Calculus, Early Transcendentals by Stewart, Section 15.4.
Introduction.
After rectangles, perhaps the single most common shape for a domain in two dimensions is a disk, and other domains with some radial symmetry are common too.
Calculations are often easiest if one uses an approach that emphasizes the symmetry of the domain: using polar coordinates based at the center of the disk. It is simple enough to chose coordinates with origin at the center of the disk, so we work with cartesian coordinates \(x\) and \(y\) and polar coordinates \(r\) and \(\theta\) as seen in Section 1.3:
\begin{equation*}
x = r \cos \theta, y = r \sin \theta,
\quad \text{so that} \quad
r^2 = x^2 + y^2, \tan\theta = y/x.
\end{equation*}
Subsection5.3.1Disks, 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: \(a=0\text{.}\)
Subsection5.3.2Integration 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
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
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,
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.
Subsection5.3.3Integrals in Polar Coordinates Over Other Domains
Polar coordinates can be useful for integrals over domains of other shapes, like annuli and sectors. Just as with the Type I and Type II domains seen in Section 5.2 with cartesian coordinates, there are two special cases where it is worth converting to polar coordinates:
For a domain \(D=\{(r,\theta) | \alpha \leq \theta \leq \beta, r_1(\theta) \leq r \leq r_2(\theta)\}\text{,}\)
\begin{equation}
\iint\limits_D f(x,y) \, dA = \int_{\theta=\alpha}^\beta \int_{r=r_1(\theta)}^{r_2(\theta)} f(r\cos\theta,r\sin\theta)\, r \, dr \, d\theta\tag{5.3.3}
\end{equation}
For a domain \(D=\{(r,\theta) | a \leq r \leq b, \theta_1(r) \leq \theta \leq \theta_2(r)\}\text{,}\)
\begin{equation}
\iint\limits_D f(x,y) \, dA = \int_{r=a}^b \int_{\theta=\theta_1(r)}^{\theta_2(r)} f(r\cos\theta,r\sin\theta) \, d\theta\, r \, dr\tag{5.3.4}
\end{equation}
The former is more common in practice.
Subsection5.3.4Calculating Areas and Volumes using Polar Coordinates
One basic use of double integrals is computing areas, and this sometimes involved polar integrals: if a region \(D\) is a polar rectangle as in Equation (5.3.1), or a more general polar analogue of a Type I or Type II region as in Equations (5.3.3) and (5.3.4), The area is \(A(D) = \iint_D dA\) is no given by the appropriate choice of
\begin{equation*}
A(D) = \int_{r=r_1}^{r_2} \int_{\theta=\theta_1(r)}^{\theta_2(r)} d\theta\ r\ dr
\end{equation*}
Also, the volume over such a region between the \(x\)-\(y\) plane and the surface \(z = f(x,y)\) is as usual given by the corresponding polar integral of function \(f\text{.}\)
Subsection5.3.5Optional 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
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.
Exercise5.3.1.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
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.
