Question

ok Here is my answer: 5pi/2. the question is below; it's simpler that way.

Answer 1

Use the method of cylindrical shells to find the volume of the solid resulting from the region enclosed by the curves when revolved around the y-axis. I'm not sure if this is classified as calc or differential equations, so here it is.\[y=\sqrt[3]{x}\], y=0, and x=1.
the cylindrical shell method is: \[\int\limits_{a}^{b} 2\pi (shell radius)(shell height) dx\]

Answer 2

So first and always draw a picture

Answer 3

I did the work. I just want to know if I got the right answer. if I didn't, than I will go back through my work. I am simply looking for confirmation or negation of my answer.

Answer 4

ok, well, tell me what you used as your radius to start

Answer 5

1, since it's around the y axis and the shell ends at x=1

Answer 6

alright and your height?

Answer 7

the curve y= the cubed root of x to y=0. essentially the cubed root of x.

Answer 8

ok, now tell, me, on your radius, why didn't you subtract something to account for the change in radius due to the hole in the middle?

Answer 9

subtract what?

Answer 10

the inner radius or x or something?

Answer 11

i am not aware of anything to subtract. the only thing I could think of to do so would be x=0, since that's where the y axis is. nothing else was specified in my problem, so that is my assumption.

Answer 12

I guess, (I am not as familiar with shell as ring, which you could use to check) i am wondering, will the radius always be 1 for each shell? or will it slowly move inwards as the function moves outwards