Multivariable-Calculus · Unit 15 · Video 1 · Interactive Practice

Double Integrals in Polar Coordinates: Why dA = r dr dθ

IKey Formulas

FormulaNameWhat it does
x=rcos⁡θ,y=rsin⁡θ,x2+y2=r2x = r\cos\theta,\quad y = r\sin\theta,\quad x^2 + y^2 = r^2Polar conversionRewrites the integrand in rr and θ\theta
ΔA≈(r Δθ) Δr  ⇒  dA=r dr dθ\Delta A \approx (r\,\Delta\theta)\,\Delta r \;\Rightarrow\; dA = r\,dr\,d\thetaPolar area elementSides Δr\Delta r (along the ray) and r Δθr\,\Delta\theta (along the arc)
∬Rf dA=∫θ1θ2 ⁣∫r1(θ)r2(θ)f(rcos⁡θ,rsin⁡θ) r dr dθ\displaystyle\iint_R f\,dA = \int_{\theta_1}^{\theta_2}\!\int_{r_1(\theta)}^{r_2(\theta)} f(r\cos\theta, r\sin\theta)\,r\,dr\,d\thetaPolar setup, rr firstFix θ\theta: the ray enters at r1r_1, leaves at r2r_2; then sweep θ\theta
∫0π/2 ⁣∫01(1−r2) r dr dθ=14⋅π2=π8\displaystyle\int_0^{\pi/2}\!\int_0^1 (1 - r^2)\,r\,dr\,d\theta = \tfrac{1}{4}\cdot\tfrac{\pi}{2} = \tfrac{\pi}{8}Quarter disk, 1−x2−y21 - x^2 - y^2Constant bounds; each ray contributes 14\tfrac14

Key Insight: The arc side of a polar piece has length r Δθr\,\Delta\theta, not Δθ\Delta\theta, because rays spread apart as rr grows — so dAdA is never dr dθdr\,d\theta alone.

IIEqual Steps, Unequal Pieces

With the same Δr\Delta r and Δθ\Delta\theta, a polar piece's area grows with rr, because its arc side is r Δθr\,\Delta\theta.

IIIInner Arc, Outer Arc, and the Limit

The inner and outer estimates of a polar piece differ by the fraction Δr/r\Delta r / r, which vanishes as Δr→0\Delta r \to 0.

IVRay by Ray: The Quarter Disk

Each fixed θ\theta is a ray from r=0r = 0 to r=1r = 1; integrate along it, then sweep θ\theta from 00 to π/2\pi/2.

VQuiz Questions

Problem 1 · Area of One Polar Piece

Given: a small polar piece with inner radius r=2r = 2, radial width Δr=0.1\Delta r = 0.1 and angular width Δθ=0.05\Delta\theta = 0.05 radians — estimate its area ΔA\Delta A using the inner arc.

✅ Correct! ΔA≈(r Δθ) Δr=(2)(0.05)(0.1)=0.01\Delta A \approx (r\,\Delta\theta)\,\Delta r = (2)(0.05)(0.1) = 0.01.
❌ That is Δr Δθ\Delta r\,\Delta\theta alone. The arc side is r Δθ=2(0.05)=0.1r\,\Delta\theta = 2(0.05) = 0.1, not Δθ=0.05\Delta\theta = 0.05 — the factor rr is missing.
❌ One factor of rr too many. Only the arc side carries rr; the radial side is just Δr\Delta r, so ΔA≈r Δr Δθ\Delta A \approx r\,\Delta r\,\Delta\theta, not r2 Δr Δθr^2\,\Delta r\,\Delta\theta.
❌ The angular width is missing. r Δr=0.2r\,\Delta r = 0.2 is not an area of the piece; the arc side is r Δθr\,\Delta\theta, so multiply by Δθ=0.05\Delta\theta = 0.05 as well.
Show solution

The piece is nearly a rectangle. Its side along the ray has length Δr=0.1\Delta r = 0.1; its side along the inner arc has length radius × angle:

r Δθ=2×0.05=0.1.r\,\Delta\theta = 2 \times 0.05 = 0.1. ΔA≈(r Δθ) Δr=0.1×0.1=0.01.\Delta A \approx (r\,\Delta\theta)\,\Delta r = 0.1 \times 0.1 = 0.01.

The exact area 12((2.1)2−22)(0.05)=0.01025\tfrac12\big((2.1)^2 - 2^2\big)(0.05) = 0.01025 lies between this inner estimate and the outer estimate (2.1)(0.05)(0.1)=0.0105(2.1)(0.05)(0.1) = 0.0105, which differ by the fraction Δr/r=0.05\Delta r / r = 0.05.

Problem 2 · Area of a Disk

Given: the disk D:x2+y2≤9D: x^2 + y^2 \le 9 — find its area ∬D1 dA\displaystyle\iint_D 1\,dA in polar coordinates.

✅ Correct! ∫02π∫03r dr dθ=92⋅2π=9π\int_0^{2\pi}\int_0^3 r\,dr\,d\theta = \tfrac{9}{2}\cdot 2\pi = 9\pi, matching πr2\pi r^2 with radius 33.
❌ That uses dr dθdr\,d\theta. ∫02π∫03dr dθ=3⋅2π\int_0^{2\pi}\int_0^3 dr\,d\theta = 3 \cdot 2\pi counts every piece as if it had area Δr Δθ\Delta r\,\Delta\theta; the area element is r dr dθr\,dr\,d\theta.
❌ The 12\tfrac12 was dropped. ∫03r dr=[12r2]03=92\int_0^3 r\,dr = \big[\tfrac12 r^2\big]_0^3 = \tfrac92, not 99.
❌ That sweeps only half the disk. The rays must cover the whole disk, so θ\theta runs from 00 to 2π2\pi, not 00 to π\pi.
Show solution

Fix θ\theta: the ray starts inside at r=0r = 0 and leaves on the circle r=3r = 3. Every direction is used, so 0≤θ≤2π0 \le \theta \le 2\pi.

