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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?

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The substitution $x=r\cos\theta,\,y=\tfrac12r\sin\theta$ of Jacobian $r/2$ writes the integral as$$\int_{\pi/4}^{\pi/2}d\theta\int_0^\infty\tfrac12re^{-r^2}dr=\tfrac{\pi}{16}.$$

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  • $\begingroup$ Can i ask you how did you find angles $\pi/4$ and $\pi/2$ ? $\endgroup$
    – simon
    Jan 10, 2021 at 15:43
  • $\begingroup$ @simon Note $\tan\theta=2y/x\ge1$. $\endgroup$
    – J.G.
    Jan 10, 2021 at 15:46
  • $\begingroup$ Sorry but what does it have to do with angles of integration? i don't get it $\endgroup$
    – simon
    Jan 10, 2021 at 15:56
  • $\begingroup$ I also thought the same $\endgroup$
    – simon
    Jan 10, 2021 at 16:02
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    $\begingroup$ @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 $\tan \theta = 1$. $\endgroup$
    – Math Lover
    Jan 10, 2021 at 16:11

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