Double integral of $e^{3+y^2}$ over a triangle

Question

Evaluate $\iint_{A}^{} e^{3+y^2}dxdy$ where $A$ is a triangle with vertices $(0,0)$, $(0,-1)$ and $(1,-1)$.

I don't know how to bite that. I tried multiplying it by $e^{x^2}$ and then changing the coordinates to polar, but I was able to compute only one boundary for $\phi$ by writing following inequalities:

$-1\leq y\leq -x \Rightarrow -1\leq r\sin\phi \leq -cos\phi$

and then

$r\sin\phi \leq -r\cos\phi \Rightarrow \tan\phi \leq -1 \Rightarrow \phi\leq -\frac{\pi}{4}$

I don't know how to find the second boundary and how to find boundaries for $r$, but I think it would simply be $0$ and $\sqrt2$ because that's the length of the longest side of the triangle.

Any hints will be greatly appreciated.

Answer 1

You don't need to change variables. If you integrate with respect to $x$ first, the upper and lower limits will be $y$ and $0$ respectively. That will give you a $y e^{3+y^2}$ when you next integrate with respect to $y$. You can then use a substitution.