Double Integrals in Polar Coordinates | Tony's Intertwingled Memex

Core Definition

A double integral in polar coordinates transforms a planar integration domain from Cartesian coordinates $(x, y)$ to polar coordinates $(r, \theta)$ through the coordinate mapping:

x = r \cos \theta, \quad y = r \sin \theta, \quad x^2 + y^2 = r^2

For a continuous function

f(x, y)

defined over a region

R

described in polar coordinates by

\alpha \le \theta \le \beta

and

g_1(\theta) \le r \le g_2(\theta)

, the double integral is evaluated as:

\iint_R f(x, y) \, dA = \int_{\alpha}^{\beta} \int_{g_1(\theta)}^{g_2(\theta)} f(r\cos\theta, r\sin\theta) \, r \, dr \, d\theta

where the extra factor of

r

is the Jacobian scaling factor required by the geometry of the polar area element:

dA = r \, dr \, d\theta

Concept Explanation

1. Why Transform to Polar Coordinates?

When evaluating double integrals in Cartesian coordinates, regions with circular, radial, or angular symmetry (such as disks, concentric rings, cylinders, and pie-shaped wedges) lead to cumbersome radical boundary curves of the form $y = \pm\sqrt{a^2 - x^2}$ . Integrating over these boundaries often demands tedious trigonometric substitutions and half-angle identities.

Furthermore, integrands containing radial combinations such as $\sqrt{x^2 + y^2}$ , $(x^2 + y^2)^n$ , or $e^{-(x^2 + y^2)}$ frequently lack elementary antiderivatives with respect to $x$ or $y$ individually. Transforming to polar coordinates aligns the coordinate grid with the natural symmetry of the domain, converting variable radical boundaries into constant limits and introducing the radial factor $r$ which serves as a natural chain-rule multiplier for algebraic substitution.

2. The Polar Area Differential Element & The Jacobian Factor

In rectangular coordinates, an infinitesimal area element is a simple rectangle with area $\Delta A = \Delta x \, \Delta y$ .

In polar coordinates, an infinitesimal sub-region $\Delta R$ is bounded by two concentric circles of radii $r_i$ and $r_i + \Delta r$ and two radial rays at angles $\theta_j$ and $\theta_j + \Delta \theta$ . This shape is a polar wedge (an annular sector):

The radial length of the wedge is $\Delta r = r_i - r_{i-1}$ .
The inner arc length is $r_i \Delta \theta$ , and the outer arc length is $(r_i + \Delta r) \Delta \theta$ .
The average radius is $\bar{r}_i = \frac{r_i + r_{i-1}}{2} = r_{i-1} + \frac{1}{2}\Delta r$ .

The exact area of this sector is:

\Delta A_i = \frac{1}{2} (r_i)^2 \Delta \theta - \frac{1}{2} (r_{i-1})^2 \Delta \theta = \frac{1}{2} (r_i + r_{i-1})(r_i - r_{i-1}) \Delta \theta = \bar{r}_i \, \Delta r \, \Delta \theta

Taking the limit as $\Delta r \to 0$ and $\Delta \theta \to 0$ , the polar area differential becomes:

dA = r \, dr \, d\theta

In multivariable calculus and general transformations, this factor $r$ is the absolute value of the determinant of the Jacobian matrix $J(r, \theta)$ :

J(r, \theta) = \frac{\partial(x, y)}{\partial(r, \theta)} = \begin{vmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} \end{vmatrix} = \begin{vmatrix} \cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta \end{vmatrix} = r\cos^2\theta - (-r\sin^2\theta) = r(\cos^2\theta + \sin^2\theta) = r

3. The 3-Step Conversion Protocol

To convert any Cartesian double integral $\iint_D f(x, y) \, dA$ into polar form:

Map the Region Boundaries: Sketch the domain $D$ in the $xy$ -plane and express the boundaries in polar parameters:
- Identify the angular sweep $\alpha \le \theta \le \beta$ .
- For any radial ray extending from the origin at angle $\theta$ , identify the entry radius $r = g_1(\theta)$ and exit radius $r = g_2(\theta)$ .
Transform the Integrand: Substitute all independent variables:
- $x = r \cos \theta$
- $y = r \sin \theta$
- Replace $x^2 + y^2$ with $r^2$ wherever it occurs.
Insert the Jacobian Differential: Replace $dx \, dy$ (or $dy \, dx$ ) with $r \, dr \, d\theta$ .

