Computing Double Integral using Degrees instead of Radians

Question

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?

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.