Calculate a double integral over a region

Question

Calculate a double integral $\iint\limits_Gf(x;y)dxdy$ over a region $$ f(x;y)=e^{(x+y)^2},\ G=\{0\leqslant x\leqslant1, 0\leqslant y\leqslant1-x\} $$

I tried to use polar coordinates, but it didn't help much. So, it would be great if someone could give me some clue, and I would take it from there.

Answer 1

Using polar coordinates is not the best way.

Make the change of variable $u=x+y,v=x$. The integral becomes $\int_0^{1}\int_0^{u} e^{u^{2}} dvdu=\int_0^{1} ue^{u^{2}} du=\frac 1 2 e^{u^{2}}|_0^{1}=\frac 1 2 (e-1)$.

The conditions $0\leq x \leq 1,0\leq y \leq 1-x$ are equivalent to the condition $0\leq v \leq u \leq 1$.

Answer 2

Yes, you can use polar coordinates. Your integral becomes$$\int_0^{\pi/2}\int_0^{1/(r\cos(\theta)+r\sin(\theta))}re^{(r\cos(\theta)+r\sin(\theta))^2}\,\mathrm dr\,\mathrm d\theta.$$Now, use the fact that$$\int_0^{1/(r\cos(\theta)+r\sin(\theta))}re^{(r\cos(\theta)+r\sin(\theta))^2}\,\mathrm dr=\frac{e-1}{2 (\sin (\theta )+\cos (\theta ))^2}.$$