Subsection 5.3.1 Disks, Annuli, Sectors, and Polar Rectangles
The disk of radius R, \(\{(x,y) : x^2 + y^2 \leq R^2\}\text{,}\) is described in polar coordinates as
\begin{equation*}
D = \{(r,\theta) | 0 \leq r \leq R, 0 \leq \theta \leq 2\pi\},
\end{equation*}
which is a type of polar rectangle.
In general, a polar rectangle is a set
\begin{equation}
\{(r,\theta) | a \leq r \leq b, \alpha \leq \theta \leq \beta\}, \text{ constants}.\tag{5.3.1}
\end{equation}
This includes annuli and sectors of disks:
An annulus is a polar rectangle covering all angles: \(\alpha=0\text{,}\) \(\beta=2\pi\text{.}\)
A sector is a polar rectangle that includes the origin: \(a=0\text{.}\)
Subsection 5.3.2 Integration Over a Polar Rectangle
Integration over a polar rectangle can be done by starting again from approximations, as in
Section 5.1. (In
Section 5.7 we will see another method: changing coordinates, as with substitution.)
Consider approximations using the midpoint rule.
First subdivide the \(r\) and \(\theta\) values into
\begin{equation*}
a= r_1 \lt r_2 \lt \cdots \lt r_n=b \text{ and } \alpha = \theta_1 \lt \theta_2 \lt \cdots
\lt \theta_m=\beta
\end{equation*}
with spacing \(r_i-r_{i-1}=\Delta r\text{,}\) \(\theta_i-\theta_{i-1}=\Delta \theta\text{.}\)
Then find the midpoints \(r_i^* = (r_{i-1} + r_i)/2\text{,}\) \(\theta_i^* = (\theta_{i-1} + \theta_i)/2\text{:}\) this divides the domain into many small polar rectangles
\begin{equation*}
R_{ij} = \{(r,\theta) | r_{i-1} \leq r \leq r_i, \theta_{j-1} \leq \theta \leq \theta_{j} \},
\end{equation*}
each with a "midpoint" \((r_i^*,\theta_j^*)\text{.}\)
Next, we can approximate the integral of \(f(r,\theta)\) over each small polar rectangle as \(f(r_i^*,\theta_j^*)A(R_{ij})\text{,}\) with \(A(R_{ij})\) the area of each polar rectangle.
The area of a sector is \(\{(r,\theta) | 0 \leq r \leq b, \alpha \leq \theta \leq \beta\}\) is \((\beta-\alpha)b^2/2\text{,}\) so by subtraction,
\begin{equation*}
A(R_{ij}) = (\theta_i-\theta_{i-1})(r_i^2-r_{i-1}^2)/2 = \Delta \theta r_i^* \Delta r
\end{equation*}
Summing all these approximations and taking the limit leads to
\begin{equation*}
\iint\limits_D f(r,\theta) dA = \lim\limits_{n,m \to \infty} \sum_{i=1}^n\sum_{j=1}^m f(r_i^*,\theta_j^*) r_i^* \, \Delta r \, \Delta \theta.
\end{equation*}
This is simply the double integral in variables \(r\) and \(\theta\) of function \(f(r,\theta)r\text{:}\)
\begin{equation}
\iint\limits_D f(r,\theta) dA = \int_{\theta=\alpha}^\beta \int_{r=a}^b f(r,\theta)\, r \, dr \, d\theta
%= \int_a^b \int_\alpha^\beta f(r,\theta) \, d\theta\, r \, dr
% COMMENTS IGNORED!\tag{5.3.2}
\end{equation}
In effect, the infinitesimal area is now expressed as \(dA = r\,dr\,d\theta\) with polar coordinates, in place of \(dA = dx\,dy\) with cartesian coordinates.
