Evaluate Double Integral sin(y^3)dA where R is the region bounded by y=x^1/2, y=2, x= 0
I've set up the integral from x^1/2 to 2 and the other integral from 0 to 4.
I've also reversed the order and got 0 to y^2 and 0^2 and both times I end up having to integrate sin(y^3) which is a massive problem. What am I doing wrong with this problem?