$\displaystyle \displaystyle{ \hdots=\int_0^\infty\int_0^\infty e^{-(u^2+v^2)}du dv }$

We write $\displaystyle u=r\cos(\theta)$ and $\displaystyle v=r\sin(\theta)$ which simplifies $\displaystyle \displaystyle{ e^{-(u^2+v^2)} }$ into $\displaystyle \displaystyle{ e^{-r^2} }$,

**but why is differential area now** $\displaystyle \mathbf{r} dr d\theta$?

$\displaystyle \frac{du}{dr}=\cos(\theta)$ and $\displaystyle \frac{dv}{d\theta}=r\cos(\theta) \Rightarrow dudv = r dr d\theta\cos^2(\theta)$

Also on a sidenote, if I rewrite $\displaystyle r$ as $\displaystyle r=\frac{u}{\cos(\theta)}=\frac{v}{\sin(\theta)}\Ri ghtarrow v =u\tan(\theta) \Rightarrow \theta=\tan^{-1}(\frac{v}{u})\stackrel{u,v\to\infty}{=}\tan^{-1}(\frac{\infty}{\infty})\stackrel{?}{=}\frac{\pi} {2}$.

(I know that $\displaystyle \tan^{-1}(\infty)=\frac{\pi}{2}$.)