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# 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?
• 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
• couldn't you use double integrals for this problem?
• @ 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.
• Couldn't he have just done washer method
• 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.
• couldn't you just use the washer method?
• Yes. Sal tends to champion the common sense approach which uses less memorization.
• 3:90
Does it matter whether the exponent (2), is in the parentheses or outside the parentheses?
(1 vote)
• 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!
• At , why is Vf not subtracted from Vg?
(1 vote)
• Because it's revolving around Y=1, so it's like the functions are flipped, with g being below f
(1 vote)
• 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?
(1 vote)
• 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
(1 vote)
• when it rotates its still being the same value? :D just saying i mean the area of R
(1 vote)
• the area is the same, but when you rotate it you solve for volume
(1 vote)
• At , 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?
(1 vote)
• Why wouldn't it just be π(ƒ(x)-g(x))^2dx
(1 vote)
• You have to remember you're rotating around y=1 (which is where the 1-f(x) comes from) and this complicates things
(1 vote)