I need help with the last one, I only have 1 try left.

Question

anonymous · Answer

attachment

anonymous · Answer

I tried $$\pi \int\limits_{0}^{\sqrt{6}}(9-y^2)^2-6^2)dy = \frac{ 81\pi \sqrt{6} }{ 5 }$$ but still got it wrong

anonymous · Answer

oh my

anonymous · Answer

So the integral should be the same just with x = 9 right?

anonymous · Answer

Correct

anonymous · Answer

Maybe you made a mistake in the integral?

anonymous · Answer

Mhm your final solution looks correct. Did you put it exactly like that as a final answer?

anonymous · Answer

IS THAT... WEBASSIGN?! 
Have you tried going from zero to nine?

anonymous · Answer

You are probably right raiichul. I square root of x should not be sqrt(6) since x =9

anonymous · Answer

your upper limit of integration is wrong. It should be 36.
and your outer radius is also wrong. It should be (9 - 6)

anonymous · Answer

sourwing, why should the upper limit be 36?

anonymous · Answer

ahhhhhhhhhhhh ........... never mind. It's sqrt(6) xD

anonymous · Answer

Here is a similar example but it uses y=sqrt(x), y=0,x=3

anonymous · Answer

so the only thing that is wrong is the outer radius

anonymous · Answer

the line x = 6 for the practice example

accessdenied · Answer

|dw:1395551459283:dw|
drawing a picture just to illustrate the situation + how the item subtracted off has to be the equivalent of a cylinder of radius 3  (volume = pi r^2 h = pi 3^2 * sqrt(6) )