📐 Concept diagram

06-10 — Double Integrals in Polar Coordinates

Phase: 6 — Calculus III: Multivariable Calculus Subject: 06-10 Prerequisites: 06-09 — Double Integrals, polar coordinates and converting between Cartesian/polar (Phase 5) Next subject: 07-01 — Triple Integrals

Learning Objectives

By the end of this subject, you will be able to:

1. Convert double integrals from Cartesian to polar coordinates using the transformation x = r cos θ, y = r sin θ
2. Explain why the area element transforms as dA = r dr dθ (the Jacobian factor r)
3. Sketch a region in the xy-plane and describe it in polar form (r bounds as functions of θ)
4. Apply polar double integrals to compute area, mass, and center of mass
5. Recognize when polar coordinates are advantageous — circular symmetry, x² + y² expressions, and angular sectors

Core Content

1. Why Polar Coordinates?

Some regions and integrands are painful in Cartesian coordinates but elegant in polar coordinates. Consider computing ∬_R (x² + y²) dA over the unit disk x² + y² ≤ 1. In Cartesian:

$∫₋₁¹ ∫_(−√(1−x²))^(√(1−x²)) (x² + y²) dy dx
$

The limits involve square roots, and the inner integral is messy. In polar coordinates this becomes simple because:

• The region is: r ∈ [0, 1], θ ∈ [0, 2π]
• The integrand: x² + y² = r²
• The area element: dA = r dr dθ

So the integral becomes:

$∫₀^(2π) ∫₀¹ r² · r dr dθ = ∫₀^(2π) ∫₀¹ r³ dr dθ
$

Which evaluates easily to π/2.

⚠️ THIS IS CRITICAL — polar coordinates are used constantly in vector calculus (Phase 7), probability theory with Gaussian integrals, and any problem with radial symmetry.

2. Polar Coordinate Transformation

Recall from Phase 5: a point (x, y) in the plane corresponds to polar coordinates (r, θ) where:

$x = r cos θ
y = r sin θ
r = √(x² + y²)        (r ≥ 0)
θ = tan⁻¹(y/x)         (watch the quadrant)
$

The ranges: r ∈ [0, ∞), θ ∈ [0, 2π) or sometimes θ ∈ [−π, π).

3. The Area Element — Why dA = r dr dθ

This is the most important concept in this subject. In Cartesian coordinates, we partition the plane into small rectangles of area Δx Δy. In polar coordinates, we partition into polar rectangles:

$       Δθ
     ←───→
    ┌─────┐  ← r + Δr
    │  ΔA │
    └─────┘  ← r
$

The area of one polar rectangle (annular sector) is:

ΔA ≈ (area of larger sector) − (area of smaller sector)
    = ½(r + Δr)² Δθ − ½r² Δθ
    = ½(r² + 2rΔr + (Δr)² − r²) Δθ
    = (r Δr + ½(Δr)²) Δθ
    ≈ r Δr Δθ        (dropping (Δr)² as negligible)

So dA = r dr dθ. The factor r is the Jacobian — it accounts for the fact that a small change in r sweeps out more area at large r than at small r. Think of it as an "area stretching factor."

Common misconception: Students often write dA = dr dθ and forget the r. Without the r, your integral will be wrong. Every polar double integral must have the r factor.

Formal Jacobian derivation (for completeness — the key result, not the derivation, is what you use day-to-day):

$J = det | ∂x/∂r  ∂x/∂θ | = det | cos θ   −r sin θ |
        | ∂y/∂r  ∂y/∂θ |       | sin θ    r cos θ  |

  = cos θ · r cos θ − (−r sin θ) · sin θ
  = r cos²θ + r sin²θ = r(cos²θ + sin²θ) = r
$

So ∬_R f(x, y) dA = ∬_R f(r cos θ, r sin θ) · r dr dθ, where R is now described in polar coordinates.

4. Setting Up Polar Double Integrals

The general recipe:

Step 1: Sketch the region in the xy-plane. Step 2: Describe the region in polar form: θ goes from α to β; for each θ, r goes from g₁(θ) to g₂(θ). Step 3: Replace x and y in the integrand with r cos θ and r sin θ. Step 4: Multiply the integrand by r (don't forget!). Step 5: Integrate: ∫(θ=α to β) ∫(r=g₁(θ) to g₂(θ)) f(r cos θ, r sin θ) · r dr dθ.

