Can't Finish Double Integral in Polar or Cartesian Alright, so I'm stuck on what I think should be a simple double integral. It is $\int_0^1\int_{\sqrt x}^1e^{y^3} \, dy \, dx$. This is just the volume between the surface $z=e^{y^3}$ and the area bounded by $y=\sqrt x$ and $y=1$, $0\le x \le 1$.
I could see pretty quickly that I can't integrate this, so I tried to switch the $dy$ with $dx$ and change limits. When I do that, though, I end up with $\int_0^1[y^2e^{y^3}] \, dy$, which I still can't integrate.
So I try by converting to polar coordinates, and I get that $0\le\theta\le\frac{\pi}{2}$ and $\frac{2\cos\theta}{1-\cos 2\theta}\le r \le 1-\sin\theta$. That radius will be a major problem when I go to integrate, so I know I went wrong somewhere.
Does someone see what I did wrong?
Thanks in advance!

Change of order of integration we have
$$\int_{0}^{1}\int_{\sqrt{x}}^{1}e^{y^3}dydx$$ 
By changing order of integration, we get area bounded by $x=y^2$, $x=0$, $0\leq y\leq 1$
$$=\int_{0}^{1}\int_{0}^{y^2}e^{y^3}dxdy$$
$$=\int_{0}^{1}\left(\int_{0}^{y^2}e^{y^3}dx\right)dy$$ 
$$=\int_{0}^{1}e^{y^3}[x]_{0}^{y^2}dy$$ $$=\int_{0}^{1}e^{y^3}\left(y^2-0\right)dy$$  $$=\int_{0}^{1}y^2e^{y^3}dy$$ 
Let $y^3=t\implies 3y^2 dy=dt$ 
$$=\int_{0}^{1}e^{t}\frac{dt}{3}$$ $$=\frac{1}{3}\int_{0}^{1}e^{t}dt$$
 $$=\frac{1}{3}[e^t]_{0}^{1}$$ $$=\frac{1}{3}(e^{1}-e^{0})$$$$=\color{blue}{\frac{1}{3}(e-1)}$$
