Polar coordinates and the Gaussian integral

Polar coordinates

A point of the plane can be given by its distance $r$ from the origin and its angle $\theta$ from the positive $x$-axis. The point at angle $\theta$ on the unit circle is $(\cos\theta, \sin\theta)$ (lesson 1.7.1), so the point at distance $r$ is $r$ times it.

The point $(1, 1)$ has $r = \sqrt{2} \approx 1.414214$ and $\theta = \frac{\pi}{4}$. The point $(-1, 1)$ has the same $r$ but $\theta = \frac{3\pi}{4} \approx 2.356194$, in the second quarter of the plane. The point $(0, -2)$ has $r = 2$ and $\theta = -\frac{\pi}{2}$; adding $2\pi$ gives the same point.

Two cases need care. Adding $2\pi$ to $\theta$ does not move the point, so a full turn is $\theta$ from 0 to $2\pi$ or from $-\pi$ to $\pi$, not both. And the origin has $r = 0$ and no angle.

The polar cell: $dA = r\,dr\,d\theta$

Polar coordinates are a map from the $(r, \theta)$ plane to the $(x, y)$ plane, as in lesson 4.7.2. A small rectangle in $(r, \theta)$, from $r$ to $r + dr$ and from $\theta$ to $\theta + d\theta$, lands on a piece of a ring.

That piece has two straight sides of length $dr$ along the rays. Its other two sides are arcs, and an arc of angle $d\theta$ on a circle of radius $r$ has length $r\,d\theta$ (lesson 1.7.1). So its area is about $r\,dr\,d\theta$: the further out, the bigger the cell.

The exact area of the piece of ring is the difference of two sectors, $\frac{1}{2}\left((r + dr)^2 - r^2\right)d\theta = \left(r + \frac{dr}{2}\right)dr\,d\theta$. Divided by the cell's area in $(r, \theta)$, $dr\,d\theta$, it tends to $r$ as the cell shrinks:

The Jacobian of lesson 4.7.2 gives the same scale. With $x = r\cos\theta$ and $y = r\sin\theta$,

by the identity $\cos^2\theta + \sin^2\theta = 1$ of lesson 1.7.1.

Discs, rings and wedges

A disc of radius $a$ about the origin is the polar rectangle $0 \le r \le a$, $0 \le \theta \le 2\pi$. Its area comes out as it should:

A ring $1 \le r \le 2$ has area $2\pi \cdot \frac{4 - 1}{2} = 3\pi \approx 9.424778$. For a function of $x^2 + y^2$, polar coordinates turn it into a function of $r$ alone. Over the unit disc,

That is about 1.570796. Without the factor $r$ the answer would be $\frac{2\pi}{3}$, which is wrong.

A wedge uses part of a turn. The right half of the disc of radius 2, where $x \ge 0$, is $-\frac{\pi}{2} \le \theta \le \frac{\pi}{2}$.

There $\iint x\,dA = \int_{-\pi/2}^{\pi/2} \cos\theta\,d\theta \cdot \int_0^2 r^2\,dr = 2 \cdot \frac{8}{3} = \frac{16}{3}$, a product as in lesson 4.7.1. Over the whole disc the two halves cancel and $\iint x\,dA = 0$.

When polar coordinates do not help

Polar coordinates suit discs, rings and wedges centred at the origin, and functions of $x^2 + y^2$. They make other regions harder.

In the square $[0, 1] \times [0, 1]$, $r$ runs out to the edge $x = 1$, which is $r = \frac{1}{\cos\theta}$ for $\theta \le \frac{\pi}{4}$. Beyond that it runs out to $y = 1$, so there are two pieces with awkward limits. For a square, keep $x$ and $y$.

Where the Gaussian integral comes from

Lesson 2.6.3 stated $\int_{-\infty}^{\infty} e^{-x^2}\,dx = \sqrt{\pi}$ without proof. The curve $e^{-x^2}$ has no antiderivative made of familiar functions, so the fundamental theorem cannot find this area. A double integral can.

Call the area $I$. The idea is to work out $I^2$ instead: as a double integral over the whole plane, it can be done in polar coordinates.

1. 1
   $$I^2 = \int_{-\infty}^{\infty} e^{-x^2}\,dx \cdot \int_{-\infty}^{\infty} e^{-y^2}\,dy$$

   Why?
2. 2
   $$I^2 = \iint_{\mathbb{R}^2} e^{-(x^2 + y^2)}\,dA$$

   Why?
3. 3
   $$I^2 = \int_0^{2\pi} \int_0^{\infty} e^{-r^2}\,r\,dr\,d\theta$$

   Why?
4. 4
   $$\int_0^{\infty} r\,e^{-r^2}\,dr = \left[-\tfrac{1}{2}e^{-r^2}\right]_0^{\infty} = \frac{1}{2}$$

   Why?
5. 5
   $$I^2 = 2\pi \cdot \frac{1}{2} = \pi, \qquad I = \sqrt{\pi}$$

   Why?

The factor $r$ from the polar cell is what makes it work. In one variable $e^{-x^2}$ has no antiderivative; in two, the area element brings in the $r$ that $r\,e^{-r^2}$ needs.

Checking it with numbers

Over a disc of radius $b$, the same steps give $\pi\left(1 - e^{-b^2}\right)$. Over the square $[-b, b] \times [-b, b]$, the product rule gives $\left(\int_{-b}^{b} e^{-x^2}\,dx\right)^2$: the square of lesson 2.6.3's table. At $b = 1$ that is $1.493648^2 = 2.230985$.

The square contains the disc of radius $b$ and lies inside the disc of radius $\sqrt{2}\,b$, which passes through its corners. The integrand is positive, so the square's volume is squeezed between the two discs' volumes.

All three settle on $\pi \approx 3.141593$, so $I^2 = \pi$, and $I = \sqrt{\pi} \approx 1.772454$, the number lesson 2.6.3 found numerically.

The bell in two dimensions

The same steps work for $e^{-(x^2 + y^2)/2}$. In polar coordinates the inner integral is $\int_0^{\infty} r\,e^{-r^2/2}\,dr = 1$, so

Its square root, $\sqrt{2\pi} \approx 2.506628$, is $\int_{-\infty}^{\infty} e^{-x^2/2}\,dx$, the number lesson 2.6.3 found by substitution. Dividing by $2\pi$ gives the surface $\varphi(x)\varphi(y) = \frac{1}{2\pi}e^{-(x^2 + y^2)/2}$, with $\varphi$ the standard normal density, and it has volume 1. Case study 4.7.5 reads that surface as the joint density of two independent normal returns.

Common mistakes

1. Keep the $r$. $dA = r\,dr\,d\theta$, not $dr\,d\theta$: $\iint (x^2 + y^2)\,dA$ over the unit disc is $\frac{\pi}{2}$, not $\frac{2\pi}{3}$.
2. A full turn is $2\pi$. $\theta$ from 0 to $\pi$ covers only the upper half; from 0 to $4\pi$ it covers the disc twice.
3. A ring of radii $a$ and $b$ has area $\pi(b^2 - a^2)$, not $\pi(b - a)^2$.
4. Find the angle with atan2. $\frac{y}{x}$ is the same for $(1, 1)$ and $(-1, -1)$; the quadrant decides the angle (practice set 1.7.4). And $r$ is never negative.
5. Use polar coordinates for round regions, centred at the origin. A square has awkward limits in $r$ and $\theta$.
6. $e^{-x^2}$ has no antiderivative, but $r\,e^{-r^2}$ does. The Gaussian integral works only through the double integral, where the area element supplies the $r$.