# Double Integrals -

1. Jan 30, 2006

### Nima

Hi, I have a question and it asks me to evaluate the volume generated by the area bounded by y = x^3 and y = x^(1/3) and the function z = x^2.y

I'm just having a few problems with setting up the ranges of my variables x and y. I drew a sketch of the area in the x-y plane but I'm not sure what my ranges should be. My guess is: 0 <= x <= 1 and 0 <= y <= [x^(1/3) - x^3].

Help appreciated.

2. Jan 30, 2006

### DMinishedCapacity

"Hi, I have a question and it asks me to evaluate the volume generated by the area bounded by y = x^3 and y = x^(1/3) and the function z = x^2.y"

To do dbl int's always choose one of the variables to put definite boundaries on (put them on the outer-most int), then put limits on the second int in terms of the first. You know that the graphs intersect at (0,0) and (1,1), so I'd suggest using 0 and 1 as the bounds of x, then choose y = x^3 and y = x^1/3 for the bounds on y. Then, integrate!

Last edited: Jan 30, 2006