$∬_R f(x, y) dA = ∫(θ=α to β) ∫(r=g₁(θ) to g₂(θ)) f(r cos θ, r sin θ) · r dr dθ
$

5. Common Polar Regions

(a) Full disk of radius a centred at origin:

$r ∈ [0, a],  θ ∈ [0, 2π]
$

(b) Annulus (washer) between r = a and r = b:

$r ∈ [a, b],  θ ∈ [0, 2π]
$

(c) Circular sector of angle φ:

$r ∈ [0, a],  θ ∈ [0, φ]
$

(d) Disk centred at (a, 0) on the x-axis, radius a:

The circle x² + y² = 2ax can be rewritten as (x − a)² + y² = a². In polar:

x² + y² = r² = 2a r cos θ  →  r = 2a cos θ (for −π/2 ≤ θ ≤ π/2)

(e) Cardioid: r = 1 + cos θ, θ ∈ [0, 2π].

(f) Lemniscate: r² = cos(2θ), θ ∈ [−π/4, π/4] ∪ [3π/4, 5π/4].

General rule: If the region description or integrand contains x² + y², polar is almost always the right choice.

6. Applications

6.1 Area

Area of region R = ∬_R 1 dA = ∬_R r dr dθ.

Example — Area of a cardioid: R = {(r, θ) : 0 ≤ θ ≤ 2π, 0 ≤ r ≤ 1 + cos θ}

$A = ∫₀^(2π) ∫₀^(1+cos θ) r dr dθ
  = ∫₀^(2π) [r²/2]₀^(1+cos θ) dθ
  = ½ ∫₀^(2π) (1 + cos θ)² dθ
$

Expand: (1 + cos θ)² = 1 + 2 cos θ + cos²θ = 1 + 2 cos θ + (1 + cos 2θ)/2 = 3/2 + 2 cos θ + (cos 2θ)/2.

$A = ½ ∫₀^(2π) (3/2 + 2 cos θ + ½ cos 2θ) dθ
  = ½ [3θ/2 + 2 sin θ + (sin 2θ)/4]₀^(2π)
  = ½ (3π) = 3π/2
$

6.2 Mass

If a lamina (thin plate) occupies region R with density ρ(x, y) (mass per unit area), then:

$Total mass M = ∬_R ρ(x, y) dA
$

In polar, M = ∬_R ρ(r cos θ, r sin θ) · r dr dθ.

6.3 Center of Mass (Centroid)

The center of mass (x̄, ȳ) is given by:

$x̄ = (1/M) ∬_R x ρ(x, y) dA      ȳ = (1/M) ∬_R y ρ(x, y) dA
$

In polar, x = r cos θ, y = r sin θ, so:

$x̄ = (1/M) ∬_R (r cos θ) · ρ(r cos θ, r sin θ) · r dr dθ
ȳ = (1/M) ∬_R (r sin θ) · ρ(r cos θ, r sin θ) · r dr dθ
$

For uniform density (ρ = constant), the center of mass equals the centroid, and the constant cancels out.

Key observation: If the region is symmetric about the x-axis and the density satisfies ρ(x, −y) = ρ(x, y), then ȳ = 0. Similarly for symmetry about the y-axis and x̄ = 0.

6.4 Moments of Inertia

$I_x = ∬_R y² ρ(x, y) dA          (moment of inertia about x-axis)
I_y = ∬_R x² ρ(x, y) dA          (moment of inertia about y-axis)
I_O = ∬_R (x² + y²) ρ(x, y) dA   (polar moment of inertia about origin)
$

In polar: I_O = ∬_R r² · ρ · r dr dθ = ∬_R ρ r³ dr dθ.

7. When to Use Polar vs. Cartesian

8. Common Pitfalls

1. Forgetting the r factor: dA ≠ dr dθ. Always include r.
2. Wrong θ range: For a full circle, use [0, 2π] (or [−π, π]). For a half-circle in the right half-plane, use [−π/2, π/2].
3. Negative r: Some texts allow negative r (meaning r in the opposite direction of θ). In double integrals, keep r ≥ 0 and adjust θ range to cover the region.
4. Trigonometric simplification: After integrating r, you'll often be left with powers of cos θ and sin θ. Use identities: cos²θ = (1 + cos 2θ)/2, sin²θ = (1 − cos 2θ)/2.
5. ∫₀^(2π) cos θ dθ = 0 and ∫₀^(2π) sin θ dθ = 0: When integrating odd powers of cos or sin over a full period, the result is zero. This can save time in symmetry arguments.

