I'm not sure how to prove the first equality. It's intuitively obvious, but I don't know how to go about proving it

I've assumed the first equality to be true. This suggests that the region of integration is a square, so the integrals can be switched around pretty easily:

$\displaystyle \lim_{b \rightarrow \infty}\int_{-b}^b \int_{-b}^b e^{-x^2-y^2} dy \ dx$

$\displaystyle \lim_{b \rightarrow \infty} \int_{-b}^b \left[ -\frac{1}{2y}e^{-x^2-y^2} \right]_{-b}^b \ dx$

$\displaystyle \lim_{b \rightarrow \infty} \int_{-b}^b -\frac{1}{2b}e^{-x^2-b^2}-\frac{1}{2b}e^{-x^2-b^2} \ dx$

The problem with the step above is that it becomes:$\displaystyle \int_{-\infty}^{\infty} 0 \ dx$ which looks hopelessly wrong.

Can anyone point me in the right direction?