Difficulty: Intermediate | Prerequisites: §15.1–15.2 (double integrals over rectangular and general regions), basic polar coordinate geometry (r, θ conversions).
Polar coordinates let you collapse messy circular or radial geometry into clean integration bounds. Any time the region of integration is a disk, annulus, sector, or cardioid, switching from dA = dx dy to dA = r dr dθ usually turns a painful integral into a routine one. You need fluency here before tackling surface area (§15.5) and cylindrical/spherical coordinates (§15.7–15.8), because the conversion logic is the same, just extended to three dimensions. If you skipped the polar coordinate review in precalculus, go back and nail the conversions x = r cos θ, y = r sin θ, r² = x² + y² before proceeding.
To integrate over a region with circular symmetry, convert x = r cos θ, y = r sin θ, and replace dA with r dr dθ. The extra factor of r is the Jacobian of the transformation and is the single most common thing students forget. Polar bounds describe the region as θ sweeping an angular range while r runs from an inner curve to an outer curve.
Polar coordinates (r, θ)
A coordinate system where every point in the plane is described by its distance r from the origin and its angle θ from the positive x-axis. Think of it as: instead of "how far right and how far up," you ask "how far out and in which direction."
Jacobian (of the polar transformation)
The scaling factor that accounts for the way area elements stretch when you change coordinates. For polar coordinates, the Jacobian is r, which is why dA = dx dy becomes dA = r dr dθ. In simple terms, wedge-shaped slivers near the origin are tiny, so the extra r shrinks their contribution appropriately.
Polar region
A region in the plane described by bounds on r and θ: typically α ≤ θ ≤ β and h₁(θ) ≤ r ≤ h₂(θ). In simple terms, you sweep a ray from the origin through an angular range, and at each angle the ray covers from the inner curve to the outer curve.
Lemniscate
A figure-eight curve whose polar equation takes the form r² = a sin(2θ) or r² = a cos(2θ). Think of it as two petals joined at the origin. The curve r = √(sin(2θ)) in Problem 2 is one loop of a lemniscate.
Annular region
The ring-shaped area between two concentric circles, such as 1 ≤ r ≤ 2. In Problem 3(b), the region between two circles (one centred at the origin, one centred at (0, 1)) is a more general version of this idea.
The substitution recipe has three parts:
Replace x with r cos θ and y with r sin θ everywhere in the integrand.
Replace dA (or dx dy) with r dr dθ. Do not forget the extra r.
Re-express the region's boundaries in terms of r and θ.
The outer integral is always over θ (the angular sweep), and the inner integral is over r (the radial reach at each angle).
Before writing any integral, sketch the region. For a curve like r = √(sin(2θ)):
Identify where r = 0: this happens when sin(2θ) = 0, i.e. θ = 0 and θ = π/2.
Identify the maximum r: sin(2θ) = 1 at θ = π/4, giving r = 1.
The curve traces a single petal in the first quadrant (0 ≤ θ ≤ π/2).
For the half-disk in Problem 3(a), the original Cartesian bounds −√(16 − y²) ≤ x ≤ √(16 − y²), 0 ≤ y ≤ 4 describe the upper half of a disk of radius 4. In polar: 0 ≤ θ ≤ π, 0 ≤ r ≤ 4.
Region: 0 ≤ r ≤ √(sin(2θ)), 0 ≤ θ ≤ π/2.
Integrand: x√(x² + y²) = (r cos θ)(r) = r² cos θ.
With the Jacobian, the full integrand becomes r² cos θ · r = r³ cos θ.
The integral is:
∫₀^{π/2} ∫₀^{√(sin 2θ)} r³ cos θ dr dθ
Inner integral: r⁴/4 evaluated from 0 to √(sin 2θ) gives sin²(2θ)/4.
Outer integral: (1/4) ∫₀^{π/2} sin²(2θ) cos θ dθ, which requires expanding sin(2θ) = 2 sin θ cos θ and simplifying.
The Cartesian integral ∫₀⁴ ∫_{-√(16-y²)}^{√(16-y²)} (y³ + x²y) dx dy covers the upper half of a disk of radius 4.
In polar: 0 ≤ θ ≤ π, 0 ≤ r ≤ 4.
Converting the integrand: y³ + x²y = y(y² + x²) = (r sin θ)(r²) = r³ sin θ.
With the Jacobian: r³ sin θ · r = r⁴ sin θ.
The integral becomes ∫₀^π ∫₀⁴ r⁴ sin θ dr dθ = (4⁵/5) ∫₀^π sin θ dθ = (1024/5)(2) = 2048/5.
D is outside the circle of radius 1 centred at the origin and inside the circle of radius 1 centred at (0, 1).
The circle centred at (0, 1) has equation x² + (y − 1)² = 1, which in polar becomes r = 2 sin θ.
The unit circle at the origin is r = 1.