Subsection 5.3.5 Optional Topic: Improper Double Integrals Using Polar Coordinates
Improper double integrals over the whole plane of radially symmetric functions (and even some others with a still simple polar form) are often best evaluated using polar coordinates. This requires a variant of the "expanding rectangles" approach seen in
Subsection 5.2.8, instead using expanding discs to cover the whole plane: with
\(B_R = B_R((0,0))\) the disc of radius
\(R\) centered at the origin
\begin{equation}
\iint\limits_{\reals^2} f(x,y) dA
= \lim_{R \to \infty} \iint_{B_R} f(x,y) dA
= \lim_{R \to \infty} \int_{r=0}^R \int_{\theta = 0}^{2\pi} f(x,y)\ r\ dr\ d\theta\tag{5.3.5}
\end{equation}
One interesting application of this is evaluating the integral associated with the Gaussian distribution,
\begin{equation}
I = \int_{-\infty}^\infty e^{-x^2}\ dx = \sqrt\pi\tag{5.3.6}
\end{equation}
First,
\begin{equation*}
I = \int_{-\infty}^\infty e^{-x^2}\ dx, = \int_{-\infty}^\infty e^{-y^2}\ dy
\end{equation*}
Multiplying these two forms and using the product rule backwards:
\begin{align*}
I^2
\amp= \int_{x = -\infty}^\infty \int_{y = -\infty}^\infty e^{-x^2} e^{-y^2}\ dy\ dx\\
\amp= \iint_{\reals^2} e^{-(x^2+y^2)}\ dA\\
\amp= \int_{r=0}^\infty \int_{\theta = 0}^{2\pi} e^{-r^2} r\ dr\ d\theta\\
\amp= \int_{r=0}^\infty e^{-r^2} r\ dr\ \int_{\theta = 0}^{2\pi} d\theta
\end{align*}
The \(\theta\) integral is simple: \(\int_{\theta = 0}^{2\pi} d\theta = 2\pi\text{.}\) The improper \(r\) integral is
\begin{align*}
\int_{r=0}^\infty e^{-r^2} r\ dr
\amp= \lim_{R \to \infty} \int_{r=0}^R e^{-r^2} r\ dr\\
\amp= \lim_{R \to \infty} \int_{u=0}^{R^2} e^{-u} \frac{du}{2} \quad \text{using } u=r^2,\ du = 2 r dr\\
\amp= \lim_{R \to \infty} \left[ \frac{-e^{-u}}{2} \right]_{u=0}^{R^2}\\
\amp= \lim_{R \to \infty} \left( \frac{-e^{-R^2} + e^{0}}{2} \right) = \frac12
\end{align*}
Thus
\(I^2 = \frac12 2\pi = \pi,\) confirming Equation
(5.3.6).
An aside on the normal (Gaussian) distribution of statistics.
A change of variables gives the form usually seen in statistics:
\begin{equation}
\int_{-\infty}^\infty e^{-x^2/2}\ dx = \sqrt{2\pi}\tag{5.3.7}
\end{equation}
so that
\begin{equation}
\int_{-\infty}^\infty \frac{e^{-x^2/2}}{\sqrt{2\pi}} dx = 1\tag{5.3.8}
\end{equation}
meaning that \(\phi(x) = \frac{1}{\sqrt{2\pi}} e^{-x^2/2}\) is a probability distribution: a non-negative function with integral over its whole domain being one.
In fact this is the standard normal distribution, or unit normal distribution; the reason for the factor of \(1/2\) can be verified in this (somewhat challenging) exercise.
Exercise 5.3.1. The standard deviation of the standard normal distribution is one.
Confirm that the standard normal distribution \(\frac{1}{\sqrt{2\pi}} e^{-x^2/2}\) has variance 1 and therefore its standard deviation (the square root of the variance) is also 1. That is, verify
\begin{equation*}
\int_{-\infty}^\infty x^2 e^{-x^2/2}\ dx = \sqrt{2\pi}
\end{equation*}
The mean of the standard normal distribution,
\begin{equation*}
\mu := \frac{\int_{-\infty}^\infty x \phi(x)\ dx}
{\int_{-\infty}^\infty \phi(x)\ dx}
= \int_{-\infty}^\infty x \phi(x)\ dx
= \frac{\int_{-\infty}^\infty x e^{-x^2/2}\ dx}{\sqrt{2\pi}}
\end{equation*}
is zero due to the oddness of the integrand, and then a change of variables shows that the more general distribution
\begin{equation}
\phi_{\mu,\sigma}(x) = \frac{1}{\sigma \sqrt{2\pi}} e^{-((x-\mu)/\sigma)^2/2}\tag{5.3.9}
\end{equation}
has integral one and so it is also probability density, with mean
\begin{equation*}
\int_{-\infty}^\infty x \phi_{\mu,\sigma}(x)\ dx = \mu
\end{equation*}
and standard deviation
\begin{equation*}
\int_{-\infty}^\infty x^2 \phi_{\mu,\sigma}(x)\ dx = \sigma
\end{equation*}
This is the Normal Distribution of mean \(\mu\) and standard deviation \(\sigma\text{,}\) sometimes denoted \(\mathcal{N}(\mu,\sigma^2)\text{;}\) it is probably the most important distribution in mathematical statistics.
