1. ## Double integral area

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?

2. Originally Posted by Calc19Matt
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?
You can integrate $\displaystyle \int_0^1 \int_0^{y^2} \sin y^3 \, dx\,dy$ with respect to $\displaystyle x$ readily.

3. Am I doing it wrong because then I get Integral sin(y^3)y^2. how do i integrate that?

4. U-substitution: if $\displaystyle u=y^3$, then $\displaystyle du=3y^2\,dy$.

--Kevin C.