Key Terms

• Formal Jacobian derivation
• Jacobian

Worked Examples

Example 1: Volume Under a Paraboloid

Problem: Find the volume under z = x² + y² and above the disk x² + y² ≤ 4.

Solution:

Step 1 — Identify region: The disk x² + y² ≤ 4 means r ∈ [0, 2], θ ∈ [0, 2π].

Step 2 — Convert integrand: z = x² + y² = r².

Step 3 — Set up integral with r factor:

$V = ∬_R (x² + y²) dA = ∫₀^(2π) ∫₀² r² · r dr dθ = ∫₀^(2π) ∫₀² r³ dr dθ
$

Step 4 — Integrate r:

$∫₀² r³ dr = [r⁴/4]₀² = 16/4 = 4
$

Step 5 — Integrate θ:

$V = ∫₀^(2π) 4 dθ = [4θ]₀^(2π) = 8π
$

Answer: Volume = 8π.

Sanity check: Compare to volume of cylinder of height 4 (the maximum z-value on the boundary), radius 2: V_cylinder = π·2²·4 = 16π. The paraboloid should have less volume, and 8π < 16π ✓. (A paraboloid of revolution has exactly half the volume of its circumscribing cylinder.)

Example 2: Area of a Petal

Problem: Find the area of one petal of the rose curve r = cos(3θ).

Solution:

Step 1 — Understand the curve: r = cos(3θ) traces a 3-petal rose. One petal occurs where cos(3θ) ≥ 0.

Solve: cos(3θ) = 0 → 3θ = π/2, 3π/2, ... → θ = π/6, π/2, ...

One petal: θ ∈ [−π/6, π/6].

Step 2 — Set up area integral:

$A = ∬_R 1 dA = ∫_(−π/6)^(π/6) ∫₀^(cos 3θ) r dr dθ
$

Step 3 — Inner integral (r):

$∫₀^(cos 3θ) r dr = [r²/2]₀^(cos 3θ) = (cos² 3θ)/2
$

Step 4 — Outer integral (θ):

$A = ∫_(−π/6)^(π/6) (cos² 3θ)/2 dθ = ½ ∫_(−π/6)^(π/6) cos² 3θ dθ
$

Use cos²φ = (1 + cos 2φ)/2 with φ = 3θ:

$cos² 3θ = (1 + cos 6θ)/2
$

$A = ½ ∫_(−π/6)^(π/6) (1 + cos 6θ)/2 dθ
  = ¼ ∫_(−π/6)^(π/6) (1 + cos 6θ) dθ
  = ¼ [θ + (sin 6θ)/6]_(−π/6)^(π/6)
  = ¼ [(π/6 + 0) − (−π/6 + 0)]
  = ¼ · (π/3) = π/12
$

Answer: Area of one petal = π/12.

Check: Total area of 3 petals = 3 × π/12 = π/4. The curve is entirely contained in the unit circle (r ≤ 1), so total area ≤ π, and π/4 < π ✓.

Example 3: Mass and Center of Mass

Problem: A lamina occupies the semi-circular region x² + y² ≤ 4, y ≥ 0, with density ρ(x, y) = x² + y² (in kg/m²). Find the mass and center of mass.

Solution:

Region in polar: r ∈ [0, 2], θ ∈ [0, π]. Density: ρ = r².

Mass:

$M = ∬_R ρ dA = ∫₀^π ∫₀² r² · r dr dθ
  = ∫₀^π ∫₀² r³ dr dθ
  = ∫₀^π [r⁴/4]₀² dθ = ∫₀^π 4 dθ = 4π  kg
$

x̄ — x-coordinate of center of mass:

$M · x̄ = ∬_R x ρ dA = ∫₀^π ∫₀² (r cos θ) · r² · r dr dθ
       = ∫₀^π ∫₀² r⁴ cos θ dr dθ
       = (∫₀² r⁴ dr) · (∫₀^π cos θ dθ)
       = [r⁵/5]₀² · [sin θ]₀^π
       = (32/5) · (0 − 0) = 0
$

So x̄ = 0 (symmetry — the region is symmetric about the y-axis).

