Questions: Double Integrals in Polar Coordinates

Question 1

A student converts ∬_D (x² + y²) dA over a disk of radius 2 to polar and writes the integral as ∫₀^{2π} ∫₀^2 r² dr dθ. What is wrong with this setup?

• A. The integrand is wrong — x² + y² does not equal r² in polar coordinates
• B. The limits on r should run from -2 to 2 to cover the full disk
• C. The area element is missing the factor r — the correct integrand is r² · r = r³
• D. The limits on θ should be 0 to π, not 0 to 2π, for a full disk

Answer: (C)

In polar coordinates x² + y² = r² is correct, and the disk 0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π is correct. The error is in the area element: dA = r dr dθ, not dr dθ. The correct integral is ∫₀^{2π} ∫₀^2 r² · r dr dθ = ∫₀^{2π} ∫₀^2 r³ dr dθ. Omitting the factor r is the single most common mistake in polar integration.

Question 2

Why does the polar area element include the factor r, making dA = r dr dθ rather than just dr dθ?

• A. The factor r converts angular measure from radians to arc length units
• B. A small wedge-shaped patch at radius r and angle dθ has arc length r dθ, so its area dr · r dθ grows with distance from the origin
• C. The factor r ensures the Jacobian determinant equals 1, preserving area under the transformation
• D. Without r, the integral would give volume instead of area

Answer: (B)

A polar 'pixel' at radius r spanning dr radially and dθ angularly is not a rectangle — it is a curved wedge. Its radial side has length dr, but its arc side has length r dθ (arc length = radius × angle). So the area is approximately dr · r dθ = r dr dθ. Near the origin r is small, so patches are tiny; far from the origin r is large, so equal parameter steps dr and dθ span much more area. This geometric fact is the Jacobian of the polar transformation.

Question 3

Near the origin, small polar patches (equal Δr and Δθ) represent much smaller areas than identical patches far from the origin.

• T. True
• F. False

Answer: True

True. The area of a polar patch is approximately r · Δr · Δθ. Since r → 0 near the origin, the factor r makes the actual area very small there, even for the same parameter increments Δr and Δθ. This is why the r factor is not optional — equal steps in the parameter space (r, θ) correspond to vastly different areas depending on where you are.

Question 4

You can convert a double integral from Cartesian to polar by substituting x = r cos θ and y = r sin θ in the integrand and writing dA = dr dθ.

• T. True
• F. False

Answer: False

False. The substitution x = r cos θ, y = r sin θ in the integrand is correct, but the area element is dA = r dr dθ, not dr dθ. The factor r is the Jacobian of the transformation and is never optional. Without it, the integral assigns equal weight to polar patches that actually have very different areas, producing an incorrect answer. For example, the area of a disk of radius R would come out wrong.

Question 5

Explain geometrically why the polar area element is r dr dθ and not simply dr dθ. What would go wrong if you omitted the factor r?

Model answer: A small patch in polar coordinates at position (r, θ) has radial width dr and arc length r dθ (not just dθ), making its area approximately dr · r dθ = r dr dθ. The factor r reflects that equal angular steps dθ sweep out more arc length — and hence more area — the farther you are from the origin. Without r, you would be treating all polar patches as having the same area, which is only true at r = 1. The result would overcount distant regions and undercount regions near the origin, giving an incorrect value for any area or integral over the disk.

The Jacobian of the coordinate change (x,y) → (r,θ) equals r, encoding this geometric distortion. The classic example: integrating 1 over a disk of radius R gives πR² correctly with the r factor; without it you get 2πR instead. The r factor is the mathematical way of saying 'account for the actual size of each coordinate patch.'