Main content

### Course: Calculus, all content (2017 edition) > Unit 8

Lesson 1: AP Calculus AB questions- Sal interviews the AP Calculus Lead at College Board
- 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
- 2011 Calculus AB free response #1a
- 2011 Calculus AB free response #2 (a & b)
- 2011 Calculus AB free response #2 (c & d)
- 2011 Calculus AB free response #3 (a & b)
- 2011 Calculus AB free response #3 (c)
- 2011 Calculus AB free response #4a
- 2011 Calculus AB free response #4b
- 2011 Calculus AB free response #4c
- 2011 Calculus AB free response #4d
- 2011 Calculus AB free response #5a
- 2011 Calculus AB free response #5b
- 2011 Calculus AB free response #5c
- 2011 Calculus AB free response #6a
- 2011 Calculus AB free response #6b
- 2011 Calculus AB free response #6c

© 2024 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# 2011 Calculus AB free response #3 (c)

Disk method to find the volume of a solid generated by rotation of a function. Created by Sal Khan.

## Want to join the conversation?

- Where does r = 1-f(x) come from?(10 votes)
- basically, the radius for that particular disk is the distance between y=1 and y=8x^3. this is effectively the difference = 1 - 8x^3. I'm pretty bad at explaining things, but i hope this helps!

Jafarr(11 votes)

- couldn't you use double integrals for this problem?(5 votes)
- @
**Carlton**: I don't see a possibility to use double integrals, but since its a volume we are looking for it might be possible. Usually double integrals are used, when a variable is defined in terms of two other variables. like in a 3-dimensional coordinate space. It's clearly easier with the rotating method.

@**Ruben**: It makes a difference whether you rotate around`y=0`

or`y=1`

. it's not so obvious with these two functions, but when you rotate the function`y=x^2`

, you can see the difference.(2 votes)

- Couldn't he have just done washer method(3 votes)
- Yeah, he could. But keep in mind what Sal likes to operate on: the intuition as compared to the math. The way he explains it is the same as the washer method, just operated without the explicit formula.(4 votes)

- couldn't you just use the washer method?(3 votes)
- Yes. Sal tends to champion the common sense approach which uses less memorization.(4 votes)

- 3:90

Does it matter whether the exponent (2), is in the parentheses or outside the parentheses?(2 votes)- YES! Overwhelmingly, yes.

When you do this, you'll use [R(x)^2 - r(x)^2]. If you put it like R(x^2), you're subbing in every x for x^2, which is wrong; if you write [R(x) - r(x)]^2, you'll actually have [R(x)^2 - 2R(x)r(x) + r(x)^2], which is not proper either. You must square each function individually!(3 votes)

- At5:25, why is Vf not subtracted from Vg?(2 votes)
- Because it's revolving around Y=1, so it's like the functions are flipped, with g being below f(2 votes)

- What would it look like if we wanted to use the shell method? Would you considered it to be easier than the washer method or not?(2 votes)
- For this problem, the washer method is much simpler. To evaluate the problem using the shell method integrate the following expression:

~ 2pi * ( 1 - y ) * ( y^( 1/3 ) / 2 - arcsin( y ) / pi ) dy

Evaluated from 0 to 1(2 votes)

- when it rotates its still being the same value? :D just saying i mean the area of R(2 votes)
- the area is the same, but when you rotate it you solve for volume(2 votes)

- Why wouldn't it just be π(ƒ(x)-g(x))^2dx(2 votes)
- You have to remember you're rotating around y=1 (which is where the 1-f(x) comes from) and this complicates things(2 votes)

- At3:22, why did he have to antidifferentiate in order to find the volume of all of the disks? He didn't use an antiderivative for the first disk, so why not just use a normal summation?(2 votes)
- He couldn't use a normal summation because there are infinite disks. Minor note: He used a definite integral, not an indefinite integral (antiderivative).(1 vote)

