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## AP®︎/College Calculus BC

### Course: AP®︎/College Calculus BC > Unit 11

Lesson 3: AP Calculus BC 2011- 2011 Calculus BC free response #1a
- 2011 Calculus BC free response #1 (b & c)
- 2011 Calculus BC free response #1d
- 2011 Calculus BC free response #3a
- 2011 Calculus BC free response #3 (b & c)
- 2011 Calculus BC free response #6a
- 2011 Calculus BC free response #6b
- 2011 Calculus BC free response #6c
- 2011 Calculus BC free response #6d

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# 2011 Calculus BC free response #3 (b & c)

Volume of a solid of rotation and Chain Rule for rates of change. Created by Sal Khan.

## Want to join the conversation?

- At6:12he mentions that thinking of (dv/dk)(dk/dt)=(dv/dt) as fractions cancelling out isn't 'rigorous' (a notion he always repeats when doing this calculation. Is there a video in which he explains a more rigorous way to think of this?(1 vote)
- I dont think so. In general, it seems that he prefers to use intuitive, geometric arguments to justify mathematical results. If you want a more rigorous discussion, try out Chris Tisdell on Youtube.(3 votes)

- Could you also solve this problem through radian integration along the angle of the z-axis? As in, the integral from z-axis angle=0 to z-axis angle=2pi of R.

That would seem to have the benefit of applying generally to any solid of revolution, you just plug in R.(1 vote)

## Video transcript

We're on part b. The region R is rotated about
the x-axis to form a solid. Find the volume, V, of
the solid in terms of k. So this is our
region R, and it's going to be rotated
around the x-axis. It's going to be rotated around
the x-axis to form a solid. So that solid is going to
look something like this. It's going to look like this. This is going to be
this end of the solid when it's rotated
around, and it's going to look
something like this. Let me do my best attempt
to draw our solid. And then that side is going
to look something like that. It'll look a little
bit like a loudspeaker. And so let me shade
it in a little bit. So that's going to
be our solid and you can imagine that maybe this
right over here is our y-axis. That is our y-axis. And then going straight through
the center of the solid, you have your x-axis. If the solid was
slightly transparent, you would see the other side of
this opening right over here, and the x-axis goes
straight through the center. So how do we find the
volume of this solid? Well, we could think about
each of the individual disks of this solid right over here. So think about it this way. Let's take a disk
right over here. Let me take this
disk right over here so that you can imagine this
little bit of a cross section. It's a little bit of a
cross section of this solid. So this disk right
over here, it's going to have some area,
some surface area, and then it's going to have some depth. I'll draw the edge. You can imagine
the depth is kind of the edge of the quarter,
while the surface area is kind of the face of the
quarter, the area of the face of the quarter. So what's going to be the
volume of this white disk right over here? Well, it's going
to be the surface area of the disk
times the depth. So what is the surface area? Well, surface area of a circle
is pi times the radius squared. So it's going to be pi times--
and what's this radius? Well, the radius is the
height between the x-axis and the function. So this radius is
e to the 2x power. So the area of each
of these disks, for this x right over
here, for this value of x, is going to be pi times
e to the 2x squared. And then if we want the
volume of this entire disk, we then multiply the
area times the depth. So times dx. So this would be the volume
of this little disk over here that has a very small
infinitesimally wide depth. What we want to do is
sum all of these disks. We want to have a
bunch of these disks, that we want to
sum up all of them. So we want to take
an infinite sum of these infinitely thin disks. So essentially we're just going
to take the sum of all of these from x is equal to 0, all
the way to x is equal to k. Right, because this
right over here, this is x is equal to k,
this is x is equal to 0, we take the sum
of all the disks. We will get our actual volume. So we can actually evaluate
this integral analytically, it's not too bad. Because either--
let me rewrite it. Pi times-- e to the 2x squared
is e to the 4x, just 2x times 2. And then, if we want
to take the integral, you might be able to do
this just by inspection, but if you want
to say, hey look, I have a little
function here, I want its derivative sitting
here, what we can do is we can multiply
this expression by 4, and we can also divide it by 4. If you multiply something
by 4 and divide by 4, you haven't changed its value. You've multiplied it by 1. But what this does
is it gives us-- let me rewrite it a little bit. We can rewrite this as pi
over 4 times 4e to the 4x. I just rewrote this. I multiplied and divided by 4,
multiplied and divided by 4. And the reason why I
did that is so that we have the derivative of 4x
sitting right over here, we have this 4. Pi over 4 is a constant, so
we can actually take it out of the integral. And so we can say that this
expression right over here is the same as this
expression right over here, from pi over 4 times the
integral from 0 to k of 4e to the 4x dx. And so the whole reason why
I multiplied and divided by 4 is I have something here,
I have its derivative. So I can really just pretend
like this is an e to the x and you can take the integral,
or the anti-derivative, with respect to 4x. So the inside is e to the 4x. And you can verify. If you take the
derivative of e to the 4x, it is the derivative of 4x,
which is 4, times e to the 4x. So this right here is the
anti-derivative of that. And we're going to evaluate it. We are going to
evaluate it from 0 to k. All of that times pi over 4. And so when you evaluate
it at k, it's e to the 4k. So this is going to be pi
over 4 times e to the 4k minus e to the 4 times 0, which is
e to the 0, which is minus 1. So this is our volume, pi over
4 times e to the 4k minus 1. So that is part b. We did that. Now part c. I think we have time for this. The volume, V, found in part
b, changes as k changes. If dk, the derivative of k
with respect to t is 1/3, determine the derivative
of V with respect to t when k is equal to 1/2. So this is straight up
from the chain rule. So the chain rule will tell us--
and if you view differentials as just very, very
small numbers, it actually seems like a
little bit of common sense-- the rate of change of V,
so a very small change in V with respect to a
very small change in t is equal to a very small
change in V with respect to a very small change in k
times a very small change in k divided by a very
small change in t. Or the derivative of
V with respect to t is equal to the derivative
of V with respect to k times the derivative of k
with respect to t. And the reason why I
said this makes sense if you view a differential as
a small number is you would-- and that's not a very
rigorous way of doing it, but it's a good kind of common
sense way of just thinking about things-- is that
these guys would cancel out if these were
numbers, and then you would have dV dt on both sides. So that's why I said that
this is slightly common sense. You can kind of
view it that way. But this essentially gives us
everything we need to know. We need to solve for dV dt. They already gave us dk dt at
1/3 when k is equal to 1/2. So they're already
telling us the rate of change of k
with respect to t. We already know that this
thing right over here is going to be 1/3. And we can figure out the
derivative of V with respect to k very easily because we
have V as a function of k right over here. So let's figure that out. So the derivative of
V with respect to k is equal to pi over 4 times
the derivative in here. And the derivative of
this is just 4e to the 4k. The derivative of
negative 1 is just 0. So this is times 4e to
the 4k, is the derivative. These guys cancel out. So it's equal to pi
e to the 4k, that's the derivative of V
with respect to k. We want to find it
when k is equal to 1/2, so we could write
V prime at 1/2, or when k is equal to 1/2 dV dk. dV dk is going to be equal to
pi times e to the 4 times 1/2 which is equal to pi e squared. So this part right over
here is pi e squared. And so the derivative
of V with respect to t when k is equal to
1/2 and the derivative of k with respect to t is
1/3 is going to be just pi e squared times 1/3. So let me just
rewrite it over here. So dV dt is going to be
equal to 1/3 times this, or we could write
pi e squared over 3. And we're done.