ȳ — y-coordinate of center of mass:

$M · ȳ = ∬_R y ρ dA = ∫₀^π ∫₀² (r sin θ) · r² · r dr dθ
       = ∫₀^π ∫₀² r⁴ sin θ dr dθ
       = (∫₀² r⁴ dr) · (∫₀^π sin θ dθ)
       = (32/5) · [−cos θ]₀^π
       = (32/5) · (−(−1) − (−1)) = (32/5) · 2 = 64/5
$

$ȳ = (64/5) / (4π) = 64/(20π) = 16/(5π) ≈ 1.019
$

Answer: M = 4π kg, (x̄, ȳ) = (0, 16/(5π)) ≈ (0, 1.02).

Check: Since the density is heaviest at the boundary (r = 2), the center of mass should be higher than the geometric centroid of a semicircle (which is at ȳ = 4r/(3π) = 8/(3π) ≈ 0.849). 16/(5π) ≈ 1.02 > 0.85 ✓.

Example 4: Switching from Cartesian to Polar

Problem: Evaluate ∫₀^√π ∫₀^(√(π−x²)) sin(x² + y²) dy dx using polar coordinates.

Solution:

Step 1 — Identify the region: x ∈ [0, √π], y ∈ [0, √(π − x²)]. This describes the quarter-circle x² + y² ≤ π in the first quadrant. In polar: r ∈ [0, √π], θ ∈ [0, π/2].

Step 2 — Convert:

$∬_R sin(x² + y²) dA = ∫₀^(π/2) ∫₀^√π sin(r²) · r dr dθ
$

Step 3 — Inner integral (r): Let u = r², du = 2r dr → r dr = du/2. When r = 0, u = 0; when r = √π, u = π.

$∫₀^√π sin(r²) · r dr = ½ ∫₀^π sin u du = ½ [−cos u]₀^π = ½ (−(−1) − (−1)) = ½ (1 + 1) = 1
$

Step 4 — Outer integral (θ):

$∫₀^(π/2) 1 dθ = π/2
$

Answer: π/2.

Lesson: Without polar coordinates, ∫ sin(x² + y²) dy dx is intractable. With polar, the substitution r dr = du/2 makes it easy.

Pitfalls

Quiz

1. Forgetting the r factor. When converting dA = dx dy to polar, the Jacobian contributes one factor of r. The most common error is omitting it entirely.
2. Using r² instead of r. The Jacobian determinant |∂(x,y)/∂(r,θ)| = r, not r². Only add r² if your integrand itself contains r² (e.g., x² + y² = r²).
3. Wrong bounds for θ. Remember that θ sweeps from the lower-angle boundary to the upper-angle boundary of the region — never go the 'long way' around.
4. Confusing r_max with the integrand. In ∫∫_D f(x,y) dA, the r in r dr dθ is the Jacobian, not part of f.
5. Using polar for non-radially-symmetric regions without checking. If the region D has a boundary described by r = g(θ), that's fine, but if it's a rectangle or triangle in Cartesian, polar may not simplify things.
6. Dropping limits in the inner integral. When integrating r from 0 to r_max(θ), ensure you substitute the correct upper limit — it's a function of θ, not a constant.
7. Forgetting to convert f(x,y) to f(r cos θ, r sin θ). The integrand must be rewritten in polar before integrating.

Q1: When converting a double integral to polar coordinates, dA becomes:

A) dr dθ B) r dr dθ C) r² dr dθ D) (1/r) dr dθ

Correct: B)

• If you chose B: The area element in polar coordinates is dA = r dr dθ. The factor r is the Jacobian determinant. Correct!
• If you chose A: Forgetting the r factor is the most common polar integration mistake.
• If you chose C: r² would overcount area — the Jacobian is r, not r².
• If you chose D: The Jacobian is r (area expands with radius), not 1/r.

Q2: Which regions are best suited for polar coordinates?