## Video transcript

Part C, write but do not evaluate an integral expression for the volume of the solid generated when R - - so that's this region right over here - when R is rotated about the horizontal line y=1. So y=1 is right over there. So, the way I like to think about it - let's think about the volume if I were to just take the bottom function - if I were to just take f(x) - if I were to just take f(x) - and if I were to rotate that function around y=x - if I were to rotate this thing around - sorry, around y=1 - what would the volume of that be, and then I'm going to subtract from that the volume if I were to take the top function - if I were to take g(x) and rotate it around. So let's first of all think about what volume I would get - and this is really the disk method, and I go into it in much more detail earlier in the calculus playlist, but let's think about the volume if f(x) is rotated around that axis. And to do that, let's imagine each of - each sliver of that volume. So this is - let me draw a little thing right over here - and you could imagine once this little silver is rotated it forms the bo- it forms the or you should - you could imagine - you could imagine - this length is the radius of a disk. And so - and just to imagine that let me draw the entire disk. So if this is rotated around - if this is rotated around it will become a disk - it will become a disk - it will become a disk that looks something like that, and I'll just call the depth of the disk - so the disk if you imagine it a coin this is kind of the side of the coin, the depth of the coin - the depth of the coin right over there. I know I can draw that better than that. So the depth of the coin is just like that. It's a fixed depth, and I'm going to call that dx - so it's just this distance right over here it is the dx - and what is going to be the area of that coin? Well the area of the surface of this is coin - so let me - let me do it like this... in a different color. So I want to do it in blue - the area of this coin is just pi times the radius of that coin, squared. And what is the radius of the coin? Well, the radius of the coin is this height - is this height right over here. And what is that height? Well it is 1-f(x). So that is equal to the radius. So the area - the surf- the area - the kind of the face of this coin is going to be pi - the area of the face of this coin is going to be pi times the radius squared, which is equal to pi times one minus f of x - one minus f of x squared - that's this blue area right over here, and if I want to find the volume of this coin I would multiply it by - by the depth of the coin. So times - so times dx. And if I wanted to find the volume of this entire solid - this entire rotated, i would want to find the sum of all of these volumes. So this is just the disk right over here but I can have another disk, a similar disk that I do right over here, I could have another disk right over here, and I want to take the sum of ALL of those disks. So I want to take - I want to take - so the volume is going to be the sum over all of those disks - so x goes from 0 - which is this bounding point - to x is equal to 1/2, times pi(1-f(x))^2 - this is the area of e- the fa- the area of the face of each of those disks and then I multiply time the depth of each of those disks - no this gives the volume of each of those disks, and I'm taking the sum of all of them. So this is the volume - this is the volume if I were to just rotate f(x) around y=1 - actually I should just write dx here - and so this right over here - this expression - I just did that so they really are equal - this is obviously just the volume of each of those disks. So this is the volume if I were to take f(x) around y=1. Let's figure out - so let me call this volume of f(x) - and by the same logic - the same exact logic - we can figure out the volume if we take g(x) - if we rotate g(x) around - if we rotate g(x) - if we construct disks like this and rotate them around y=1. And so the volume - if I take g(x) around y=1 would be 0 to 1/2 times pi(1-g(x) - cause 1-g(x) is each of these radius' right over here - each of these radius' - that squared, dx. And so the volume of what they're asking us - the volume of the solid generated when R is rotated - well R is kind of the space in between f(x) and g(x) - so so what's going to be - is g- the volume is going to be the difference between these volumes - it's going to be this volume - this is kind of the outer volume, and we're going to take out its hollow core - we're going to hollow it out be subtracting this volume. So the volume of that region is going to be the integral - I'll do this in a new color - integral from 0 to 1/2 of pi(1-f(x))^2 dx MINUS the integral from 0 to 1/2 of pi(1-g(x))^2 dx and this is completely valid answer but you might want to simplify it we have the same bounds of integration - we have the same variable of integration - and actually we have this pi over here so we could just factor that out - and so this is the same thing as pi times the integral from 0 to 1/2 of (1-f(x))^2 MINUS (1-g(x))^2 - and then all of that dx. And then - and actually you probably would want to do the - you probably would want to do this while taking the AP Exam, not just leaving it in terms of f(x) and g(x) - you would actually want to write the expression for what f(x)... f(x) and g(x) are. So really the best answer would probably be: pi times the integral from 0 to 1/2, times [1(1-... well f(x) is 8x to the third power - 8x to the third power, squared. Minus, (1-g(x)... and g(x) is sin of pi x. That squared - that squared times dx. And that would be our answer and you could see why they didn't want us to go through the trouble of evaluating it - they just wanted us to set up this integral.