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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?
• 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?
• Because it's revolving around Y=1, so it's like the functions are flipped, with g being below f
• 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?
• 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
• when it rotates its still being the same value? :D just saying i mean the area of R
• the area is the same, but when you rotate it you solve for volume