A) Rectangular regions B) Circular, annular, or sector-shaped regions C) Triangular regions D) Regions bounded by straight lines only

Correct: B)

• If you chose B: Polar coordinates excel when the region or integrand involves circles, disks, or expressions like x² + y². Correct!
• If you chose A: Rectangular regions are easiest in Cartesian coordinates.
• If you chose C: Triangles are generally easier in Cartesian unless they have circular symmetry.
• If you chose D: Polar coordinates are for curved (circular) boundaries, not straight lines.

Q3: To integrate over a full disk of radius R, the polar limits are:

A) θ: 0 to π, r: 0 to R B) θ: 0 to 2π, r: 0 to R C) θ: 0 to 2π, r: −R to R D) θ: 0 to π/2, r: 0 to R

Correct: B)

• If you chose B: A full disk requires θ from 0 to 2π and r from 0 to R. Correct!
• If you chose A: That covers only a semicircle.
• If you chose C: r is nonnegative in standard polar coordinates.
• If you chose D: That covers only the first quadrant.

Q4: ∫₀^{2π} ∫₀^R e^{-r²} r dr dθ evaluates to:

A) 2π(1 − e^{-R²}) B) π(1 − e^{-R²}) C) 2πe^{-R²} D) πe^{-R²}

Correct: B)

• If you chose B: Inner: ∫₀^R e^{-r²} r dr = [−½e^{-r²}]₀^R = ½(1 − e^{-R²}). Outer integral over 2π gives π(1 − e^{-R²}). Correct!
• If you chose A: Forgot the ½ from integrating r e^{-r²}.
• If you chose C, D: Sign error in antiderivative evaluation.

Q5: When converting f(x, y) = x² + y² to polar coordinates, it becomes:

A) r B) r² C) 2r D) r² cos θ sin θ

Correct: B)

• If you chose B: x² + y² = r² cos²θ + r² sin²θ = r²(cos²θ + sin²θ) = r². Correct!
• If you chose A: That's √(x² + y²) = r, not x² + y².
• If you chose C: No factor of 2 appears.
• If you chose D: This would be xy, not x² + y².

Q6: To find the area inside the cardioid r = 1 + cos θ, the correct setup is:

A) ∫₀^{2π} ∫₀^{1+cos θ} r dr dθ B) ∫₀^{2π} ∫₀^{1+cos θ} dr dθ C) ∫₀^{π} ∫₀^{1+cos θ} r dr dθ D) ∫₀^{2π} (1 + cos θ) r dθ

Correct: A)

• If you chose A: Area = ∬ 1 dA = ∫₀^{2π} ∫₀^{1+cos θ} r dr dθ. The r in the integrand is essential. Correct!
• If you chose B: Missing the r factor in the area element.
• If you chose C: The cardioid is traced for θ ∈ [0, 2π].
• If you chose D: This misses the r dr integration.

Practice Problems

Solve each problem, then check your answers at the end.

P1: Evaluate ∬_R (x² + y²) dA where R is the annular region 1 ≤ x² + y² ≤ 4.

P2: Find the volume under z = 9 − x² − y² and above the xy-plane.

P3: Find the area of the region inside the cardioid r = 1 + sin θ and outside the circle r = 1.

P4: A uniform lamina occupies the region inside r = 2 cos θ (a circle). Find its centroid.

P5: Evaluate ∫₀^√2 ∫_y^(√(2−y²)) (x² + y²) dx dy by converting to polar coordinates.

P6: Find the mass of a lamina in the shape of the disk r ≤ 3 with density ρ(r, θ) = e^(−r²).

Summary

1. The polar transformation for double integrals is: x = r cos θ, y = r sin θ, and critically, dA = r dr dθ — the r factor comes from the Jacobian and must NEVER be omitted
2. Regions with circular symmetry (disks, annuli, sectors, cardioids, roses) are described simply in polar: r bounds as functions of θ, then θ ranges over an interval
3. Integrands containing x² + y² become r² in polar, enabling simplification and integration that is often impossible in Cartesian coordinates
4. Applications like area, mass, center of mass, and moments of inertia follow the same double-integral formulas with the polar area element substituted
5. When integrating powers of sin θ and cos θ over [0, 2π], use symmetry (odd powers integrate to zero) and the half-angle/power-reduction identities for even powers

Next Steps

Move on to 07-01 — Triple Integrals to extend integration to three dimensions — setting up iterated integrals over general 3D regions, computing volume, mass, and moments, and mastering the skill of determining integration bounds from region descriptions.