Cylindrical Shell Volume.
y^2=x ; x=2y; about the y-axis.
I don't know what I'm doing wrong.

Question

Cylindrical Shell Volume.
y^2=x ; x=2y; about the y-axis. 
I don't know what I'm doing wrong.

phi · Answer

what did you get so far ?

anonymous · Answer

I actually worked this one out. Sorry, I was using 2y as my inner radius instead out outer. After intergrating i ended up with (4y^3)/3 - y^5/5 inbetween 2 and 0.

anonymous · Answer

all times pi.

anonymous · Answer

I'm actually having a hard time with a new one. y=x^2; x=y^2, about y=1. @phi

abb0t · Answer

Did you find your bounds first? Set them equal to each other.

anonymous · Answer

1 and 0 ?

abb0t · Answer

Yes. Those are your bounds. Now, if you graph them, youmight be able to visualize the graph. I believe it's $\int\limits (outer) - (inner) dx$ or dy, if you're using y.

abb0t · Answer

Then, evaluate the integral from 0<x<1 

I chose x b/c I feel more comfortable with "x". Just a personal choice :)

anonymous · Answer

Alright so its from y=1, would I do (1+sqrt(y))^2 - (1+y^2)^2