= Polar-coordinate proof of the Gaussian integral
{title2=$I(1)^2=2\pi\int_0^\infty re^{-r^2}\,dr=\pi$}
Put $I=I(1)$. The nonnegative <integrand> allows <Tonelli's theorem> to write the square as a <double integral>:
$$
I^2=\int_{\mathbb R^2}e^{-(x^2+y^2)}\,dx\,dy.
$$
In <polar coordinates>, $x=r\cos\theta$, $y=r\sin\theta$, and the <Jacobian determinant> supplies the area factor $r$:
$$
I^2=\int_0^{2\pi}\int_0^\infty e^{-r^2}r\,dr\,d\theta
=2\pi\left[-\frac12e^{-r^2}\right]_0^\infty
=\pi.
$$
Since $I>0$, take the positive <square root> to get $I=\sqrt\pi$. The computation proves finiteness as well as the value. For an explicit limiting argument, first integrate over disks of radius $R$; their integrals are $\pi(1-e^{-R^2})$, which increase to $\pi$ as $R\to\infty$. https://openstax.org/books/calculus-volume-3/pages/5-3-double-integrals-in-polar-coordinates[Strang and Herman, Calculus Volume 3, §5.3: the polar area element]{ref}
\Image[/gaussian-integral-polar-annulus.png]
{title=Gaussian height times the circumference of a ring gives its radial contribution}
{description=Left: $e^{-(x^2+y^2)}$ is constant around each circle; the marked annulus illustrates a finite-width shell. Right: the radial contribution $2\pi r e^{-r^2}$ includes the circumference factor and integrates to $\pi$. The shell in the picture has finite width; $2\pi r\,dr$ is the infinitesimal area element.}
The geometric surprise is that radial shells get longer as $r$ increases, even while the <Gaussian function> gets smaller. Its central height is largest at zero, but the radial contribution $2\pi r e^{-r^2}$ vanishes there and peaks at $r=1/\sqrt2$. Differentiating gives $2\pi e^{-r^2}(1-2r^2)$, which changes sign at that radius. Keeping the <Jacobian determinant> is what makes the one-dimensional <integral> and the plane geometry agree.
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