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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?

DreamConspiracy
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  • 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: https://en.wikipedia.org/wiki/Radian . See advantages of radians section – user160738 Jun 11 '17 at 18:12
  • @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? – DreamConspiracy Jun 11 '17 at 18:16
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    @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 $$ \frac{\partial (x,y)}{\partial (r,\theta)} = r $$ – Ben Grossmann Jun 11 '17 at 18:16
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    @DreamConspiracy for more on why radians are special, see this post – Ben Grossmann Jun 11 '17 at 18:20
  • 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 $2\pi re^{-r^2}dr$. – Thomas Andrews Jun 11 '17 at 18:28
  • @Omnomnomnom thanks for the link, even though I would argue that radians are practical rather than special. – DreamConspiracy Jun 12 '17 at 20:38
  • @DreamConspiracy not sure what that's supposed to mean. I would say that radians are the "natural"/"correct" choice of units for calculus as opposed to a mere convention that mathematicians have agreed upon (if that's what you meant). – Ben Grossmann Jun 12 '17 at 22:50
  • @Omnomnomnom I meant that rather than being fundamentally different, as "special" might imply, radians are just far more practical and easier to work with than any other measure of angle. But we seem to be on the same page. – DreamConspiracy Jun 12 '17 at 23:28

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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.

Emilio Novati
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  • 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. – DreamConspiracy Jun 12 '17 at 20:37