 Let's look at another example of finding the volume of solid by this technique of cross-sectional slicing. Let's find the volume of the solid, which we're going to call S, to describe in the following way. We're going to take the base of S to be a circle. Just a standard circle, its radius will be R. Again, nothing too fancy right there. What we're going to do is we're going to stack on top of this circle parallel cross-sections, which are perpendicular to the base, and they themselves are squares. We get a square that's stacked. Whoops, try that again. We take a square that's stacked here. That's a square. Ooh, that's the hideous drawing. Let me try that one more time. We take a square that looks like something like that. That looks a little better. We take another square, maybe right here. We have just a whole bunch of squares that are going to be stacked on top of this solid here. You can see that I'm having a hard time drawing a picture. What I'm going to do is actually go to a web-based app online that you can actually take a look at this solid right here. The link to this website you can find in the description of this video if you want to play around with yourself. What you see here is this solid where the bottom of the solid is a circle. It's a circle. We've stacked these squares on top of it. Look at the bottom left right here to see the skeleton of this object right here. As you spin this around, you can see the circle that's on the bottom. We're rotating around the center of that circle. There's all these different squares that are placed on the base of the circle. The base circle is going to determine how big the square is. The closer you are to the middle of the square is going to tell you how big the square is going to be bigger. When you're to the far edges, you're going to get smaller and smaller squares. This one right here can show you one by one. We have a bigger square, a bigger square, a bigger one, bigger until you get the middle. Then they're going to start shrinking as they get closer and closer to the edge. You get these corners that come up here. Then this picture on the top, you see it's colored and shaded, so we get a little bit more of a three-dimensional perspective to this thing. It's kind of an interesting little shape here. What would be the volume of shape like this? It turns out that using our technique of cross-sectional slicing, we can actually find the volume of this thing fairly nicely. Let's first think about the unit circle. We're going to put that in the XY plane. We have this, not the unit circle, but the circle of radius R. It's center is going to be at zero, zero, right? This equation of the circle is given by X squared plus Y squared equals R squared. That's going to be important to us. To find the volume using these cross-sectional slicing, we want to integrate our area function, A of X dx. The dx is the thickness of these prisms. The area of X is the area of the function. Now, because we have a square, I'm going to draw the square in front of you, the square, if we say that the side length is S, then our area function is just going to be S squared. Now, that's a good place to get started, but we have to do a little bit better than that because as we're integrating the respect to X, not S, we're going to have to somehow represent S as a function of X. We'll get back to that in just a moment. What I want to next think about, what are the bounds of integration? What are these values over here? Well, let's look at the circle. In terms of our X coordinates, our Xs are going to be anywhere from this side of the circle all the way to the other side of the circle. Now, this point on the right is R comma zero, and this point on the left is going to be negative R zero. Those are going to represent the bounds we're considering. X can range from negative R up to R. Now, we have to represent this value S in terms of the variable X. Thinking about our squares here, a typical cross-section would look something like this, where S is the length of this side right here. What we do know about this side length of the square is we do have this point right here, X comma Y, where if we solve the equation of the circle in terms of Y, we get Y equals the square root of R squared minus X squared. The side length S is going to equal two times this Y coordinate, and if we make this substitution in for Y, we can then square S and we get the following. We're going to get four times Y squared, so we squared the two, we squared the Y, and as Y squared is the square of the square root of R squared minus X squared, that's what we get in there, R squared minus X squared. We're going to make this substitution in for the area function right there. Then the volume is going to look like the integral from negative R to R. We had four times R squared minus X squared dx. Now we're going to use symmetry again here, because we're integrating from negative R to R, and this thing is symmetric, like if we come back to the picture here, right? This thing does have a nice symmetry along the X axis, which would be this value in the middle there. And so, using symmetry, we can actually write this integral as two times the integral from zero to R. I'm actually going to, so we have the four as well, I'm going to bring the four out. So we're going to get two times four, which is actually an eight, and that leaves behind an R squared minus X squared on the inside, which we can integrate that fairly simply. Now be aware that X is our variable, R is a constant, just an unspecified constant. So as we integrate, the anti-derivative R squared is going to be R squared X. Make sure you don't say R cubed over three. We're not taking the integral with respect to R, we're taking respect to X. On the other hand, the anti-derivative X squared will be an X cubed over three, and we plug in zero and R. And so, when you plug that in, you're going to get an R cubed minus R cubed over three, like here. Factory out the R cubed, you're going to get eight R cubed, and then you get one minus a third. One minus a third, of course, is two thirds, and so we see that the area is going to be 16 thirds R cubed. It seems somewhat fantastic to me that the answer does not have the number pi in it whatsoever, because after all, the shape was defined using some type of circle on the base, right? So you would think that the area of the circle is pi R squared, but be aware that the base is just a constraint that the volume is actually a string of squares. Squares having area S squared have nothing to do with pi. And so the final volume of this solid doesn't involve pi at all. It involves just the number 16 thirds times the radius of the circle cubed.