I wanted to post the following answer to a question on double integral on Math.SE, but someone had submitted his work before I finished typing. As a result, I’m posting this on my personal blog.

Let $r = \sqrt{x^2+4y^2}$ and $t = \begin{cases} \tan^{-1}(2y/x) &\text{ if } x > 0 \ \pi/2 &\text{ if } x = 0. \end{cases}$ Then $\begin{cases} x &= r \cos t \ y &= (r \sin t)/2 \end{cases}$ and $D = { (r,t) \mid r \ge 0, t \in [\pi/4, \pi/2] }$. Calculate the Jacobian $$\begin{vmatrix} \dfrac{\partial x}{\partial r} & \dfrac{\partial x}{\partial t} \ \dfrac{\partial y}{\partial r} & \dfrac{\partial y}{\partial t} \end{vmatrix} = \begin{vmatrix} \cos t & -r \sin t \ (\sin t) / 2 & (r \cos t) / 2 \end{vmatrix} = r/2$$ $$\therefore \iint_D e^{-x^2-4y^2} , dxdy = \int_{\pi/4}^{\pi/2} dt \int_0^\infty e^{-r^2} \frac12 r , dr = \frac{\pi}{4} \cdot \frac12 \cdot \frac12 \cdot 1 = \frac{\pi}{16}$$