The two circles intersect where r = 1 = 2 sin θ, giving θ = π/6 and θ = 5π/6.
The region splits by angle:
For π/6 ≤ θ ≤ 5π/6, r ranges from 1 to 2 sin θ (inside the upper circle, outside the unit circle).
Integrand: y/(x² + y²) = (r sin θ)/r² = sin θ/r.
With the Jacobian: (sin θ/r) · r = sin θ.
The integral becomes ∫_{π/6}^{5π/6} ∫₁^{2 sin θ} sin θ dr dθ.
Coordinate conversion
x = r cos θ, y = r sin θ, r² = x² + y², tan θ = y/x
Area element
dA = r dr dθ (the r is the Jacobian, not optional)
General double integral in polar
∫∫_D f(x, y) dA = ∫α^β ∫{h₁(θ)}^{h₂(θ)} f(r cos θ, r sin θ) · r dr dθ
Common polar curves
Circle centred at origin, radius a: r = a
Circle through origin, centred at (0, a): r = 2a sin θ
Circle through origin, centred at (a, 0): r = 2a cos θ
Lemniscate: r² = a sin(2θ) or r² = a cos(2θ)
Cardioid: r = a(1 + cos θ) or r = a(1 + sin θ)
Polar integrals show up whenever a physical quantity is distributed over a circular or radial domain. Radar and sonar compute signal intensity over circular sweep regions using exactly this kind of integration. Engineers designing circular plates or disks compute mass, moment of inertia, and deflection with polar double integrals.
Students often forget the extra factor of r in dA = r dr dθ. If your answer is off by a power of r, this is almost certainly why.
Students sometimes write θ bounds as 0 to 2π without thinking. Many polar regions only occupy part of the plane. Sketch first.
When a circle is not centred at the origin, students try to use r = constant as the boundary. A circle centred at (0, 1) with radius 1 is r = 2 sin θ, not r = 1. Convert the Cartesian equation.
Students confuse the integrand conversion with the area element. The r in the integrand (from converting x or y) and the r from the Jacobian are separate. Both must appear.
⚠️ Converting a Cartesian integral to polar and evaluating is a near-certain exam question. Know the full recipe cold.
⚠️ Sketching the polar region is often worth marks on its own. Exams frequently ask you to sketch D.
⚠️ Problems with integrands containing x² + y² or √(x² + y²) are strong signals to use polar coordinates.
⚠️ Regions described by overlapping circles (Problem 3b) require careful intersection analysis. Finding the correct θ bounds is where most students lose marks.
True or false: dA in polar coordinates is simply dr dθ.
False. dA = r dr dθ. The factor of r is essential.
Fill in the blank: The polar equation of a circle of radius 1 centred at (0, 1) is r = ___.
r = 2 sin θ
True or false: To convert ∫∫ f(x, y) dA to polar, you only need to substitute x = r cos θ and y = r sin θ in the integrand.
False. You must also multiply by the Jacobian r and convert the region bounds.
Fill in the blank: The curve r = √(sin 2θ) for 0 ≤ θ ≤ π/2 traces one loop of a ___.
Lemniscate
Q: Convert the integral ∫₀² ∫₀^{√(4 − x²)} (x² + y²) dy dx to polar coordinates and evaluate.
A: The region is the quarter-disk x² + y² ≤ 4 in the first quadrant, so 0 ≤ θ ≤ π/2, 0 ≤ r ≤ 2. The integrand x² + y² = r². With the Jacobian: ∫₀^{π/2} ∫₀² r² · r dr dθ = ∫₀^{π/2} ∫₀² r³ dr dθ = (π/2)(2⁴/4) = 2π.
Q: Sketch the region where 0 ≤ r ≤ 2 sin θ and set up the integral of f(r, θ) over this region.
A: r = 2 sin θ is a circle of radius 1 centred at (0, 1). For θ from 0 to π, r goes from 0 to 2 sin θ. The integral is ∫₀^π ∫₀^{2 sin θ} f(r, θ) · r dr dθ.
Q: Why does the integrand in Problem 3(b) simplify so cleanly after conversion to polar?
A: Because y/(x² + y²) = (r sin θ)/r² = sin θ/r. When multiplied by the Jacobian r, the r cancels entirely, leaving just sin θ. The radial integration then becomes trivial.
Q: For which types of regions and integrands should you strongly consider switching to polar coordinates?
A: Regions bounded by circles, arcs, or curves most naturally described by distance from the origin. Integrands containing x² + y², √(x² + y²), or expressions that simplify when written in terms of r and θ.
Polar double integrals are the two-dimensional warm-up for cylindrical coordinates (§15.7), which add a z-axis to the same r, θ setup. If you can set up polar bounds comfortably, cylindrical integrals are just one more variable. The Jacobian idea generalises to spherical coordinates (§15.8) where the area element is r² sin φ. Surface area calculations (§15.5) also frequently use polar parameterisations when the surface projects onto a circular region.
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