Volume with cross sections: intro | Applications of integration | AP Calculus AB | Khan Academy

You are likely already familiar with finding the area between curves, and in fact, if you're not, I encourage you to review that on Khan Academy. For example, we could find this yellow area using a definite integral.

But what we're going to do in this video is something even more interesting. We're going to find the volume of shapes where the base is defined in some way by the area between two curves.

In this video, we're going to think about a shape, and I'm going to draw it in three dimensions. So let me draw this over again but with a little bit of perspective.

So let's make this the y-axis, so that's the y-axis. This is my x-axis; that is my x-axis. This is the line y is equal to 6, right over there, y is equal to 6. This dotted line we could just draw it like this, and so this would be the point x equals 2.

Then the graph of y is equal to 4 times the natural log of 3 minus x would look something like this—look something like this. And so this region is this region, but it's going to be the base of a three-dimensional shape where any cross-section, if I were to take a section right over here, is going to be a square.

So, whatever this length is, we also go that much high. And so the cross section is a square right over there. The cross section right over here is going to be a square. Whatever the difference between these two functions is, that's also how high we are going to go.

This length, which is 6 at this point, is also going to be the height. It is going to be a square; it's going to be quite big. I have to scroll down so we can draw the whole thing roughly at the right proportion.

So it looks something like this; this should be a square. It's going to look something like this, and so the whole shape would look something like this—would look something like that. Try to shade that in a little bit so that you can appreciate it a little bit more, but hopefully, you get the idea.

Some of you might be excited, and some of you might be a little intimidated. Well, hey, I've been dealing with the two dimensions for so long; what's going on with these three dimensions? But you'll quickly appreciate that you already have the powers of integration to solve this.

To do that, we just have to break up the shape into a bunch of these. You could view them as these little square tiles that have some depth to them. So let's make that into a little tile; this one into it that also has some depth to it.

You could even—I could draw it multiple places—you could view this as breaking it up into these things that have a very small depth that we could call dx. And we know how to figure out what their volume is.

What would be the volume of one of these things? Well, it would be the depth times the area times the surface area of this cross-section right over here. Let me do that a different color.

So what would be the area that I am shading in, in pink, right over here? Well, that area is going to be the base length squared. What's the base length? What's the difference between these two functions? It is going to be 6 minus our bottom function, which is 4 times the natural log of 3 minus x.

And so that would just give us that length. But if we square it, we get this entire area. You square it, and then you multiply it times the depth; you multiply it times the depth. Now you have the volume of just this little section right over here, and I think you might see where this is going.

Now, what if you were to add up all of these from x equals zero to x equals two? Well, then you would have the volume of the entire thing. This is the power of the definite integral.

So we could just integrate from x equals zero to x equals two. From x equals zero to x equals two, if you drew where these intersect our base, you would say, “All right, this thing right over here would be this thing right over here,” where it's dx.

Instead of just multiplying dx times the difference between these functions, we're going to square the difference of these functions because we're visualizing this three-dimensional shape—the surface area of this three-dimensional shape— as opposed to just the height of this little rectangle.

If you were to evaluate this integral, you would indeed get the volume of this kind of pedestal horn-looking thing. This is not an easy definite integral to evaluate by hand, but we can actually use a calculator for that.

So we can hit math and then hit choice number nine for definite integral, and then we just have to input everything. We're going from 0 till 2 of, and then we have—let me open parentheses because I'm going to square everything—6 minus 4 times the natural log of x—or actually the natural log of 3 minus x.

And so let me close the parentheses on the natural log part, and then if I close the parentheses on this whole thing, I want to then square it, and then I'm integrating with respect to x.

Enter. I got approximately 26.27. So, approximately 26.27, and this is a volume. Here, if we thought about units, it would be in our units cubed or cubic units.