∬D1 dA=∫02π ⁣∫03r dr dθ=∫02π[12r2]03dθ=∫02π92 dθ=9π.\iint_D 1\,dA = \int_0^{2\pi}\!\int_0^3 r\,dr\,d\theta = \int_0^{2\pi} \Big[\tfrac12 r^2\Big]_0^3 d\theta = \int_0^{2\pi} \tfrac92\,d\theta = 9\pi.

This is π⋅32\pi \cdot 3^2, as it must be.

Problem 3 · Half Annulus

Given: RR is the upper half of the ring 1≤x2+y2≤41 \le x^2 + y^2 \le 4, y≥0y \ge 0 — evaluate ∬R(x2+y2) dA\displaystyle\iint_R (x^2 + y^2)\,dA.

✅ Correct! ∫0π∫12r2⋅r dr dθ=154⋅π\int_0^{\pi}\int_1^2 r^2\cdot r\,dr\,d\theta = \tfrac{15}{4}\cdot\pi.
❌ The area element is missing its rr. ∫0π∫12r2 dr dθ=73π\int_0^\pi\int_1^2 r^2\,dr\,d\theta = \tfrac73\pi uses dr dθdr\,d\theta; with dA=r dr dθdA = r\,dr\,d\theta the integrand becomes r3r^3.
❌ That sweeps the whole ring. y≥0y \ge 0 keeps only the upper half, so θ\theta runs from 00 to π\pi, not 2π2\pi.
❌ The hole was filled in. Each ray enters the region at r=1r = 1, not r=0r = 0: ∫12r3 dr=154\int_1^2 r^3\,dr = \tfrac{15}{4}, while ∫02r3 dr=4\int_0^2 r^3\,dr = 4 integrates over the full half disk.
Show solution

Bounds: fix θ\theta; the ray enters RR on the circle r=1r = 1 and leaves on r=2r = 2. The rays with y≥0y \ge 0 are 0≤θ≤π0 \le \theta \le \pi.

Integrand: x2+y2=r2x^2 + y^2 = r^2, and dA=r dr dθdA = r\,dr\,d\theta:

∬R(x2+y2) dA=∫0π ⁣∫12r2⋅r dr dθ.\iint_R (x^2 + y^2)\,dA = \int_0^{\pi}\!\int_1^2 r^2 \cdot r\,dr\,d\theta. ∫12r3 dr=[14r4]12=164−14=154.\int_1^2 r^3\,dr = \Big[\tfrac14 r^4\Big]_1^2 = \tfrac{16}{4} - \tfrac14 = \tfrac{15}{4}. ∫0π154 dθ=15π4.\int_0^{\pi} \tfrac{15}{4}\,d\theta = \tfrac{15\pi}{4}.

Problem 4 · The Trade-off: Integrand x

Given: RR is the quarter of the unit disk in the first quadrant — evaluate ∬Rx dA\displaystyle\iint_R x\,dA in polar coordinates.

✅ Correct! ∫0π/2∫01r2cos⁡θ dr dθ=13⋅1=13\int_0^{\pi/2}\int_0^1 r^2\cos\theta\,dr\,d\theta = \tfrac13 \cdot 1 = \tfrac13.
❌ The area element is missing its rr. x=rcos⁡θx = r\cos\theta, and dA=r dr dθdA = r\,dr\,d\theta adds a second factor of rr: the integrand is r2cos⁡θr^2\cos\theta, not rcos⁡θr\cos\theta.
❌ Sign error in the θ\theta-integral. An antiderivative of cos⁡θ\cos\theta is sin⁡θ\sin\theta, so ∫0π/2cos⁡θ dθ=1\int_0^{\pi/2}\cos\theta\,d\theta = 1; x≥0x \ge 0 on RR, so the integral cannot be negative.
❌ The cos⁡θ\cos\theta was dropped. x=rcos⁡θx = r\cos\theta, not rr; ∫0π/2∫01r2 dr dθ=π6\int_0^{\pi/2}\int_0^1 r^2\,dr\,d\theta = \tfrac{\pi}{6} integrates x2+y2\sqrt{x^2 + y^2} instead.
Show solution

Bounds: 0≤r≤10 \le r \le 1, 0≤θ≤π/20 \le \theta \le \pi/2. Integrand: x=rcos⁡θx = r\cos\theta, times the area element r dr dθr\,dr\,d\theta:

∬Rx dA=∫0π/2 ⁣∫01r2cos⁡θ dr dθ.\iint_R x\,dA = \int_0^{\pi/2}\!\int_0^1 r^2\cos\theta\,dr\,d\theta.

Inner integral (θ fixed): ∫01r2cos⁡θ dr=13cos⁡θ\displaystyle\int_0^1 r^2\cos\theta\,dr = \tfrac13\cos\theta — it now depends on θ\theta.

∫0π/213cos⁡θ dθ=13[sin⁡θ]0π/2=13.\int_0^{\pi/2} \tfrac13\cos\theta\,d\theta = \tfrac13\Big[\sin\theta\Big]_0^{\pi/2} = \tfrac13.

Check in rectangular coordinates: ∫01x1−x2 dx=[−13(1−x2)3/2]01=13\displaystyle\int_0^1 x\sqrt{1 - x^2}\,dx = \Big[-\tfrac13(1 - x^2)^{3/2}\Big]_0^1 = \tfrac13.

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