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## AP®︎ Calculus AB (2017 edition)

### Course: AP®︎ Calculus AB (2017 edition) > Unit 12

Lesson 2: AP Calculus AB 2015 free response- 2015 AP Calculus AB/BC 1ab
- 2015 AP Calculus AB/BC 1c
- 2015 AP Calculus AB/BC 1d
- 2015 AP Calculus AB 2a
- 2015 AP Calculus AB 2b
- 2015 AP Calculus 2c
- 2015 AP Calculus AB/BC 3a
- 2015 AP Calculus AB/BC 3b
- 2015 AP Calculus AB/BC 3cd
- 2015 AP Calculus AB/BC 4ab
- 2015 AP Calculus AB/BC 4cd
- 2015 AP Calculus AB 5a
- 2015 AP Calculus AB 5b
- 2015 AP Calculus AB 5c
- 2015 AP Calculus AB 5d
- 2015 AP Calculus AB 6a
- 2015 AP Calculus AB 6b
- 2015 AP Calculus AB 6c

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# 2015 AP Calculus AB 2b

Volume of solid.

## Want to join the conversation?

- For part b at6:08, Shouldn't the answer be multiplied by π for Volume?(1 vote)
- No, I don't think so because it is not volume with circle. Its a volume with a square.(3 votes)

- For part a, f is not always above g, should we have two integrals?(2 votes)
- You could use two integrals, where you would need to find the coordinate, where the two functions meet each other. An easier way to do this is, where you take the absolute value, which makes both areas positive, without you needing to split the area in to two integrals. But you could spilt it in to two integrals, it would just take some time.(3 votes)

- For part a, f is not always above g, should we have two integrals?(2 votes)
- Well the reason why the he did the absolute value was because the equation asked for a sum of positive values (i.e. area). To be more specific with these equations, yes.(3 votes)

- Can this be done without a calculator?(2 votes)
- Yes, it can. But the mathematics behind the functions with e^x and stuff like that is really time consuming, and on a timed test like the AP Calc test, it's better to use a calculator (hopefully, one like this won't show up on the non-calc portion!). You would have to plug the definite integral boundaries into the equation and solve.(2 votes)

## Video transcript

- [Voiceover] Part Two, B. Region S is the base of a
solid whose cross sections perpendicular to the x-axis are squares. Find the volume of the solid. Alright. 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. Now, you could try out
some values and think about how'd they behave, or you
could get a sense that, hey, like a fourth degree
quadratic's gonna have these ups and downs like that
while the, 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 upward trend also as x becomes more and more negative. Because this exponent is
going to become positive. So, you can think about 'em
in different ways like that. But what's important is region S. And there's a couple 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 two 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 inputed 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 'cause obviously the absolute value of
zero is going to be zero. But we've already inputed
this into our calculator. So, it's going to be useful. So, we can solve that thing we
inputed into our calculator. When does that equal zero? And actually, let's just do
that right from the get go. So, on our calculator, we
already inputed right over, let's make sure it's on. In our calculator, we already
inputed the absolute value 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
into where we see 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. Scroll down to the Solver, the Solver. And let's see, we want to redefine. Okay, so we want to see, so
we could go to the orginal screen of the Solver. And actually, I did this
in the previous problem, although we redefined y sub one. But you say, okay, when
does zero equal y sub one, or when does y sub one,
that's this here, equal zero? And so, you press Enter. And then, we can put in an initial guess. Let's look some place
in-between zero and two. So, I'll put in an initial guess of one. Now, 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 interval 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 gonna 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 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 dxs, you get that volume of each of those little sections. And you sum 'em all up, and
then, you are going to 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 interval from x equals a all the way,
so that's approximately, well, I'll just write a
over here to x equals two. Of this business, this is
the length of one side of the square. But we want to square it
'cause that's gonna give us the area of that cross section. And then, so 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 inputed
into my calculator. And, so, that's gonna give
the area of each of those cross sections
perpendicular to the x-axis. And you multiply it times d
x, you get the volume of that little section. And then, you add 'em all up 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 interval
or a function interval. Select that. And now, I'm not gonna do y sub one. I'm gonna do y sub one squared. So, let me go variables, y variables. It's a function, y sub one. And I am going to, let me 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 short cuts 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.