I need help solving this double integral:
$$\int_{0}^\infty\int_{0}^\infty e^{-(x+y)^2-3y^2} dxdy$$
The result is: $$\frac{\pi}{{6\sqrt{3}}}$$
I know that I need to substitute:
$$u = x+y$$ $$v = y$$
And then:
$$u = r \cos(\phi)$$ $$v = \frac{1}{\sqrt3}r\sin(\phi)$$
However, I am not entirely sure what my new limits are supposed to become after each substitution.
Under new substitution, we have $${0< v=y< \infty\\0<x=u-v<\infty}$$hence, the new bounds are $${v\in(0,\infty)\\u\in (v,\infty)}$$In terms of $r$ and $\phi$, we can write $$v>0\implies r\sin\phi>0\implies\sin\phi>0\implies 0<\phi<\pi$$and $$u>v\implies r\cos\phi>{1\over\sqrt 3}r\sin\phi\implies \tan\phi<\sqrt 3\implies 0<\phi<{\pi\over 3}$$therefore $$ {0<r<\infty\\0<\phi<{\pi\over 3}} $$