The region between the graphs of x=y^2 and x=3y is rotated around the line y=3. The volume of the resulting

The region between the graphs of x=y^2 and x=3y is rotated around the line y=3.
The volume of the resulting solid is ?

I found the equation integral from 0 to 3 of (3-(x/3))^2 - (3-x^(1/2))^2 = 8.28461 but that's not the right answer. What am I doing wrong?!

y1=x^(1/2)
y2=x/3

I'm assuming you were trying to solve this using "washers" (disks with a hole taken out of the middle)...

For the two radii I got r1= 3-(x/3) and r2=3-x^(1/2).... we're still on the same page here

Next, I found dV= pi ((r1^2)-(r2^2)) dx.... I think you forgot the pi in your equation (area of a circle = pi r^2)

So, V= pi (integral from 0 to 9) ((3-(x/3))^2 - (3-x^(1/2))^2) dx... Also, you took the integral from 0 to 3, it should be 0 to 9 because we are using dx and the graphs intersect when x=9
I got 42.4115025 for my answer.

Hope that helped!

Volume = π ∫ [(3-x/3)^2 - (3-sqrt(x))^2] dx --- integrate from 0 to 9
=27π/2

y goes from 0 to 3
x=3y, so x goes from 0 to 9