<s>[INST] Double integral - transformation I'm trying to calculate $$\iint_{\Omega } e^{(x+y^2)^{3/2}} \,\mathrm{d}A,$$ where $$\Omega =\{x,y>0 : x+y\leq 2\}. $$ 
Not sure where to go with it. I need to find a transformation and then calculate the integral. Any pointers?
 [/INST] With the substitution
\begin{align*}
x &= (r\cos\theta)^2 \\
y &= r\sin\theta 
\end{align*}
we find
\begin{align*}
\iint_{\Omega } e^{(x+y^2)^{3/2}} \,d A
&= \int_0^{\pi/2}\int_0^{R(\theta)} 2 e^{r^3}r^2\cos\theta \,dr\,d\theta \\
&= \frac{2}{3} \int_0^{\pi/2}(e^{R(\theta)^3}-1)\cos\theta\,d\theta,
\end{align*}
where
$$R(\theta) = \frac{-\sin\theta+\sqrt{8-7\sin^2\theta}}{2\cos^2\theta}.$$
($R(\theta)$ is found by solving $x(r,\theta)+y(r,\theta) = 2$ for $r$.)
Letting $\sin\theta = t$ we find
\begin{align*}
\iint_{\Omega } e^{(x+y^2)^{3/2}} \,\mathrm{d}A
&= \frac{2}{3} \int_0^1 \left(\exp\left[\left(\frac{-t+\sqrt{8-7t^2}}{2(1-t^2)}\right)^3\right] -1\right)dt \\
&= -\frac 2 3 + \frac{2}{3} \int_0^1 \exp\left(\frac{64}{(t+\sqrt{8-7t^2})^3}\right) dt.
\end{align*}
The final one-dimensional integral may be found numerically.
</s>