use polar coordinates to evaluate the following integral $\int^{a}_{-a} \int^{\sqrt{a^2-x^2}}_0 e^{-(x^2+y^2)}dydx$

i did in my paper was my limit are after converting it to polar coordinates $\int^{\pi}_0 \int^{{a}}_0 e^{-r^2}rdrd\theta$

Are the limits in the above integral true?

  • 2
    $\begingroup$ $x^2+ y^2= r^2$, not $r$. Your integral should have $e^{-r^2}$ rather than $e^{-r}$. Then use the substitution $u= r^2$. $\endgroup$
    – user247327
    Feb 21, 2019 at 12:22
  • $\begingroup$ Shouldnot the value be $x^2+y^2=a^2$? $\endgroup$ Feb 21, 2019 at 14:08

1 Answer 1


The domain is a half-disk of radius $a$, on the side of the positive $y$.

In Cartesian coordinates, $x$ runs from $-a$ to $a$ and $y$ is between $0$ and $\sqrt{a^2-x^2}$ (where $x^2+y^2=a^2$).

In polar coordinates, $\theta$ runs from $0$ to $\pi$ and $r$ from $0$ to $a$.

In both cases,

$$x^2+y^2\le a^2\land y\ge0.$$

Indeed $$x^2\le a^2\implies x^2+(a-x^2)\le a^2,$$


$$r^2\le a^2\land \theta\in[0,\pi]\implies\sin\theta\ge0.$$

  • $\begingroup$ yes thanks i exatly did it like this $\endgroup$
    – Hamza.S
    Feb 21, 2019 at 12:58

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