It is well known that $$\int_{-\infty}^\infty{e^{-x^2}}dx=\sqrt{\pi}$$ This integral is usually computed by converting the integral to a double integral in cylindrical coordinates and then solving. However, I attempted to calculate this same integral using degrees instead of radians, and I got an entirely different answer: $$I=\int_{-\infty}^\infty{e^{-x^2}}dx$$ $$I^2=\int_{-\infty}^\infty{e^{-x^2}}{e^{-y^2}}dydx$$ $$I^2=\int_0^{360}\int_0^\infty{r{e^{-r^2}}}drd\theta$$ $$I^2=360\left(-\frac12\right)\left(e^{-r^2}\right)_0^\infty$$ $$I^2=-180(0-1)$$ $$I=\sqrt{180}$$ However, this is not the same result as I got originally. Since the area under the original function is independent of how we convert to polar coordinates (or in general independent of what we use as a measure of angle) the answer should be the same. Why isn't it? Did I make a mistake? Or are radians special for some reason?

  • $\begingroup$ I remember reading some phrases like "differentiation with respect to angles in degress does not make sense". So integrating also doesn't make sense. This may be helpful: en.wikipedia.org/wiki/Radian . See advantages of radians section $\endgroup$
    – user160738
    Jun 11, 2017 at 18:12
  • $\begingroup$ @user160738 but then my question still stands. Why are radians special? That is why should a circle have a measure of 2*pi units, and not 360, or even just one? $\endgroup$ Jun 11, 2017 at 18:16
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    $\begingroup$ @DreamConspiracy If $\hat \theta$ is in degrees, then the Jacobian is given by $$ \frac{\partial (x,y)}{\partial (r,\hat \theta)} = \frac {\pi}{180} r $$ instead of the usual $$ \frac{\partial (x,y)}{\partial (r,\theta)} = r $$ $\endgroup$ Jun 11, 2017 at 18:16
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    $\begingroup$ @DreamConspiracy for more on why radians are special, see this post $\endgroup$ Jun 11, 2017 at 18:20
  • $\begingroup$ Another way of looking at it has nothing to do with radians, per se, but the area of the region $r<\sqrt{x^2+y^2}<r+\Delta r$ is close to $2\pi r\Delta r$. So if you think of the double integral as being a "volume", then the value added to the volume by the circle of radius $r$ is $2\pi re^{-r^2}dr$. $\endgroup$ Jun 11, 2017 at 18:28

1 Answer 1


You can use the degree to measure the angles, but this means that you are using a new variable $\alpha$ that is linked to $\theta$ (the variable that measure the angle in radians) by: $$ \alpha=\frac{180}{\pi}\theta $$

so $$ d \theta= \frac{\pi}{180}d \alpha $$

substitue this in the integral with the new limits in degrees and you have the right result.

  • $\begingroup$ Yes, I realized that the Jacobian of $r$ was no longer correct after omnomnomnom's, and this is another way of looking at that same issue. Thanks for your response though. $\endgroup$ Jun 12, 2017 at 20:37

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