus 2r. And since I need r in terms of h, I'll solve this equation for r, and I could substitute that into my integral. And so in my integrand, the only variable is h, and we can proceed with the calculus. Here we'll do a u-substitution. We'll let u equals 6 minus h over 2. So du will be minus one-half dh. We'll manipulate our integrand so we get a minus one-half dh in it. This constant minus 2 can be moved to the front. We can do our u-substitution and our du-substitution. And we get an integral that we can evaluate. And putting everything backward belongs, we get our antiderivative. Evaluating the antiderivative gives us our final volume, 18 pi. I don't know about this. This is awfully complicated. Maybe this is fake news, and the volume isn't really 18 pi, that there's some sort of conspiracy to make us think that it's 18 pi. Well, let's check it out. We do have a volume formula for a cone. Volume equals one-third pi r squared h. So we can calculate the volume directly. We know the radius of the cone. We know the height of the cone, so we'll substitute those into our volume formula. And find that we do get 18 pi as our volume.