2015 AP Calculus AP 2b | AP Calculus AB solved exams | AP Calculus AB | Khan Academy

Part 2b: Region S is the base of a solid whose cross sections perpendicular to the x-axis are squares. Find the volume of the solid.

All right, so region S we see it right over here. In the last part, we already said that this function is f and this function is g. You could try out some values and think about how they behave; or you could, uh, get a sense that hey, look, a fourth degree quadratic is going to have these ups and downs like that. While the something, well, this is it has an x term, but then it has an exponential term right here that might have more of a kind of an upward trend, especially as x increases and then an upper trend also as x becomes more and more negative because this exponent is going to become positive.

So you can think about them in different ways like that, but what's important is region S. And there's a couple of things we have to think about. We have to think about the bounds of region S. We know that the upper bound of region S, x is equal to... we know x is equal to 2 here, but we don't know this value right over here. We don't know, let's just call that a; I'll put a question mark over there. We don't know what this x value is, but we do know that at that x value, the functions are going to be equal to each other.

So we know that f of a is going to be equal to g of a. Or another way, in the last section, we talked about the absolute value of f of x minus g of x, which we've already inputted into our calculator and saved it. We could also say that the absolute value of f of a minus g of a is equal to zero. And you don't need the absolute value there because obviously, the absolute value of zero is going to be zero. But we've already inputted this into our calculator so it's going to be useful so we can solve that thing.

We inputted into our calculator when does that equal zero. And actually, let's just do that right from the get-go. So in our calculator, we already input it right over... let’s make sure it’s on. In our calculator, we already inputted the absolute value of f of f of x minus g of x, and so we can use our solver now to figure out when does that equal zero. It actually equals zero just in the where we see it; it equals it one time, two times, and three times. So we want to find this place right over here; we want to figure out what x is or what a is.

So let's look at the math functions. Actually, let's clear this and actually let me quit that mode. And so let's go to math and scroll down to the solver. The solver. And let’s see, we want to redefine. Okay, so we want to see when does 0 equal y sub 1, or when does y sub 1—that's this here—equal 0. And so you press enter, and then we could put an initial guess. This looks someplace in between zero and two, so I'll put an initial guess of one, and let’s see what it gravitates to.

Oh whoops! And then I could do alpha solve. Alpha solve; let's let it munch on it a little bit. Let's let it munch, it's going... and there you go! It got to 1.03, which is the one we want. It didn't go to x equals zero or x equals two; it went to 1.03. So a is approximately equal to 1.03.

So a is approximately equal to 1.03. And so what integral do we take? We don't want just the area of S. We're saying that S is the base, and if we take a cross section, we're going to have squares. So this is going to be a base of a square. So this length right over here is going to be the absolute value of f of f of x minus g of x; that's the length of the base.

But if we want the area of that cross section, you can imagine—let me draw it at a little bit of an angle here—so you have a cross section of a square; there you have a cross section of a square here. And so the area of each of those cross sections is going to be the length of the base squared. And then you multiply that times the little dx's; you get that volume of that craw of each of those little sections and you sum them all up. And then you are going to get, you already get the volume of that figure.

So the volume of that figure is going to be equal to... so the volume is going to be equal to the integral from x equals a all the way... so that’s approximately, you know, approximate... well, I’ll just write a over here to x equals 2 of this business. This is the length of one side of the square but we want to square it because that's going to give us the area of that cross section.

And then let's do that. So times, I could say the absolute value of f of x minus g of x; I could just square that. Or if I want, I could just put parentheses there. I'll actually leave it as absolute value since that's what I've already inputted into my calculator. And so that's going to give the area of each of those cross sections perpendicular to the x-axis. And you multiply it times dx; you get the volume of that little section.

And then you add them all up from a equal from x equals a to x equals two. And luckily, we've already done a lot of the work here on our calculator. Our value of a is stored in the variable x already, so let me quit this and then... so let me go back to my math functions. Let me go to definite integral or function integral: select, select that. And now I'm not going to do y sub 1; I'm going to do y sub 1 squared.

So let me do variables: y variables, it's a function y sub 1, and I'm going to make sure I'm going to square it. That's what I do right over here. My variable of integration is x. I'm going to do it from x. The value of a is stored in the variable x in my calculator right now; it's a little confusing. This first x is a variable of integration; the next x is my lower bound of integration, and then my upper bound is two.

And so the most complex part here is some of the shortcuts I'm doing with the calculator, and we let the calculator munch on it, and it's approximately 1.283. So it's approximately 1.283 cubic units, and we are done.