# compute $\iint_De^{-x^2-4y^2} \ dxdy$

The question is: $$\iint_De^{-x^2-4y^2} \ dxdy, \quad D=\{(x,y):0\leq x\leq2y\}$$ This is how i've tried to attack this but i was getting nowhere with it : $$\int_{0}^{\infty}e^{-4y} \ dy \int_{0}^{2y}e^{-x^2}dx$$ Am i on the right path? how should i proceed

Any suggestion?

The substitution $$x=r\cos\theta,\,y=\tfrac12r\sin\theta$$ of Jacobian $$r/2$$ writes the integral as$$\int_{\pi/4}^{\pi/2}d\theta\int_0^\infty\tfrac12re^{-r^2}dr=\tfrac{\pi}{16}.$$
• Can i ask you how did you find angles $\pi/4$ and $\pi/2$ ? Jan 10, 2021 at 15:43
• @simon Note $\tan\theta=2y/x\ge1$.
• @simon as far as the limits, please note that your original integral is over the line $x = 2y$ and with the substitution it would mean $\tan \theta = 1$. Jan 10, 2021 at 16:11