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Separability in Polar Integrals

Formulas & Definitions
Core Definition

If a double integral in polar coordinates is evaluated over a polar rectangle R={(r,θ)∣a≤r≤b, α≤θ≤β}R = \{(r, \theta) \mid a \le r \le b, \, \alpha \le \theta \le \beta\}R={(r,θ)∣a≤r≤b,α≤θ≤β} and its integrand can be factored completely into a single-variable function of rrr multiplied by a single-variable function of θ\thetaθ:

f(r,θ)=g(r)⋅h(θ)f(r, \theta) = g(r) \cdot h(\theta)f(r,θ)=g(r)⋅h(θ)
then by Fubini's Theorem, the double integral factors into the direct algebraic product of two independent single-variable definite integrals:
∬Rg(r)h(θ) dA=(∫αβh(θ) dθ)×(∫abg(r)⋅r dr)\iint_R g(r) h(\theta) \, dA = \left( \int_{\alpha}^{\beta} h(\theta) \, d\theta \right) \times \left( \int_{a}^{b} g(r) \cdot r \, dr \right)∬R​g(r)h(θ)dA=(∫αβ​h(θ)dθ)×(∫ab​g(r)⋅rdr)
(Note: The Jacobian factor rrr is always grouped with the radial function g(r)g(r)g(r)).

Concept Explanation

1. The Two Non-Negotiable Conditions for Separability

To apply the product separability shortcut, both of the following criteria must hold simultaneously:

  1. Constant Polar Boundaries (Polar Rectangle): All four limits of integration (r∈[a,b]r \in [a, b]r∈[a,b] and θ∈[α,β]\theta \in [\alpha, \beta]θ∈[α,β]) must be strictly numeric constants. If the radial bound depends on the angle (e.g., 0≤r≤1+cos⁡θ0 \le r \le 1 + \cos\theta0≤r≤1+cosθ or r≤2sin⁡θr \le 2\sin\thetar≤2sinθ), the inner variable cannot be isolated from the outer boundary, and the shortcut cannot be used.
  2. Multiplicative Integrand Structure: The function f(r,θ)f(r, \theta)f(r,θ) must be a pure product of radial and angular terms. If the integrand contains un-factorable sums (such as r2+sin⁡θr^2 + \sin\thetar2+sinθ or r+cos⁡θ\sqrt{r + \cos\theta}r+cosθ​), the integral must be evaluated sequentially from the inside out using nested integration.

2. Efficiency Advantage in Multivariable Engineering

Evaluating nested iterated integrals requires carrying outer variables as constants through antiderivative substitutions, which increases algebraic manipulation and the risk of sign errors.

When separability applies, the two single-variable integrals can be evaluated simultaneously and in parallel. Furthermore, if either independent factor evaluates to zero (for instance, integrating sin⁡θ\sin\thetasinθ or cos⁡θ\cos\thetacosθ over an entire symmetric period [0,2π][0, 2\pi][0,2π] or [−π/2,π/2][-\pi/2, \pi/2][−π/2,π/2]), the entire double integral immediately vanishes to zero without requiring computation of the radial component.

3. Contrast: Separable vs. Non-Separable Cases

Case Formulation Valid Method
Separable ∫02π∫13r3cos⁡2θ dr dθ\int_{0}^{2\pi} \int_{1}^{3} r^3 \cos^2\theta \, dr \, d\theta∫02π​∫13​r3cos2θdrdθ Product: (∫02πcos⁡2θ dθ)(∫13r3 dr)\left(\int_0^{2\pi} \cos^2\theta \, d\theta\right) \left(\int_1^3 r^3 \, dr\right)(∫02π​cos2θdθ)(∫13​r3dr)
Additive Integrand ∫0π∫02(r2+sin⁡θ) r dr dθ\int_{0}^{\pi} \int_{0}^{2} (r^2 + \sin\theta) \, r \, dr \, d\theta∫0π​∫02​(r2+sinθ)rdrdθ Must evaluate nested inside-out, or split by linearity into sum of two integrals
Variable Radial Bound ∫02π∫01+cos⁡θr dr dθ\int_{0}^{2\pi} \int_{0}^{1+\cos\theta} r \, dr \, d\theta∫02π​∫01+cosθ​rdrdθ Strictly sequential: Inner rrr-evaluation produces functions of θ\thetaθ that feed the outer integral

