Double Integrals in Polar Coordinates

This guide covers the techniques for evaluating double integrals by transforming Cartesian coordinates $(x, y)$ into polar coordinates $(r, \theta)$. This method is particularly powerful for regions with circular symmetry.

Learning Objectives

• Recognize regions and integrands that are simplified by polar conversion.
• Apply the transformation $dA = r \, dr \, d\theta$ correctly in iterated integrals.
• Evaluate double integrals over polar rectangular and general polar regions.
• Convert limits of integration from rectangular to polar form.

Key Terms & Glossary

• Polar Rectangle: A region $R$ defined by $a \le r \le b$ and \alpha \le \theta \le eta.
  • Example: A semi-annulus (half-donut shape) where $1 \le r \le 2$and $0 \le \theta \le \pi$.
• Area Element ($dA$): The infinitesimal area $dx \, dy$ which transforms to $r \, dr \, d\theta$ in polar coordinates.
  • Example: When integrating over a disk, the "extra $r$" accounts for the widening of sectors as $r$ increases.
• General Polar Region: A region bounded by two rays ($\theta = \alpha, \theta = \beta$) and two continuous functions of $\theta$ ($r = h_1(\theta), r = h_2(\theta)$).
  • Example: The interior of a cardioid $r = 1 + \cos(\theta)$.

The "Big Idea"

In Cartesian coordinates, circular boundaries result in integration limits involving square roots (e.g., $y = \sqrt{a^2 - x^2}$), which are often difficult to compute. By switching to polar coordinates, circular boundaries become constant limits ($r = a$), turning complex regions into "rectangles" in the $r\theta$-plane. This effectively "straightens out" the geometry of the problem.

Formula / Concept Box

[!IMPORTANT] Never forget the extra $r$ in the integrand! $dx \, dy \neq dr \, d\theta$. The correct substitution is $dx \, dy = r \, dr \, d\theta$.

Hierarchical Outline

• I. Motivation for Polar Integration
  • Circular Symmetry: Use when the region $D$ is a disk, ring, or sector.
  • Integrand Simplification: Use when $f(x,y)$ contains the term $x^2 + y^2$.
• II. Integration over Polar Rectangles
  • Constant Limits: Bounds are $a \le r \le b$ and $\alpha \le \theta \le \beta$.
  • Iterated Setup: Outer integral usually $\theta$, inner integral $r$.
• III. Integration over General Polar Regions
  • Functional Limits: $r$ varies between two polar curves $r = g( heta)$ and $r = h( heta)$.
  • Radial Arrows: Visualize by drawing a ray from the origin through the region.

Visual Anchors

Decision Flow: When to use Polar Coordinates?

Geometry of the Polar Area Element

Definition-Example Pairs

• Definition: Annular Region — The area between two concentric circles.
  • Real-World Example: Finding the mass of a circular metal washer with a hole in the center. The density might vary with distance from the center.
• Definition: Radial Bound — The function $r(\theta)$ that defines the outer edge of a shape.
  • Real-World Example: A lighthouse beam rotating; the area swept over a specific time is a sector defined by the range of the beam ($r$) and the angle of rotation ($\theta$).

Worked Examples

Example 1: Evaluating over a Disk

Evaluate $\iint_D e^{x^2+y^2} \, dA$ where $D$ is the unit disk $x^2 + y^2 \le 1$.

1. Convert Region: The unit disk is $0 \le r \le 1$and $0 \le \theta \le 2\pi$.
2. Convert Integrand: $e^{x^2+y^2} \rightarrow e^{r^2}$.
3. Setup Integral: $\int_0^{2\pi} \int_0^1 e^{r^2} (r \, dr \, d\theta)$
4. Evaluate Inner ($r$): Let $u = r^2, du = 2r \, dr$. $\int_0^1 r e^{r^2} \, dr = \left[ \frac{1}{2} e^{r^2} \right]_0^1 = \frac{1}{2}(e - 1)$
5. Evaluate Outer ($\theta$): $\int_0^{2\pi} \frac{1}{2}(e - 1) \, d\theta = \frac{1}{2}(e - 1) [2\pi] = \pi(e - 1)$

Example 2: General Polar Region

Find the area of the region $D$ inside the cardioid $r = 1 + \cos \theta$.

1. Setup: Area $A = \iint_D 1 \, dA = \int_0^{2\pi} \int_0^{1+\cos \theta} r \, dr \, d\theta$.
2. Inner Integral: $\int_0^{1+\cos \theta} r \, dr = \left[ \frac{1}{2} r^2 \right]_0^{1+\cos \theta} = \frac{1}{2}(1 + \cos \theta)^2 = \frac{1}{2}(1 + 2\cos \theta + \cos^2 \theta)$
3. Use Identity: $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$.
4. Outer Integral: $\int_0^{2\pi} \frac{1}{2} (1 + 2\cos \theta + \frac{1}{2} + \frac{1}{2}\cos 2\theta) \, d\theta = \frac{1}{2} [ \frac{3}{2}\theta + 2\sin \theta + \frac{1}{4}\sin 2\theta ]_0^{2\pi} = \frac{3\pi}{2}$

Checkpoint Questions

1. What is the Jacobian (the scaling factor) when moving from Cartesian to Polar double integrals?
2. If you are integrating over a region in the first quadrant bounded by $y=x$, $y=0$, and $x^2+y^2=9$, what are the $\theta$ limits?
3. True or False: $\int_0^1 \int_0^1 f(r, \theta) \, dr \, d\theta$ represents the integral over a square of side length 1 in the $xy$-plane.

▶Click to see answers

1. The factor is $r$.
2. $\theta$ goes from 0 to $\pi/4$.
3. False. It represents a unit sector (a quarter-circle with radius 1).