Engineering Applications

Mechanical & Aerospace Engineering: Calculating mass moments of inertia $I_z = \iint_D r^2 \, \rho(r, \theta) \, dA$ for spinning turbines, satellite reaction flywheels, and circular brake rotors.
Civil & Fluids Engineering: Determining volumetric flow discharge $Q = \iint_D v(r) \, dA$ for viscous fluids moving through cylindrical pipes and circular storm culverts.
Electrical & RF Engineering: Modeling radar detection sweeps, circular phased-array antenna radiation patterns, and electric field flux across circular capacitor plates.
Structural Engineering: Calculating polar moments of inertia $J = \iint_D (x^2 + y^2) \, dA = \iint_D r^3 \, dr \, d\theta$ to determine torsional shear stress distributions in solid and hollow cylindrical shafts.

Example Problems

Example 1: Converting a Non-Elementary Gaussian Integral

Evaluate the double integral:

I = \int_{0}^{2} \int_{0}^{\sqrt{4 - x^2}} e^{-(x^2 + y^2)} \, dy \, dx

Solution

Step 1: Map the Cartesian Domain to Polar Coordinates
Formulate the boundary inequalities from the Cartesian limits:

0 \le x \le 2, \quad 0 \le y \le \sqrt{4 - x^2}

The equation

y = \sqrt{4 - x^2}

represents the upper half of the circle

x^2 + y^2 = 4

of radius

R = 2

. Because

x

ranges from

0

2

and

y \ge 0

, the domain

D

is the quarter-disk located entirely in the first quadrant. In polar coordinates, the region is a polar rectangle:

0 \le r \le 2, \quad 0 \le \theta \le \frac{\pi}{2}

Step 2: Transform Integrand and Differential
State the Polar Coordinate Transformation Law:

x^2 + y^2 = r^2, \quad dy \, dx = r \, dr \, d\theta

Substitute these relations into the integral:

I = \int_{0}^{\frac{\pi}{2}} \int_{0}^{2} e^{-r^2} \, r \, dr \, d\theta

Step 3: Evaluate via Substitution
Let $u = -r^2$ , so $du = -2r \, dr \implies r \, dr = -\frac{1}{2} du$ . When $r = 0$ , $u = 0$ ; when $r = 2$ , $u = -4$ . Evaluate the inner $r$ -integral:

\int_{0}^{2} e^{-r^2} r \, dr = \int_{0}^{-4} -\frac{1}{2} e^u \, du = \left[ -\frac{1}{2} e^u \right]_{0}^{-4} = -\frac{1}{2} e^{-4} - \left(-\frac{1}{2} e^0\right) = \frac{1}{2}\left(1 - e^{-4}\right)

Evaluate the outer $\theta$ -integral:

I = \int_{0}^{\frac{\pi}{2}} \frac{1}{2}\left(1 - e^{-4}\right) d\theta = \frac{1}{2}\left(1 - e^{-4}\right) [\theta]_0^{\frac{\pi}{2}} = \frac{\pi}{4}\left(1 - e^{-4}\right)

Example 2: Semicircular Coordinate Transformation

Evaluate the integral:

I = \int_{-3}^{3} \int_{0}^{\sqrt{9 - x^2}} y \, dy \, dx

Solution

Step 1: Map the Boundary to Polar Limits
The upper limit $y = \sqrt{9 - x^2}$ defines the circle $x^2 + y^2 = 9$ of radius $R = 3$ . The bounds $-3 \le x \le 3$ and $0 \le y \le \sqrt{9 - x^2}$ describe the entire upper semicircle. In polar coordinates:

0 \le r \le 3, \quad 0 \le \theta \le \pi

Step 2: Substitute Polar Equations
Substitute $y = r \sin \theta$ and $dA = r \, dr \, d\theta$ :

I = \int_{0}^{\pi} \int_{0}^{3} (r \sin \theta) \, r \, dr \, d\theta = \int_{0}^{\pi} \int_{0}^{3} r^2 \sin \theta \, dr \, d\theta

Step 3: Evaluate the Iterated Integral
Apply Fubini's Product Separability Rule (defined in Separability in Polar Integrals):

I = \left( \int_{0}^{\pi} \sin \theta \, d\theta \right) \left( \int_{0}^{3} r^2 \, dr \right)

Evaluate the angular integral:

\int_{0}^{\pi} \sin \theta \, d\theta = [-\cos \theta]_{0}^{\pi} = (-\cos \pi) - (-\cos 0) = -(-1) - (-1) = 2

Evaluate the radial integral:

\int_{0}^{3} r^2 \, dr = \left[ \frac{r^3}{3} \right]_{0}^{3} = \frac{27}{3} = 9

Multiply the results:

I = 2 \times 9 = 18

Connections & References

Parent Concepts: Multiple Integrals, Double Integrals over a General Region
Sub-Concepts & Techniques: Polar Rectangles and Annular Sectors, Separability in Polar Integrals, Volume and Surface Integration in Polar Coordinates
Course Sessions: Session 15 - Double Integrals in Polar Coordinates