Find the volume formed by rotating the region enclosed by:

Question

x=2y and y^3=x with y greater than or equal to 0 about the y-axis

Answer 1

There are two regions between the curves, from (0,0) to (2 sqrt 2, sqrt 2) and the same in the third quadrant,
So do the integral in quadrant 1 and double.
We could do little vertical cylinders and integrate over x or horizontal rings and integrate over y.
For the rings;
integral of
dy pi (x outer^2 - x inner^2) from y =0 to y = sqrt 2
where x outer = 2y and x inner = y^3

pi dy (4 y^2 -y^6)
then
pi [ (4/3)y^3 -(1/7)y^7 ]
pi [ (4/3) 2^(3/2) - (1/7) 2^(7/2) ]
check my numbers !!!

Answer 2

I often found that students were more comfortable and familiar with rotations around the x-axis.
Notice we could just switch the x and y variables, and then rotate around the x-axis.
All shapes and volumes would be retained.

Then volume
= 2pi[integral] (4x^2 - x^6)dx from 0 to √2

the actual work and calculations would be same as Damon showed you.
Just remember to double Damon's final answer because we have two identical solids.