Engineering Applications

  • Structural Mechanics: Calculating centroid coordinates (xˉ,yˉ)(\bar{x}, \bar{y})(xˉ,yˉ​) and second moments of area Ix=∬y2 dA=∬r3sin⁡2θ dr dθI_x = \iint y^2 \, dA = \iint r^3 \sin^2\theta \, dr \, d\thetaIx​=∬y2dA=∬r3sin2θdrdθ for symmetric annular circular columns.
  • Aerodynamics & Turbomachinery: Evaluating axisymmetric velocity distributions v(r,θ)=v(r)v(r, \theta) = v(r)v(r,θ)=v(r) over circular rotor disks, where the angular integral trivially reduces to a factor of 2π2\pi2π.
  • Electromagnetics: Computing electric flux ΦE=∬DE(r)cos⁡θ dA\Phi_E = \iint_D E(r) \cos\theta \, dAΦE​=∬D​E(r)cosθdA across aperture antennas and microwave waveguide cross-sections.

Example Problems

Example 1: Evaluating via the Product Separability Shortcut

Evaluate the double integral:

I=∬Rr2sin⁡θ dAI = \iint_R r^2 \sin\theta \, dAI=∬R​r2sinθdA
over the semicircular region R={(r,θ)∣0≤r≤2, 0≤θ≤π}R = \{(r, \theta) \mid 0 \le r \le 2, \, 0 \le \theta \le \pi\}R={(r,θ)∣0≤r≤2,0≤θ≤π}.

Solution

Step 1: Check Separability Conditions

  1. Bounds are constants: r∈[0,2]r \in [0, 2]r∈[0,2] and θ∈[0,π]\theta \in [0, \pi]θ∈[0,π].
  2. Full integrand with the Jacobian rrr:
    f(r,θ)⋅r=(r2sin⁡θ)⋅r=r3sin⁡θf(r, \theta) \cdot r = (r^2 \sin\theta) \cdot r = r^3 \sin\thetaf(r,θ)⋅r=(r2sinθ)⋅r=r3sinθ
    This is purely multiplicative: g(r)=r3g(r) = r^3g(r)=r3 and h(θ)=sin⁡θh(\theta) = \sin\thetah(θ)=sinθ.

Step 2: Split into Independent Factors
State Fubini's Separability Theorem in Polar Coordinates:

I=(∫0πsin⁡θ dθ)(∫02r3 dr)I = \left( \int_{0}^{\pi} \sin\theta \, d\theta \right) \left( \int_{0}^{2} r^3 \, dr \right)I=(∫0π​sinθdθ)(∫02​r3dr)

Step 3: Evaluate Each Integral Independently
Evaluate the angular integral:

∫0πsin⁡θ dθ=[−cos⁡θ]0π=(−cos⁡π)−(−cos⁡0)=−(−1)−(−1)=1+1=2\int_{0}^{\pi} \sin\theta \, d\theta = [-\cos\theta]_0^\pi = (-\cos\pi) - (-\cos 0) = -(-1) - (-1) = 1 + 1 = 2∫0π​sinθdθ=[−cosθ]0π​=(−cosπ)−(−cos0)=−(−1)−(−1)=1+1=2
Evaluate the radial integral:
∫02r3 dr=[r44]02=164−0=4\int_{0}^{2} r^3 \, dr = \left[ \frac{r^4}{4} \right]_0^2 = \frac{16}{4} - 0 = 4∫02​r3dr=[4r4​]02​=416​−0=4

Step 4: Compute the Final Product

I=2×4=8I = 2 \times 4 = 8I=2×4=8

Example 2: Non-Separable Nested Evaluation (Cardioid Enclosed Area)

Find the area of the region enclosed by the cardioid r=1+cos⁡θr = 1 + \cos\thetar=1+cosθ for 0≤θ≤2π0 \le \theta \le 2\pi0≤θ≤2π.

Solution

Step 1: Formulate the Region Bounds
The upper radial limit depends on θ\thetaθ:

0≤r≤1+cos⁡θ,0≤θ≤2π0 \le r \le 1 + \cos\theta, \quad 0 \le \theta \le 2\pi0≤r≤1+cosθ,0≤θ≤2π
Because the upper bound is a function of θ\thetaθ, the integral is non-separable and must be integrated sequentially from the inside out.

Step 2: Set Up and Evaluate the Inner Integral (drdrdr)
State the General Polar Area Integral:

Area=∫02π∫01+cos⁡θr dr dθ\text{Area} = \int_{0}^{2\pi} \int_{0}^{1+\cos\theta} r \, dr \, d\thetaArea=∫02π​∫01+cosθ​rdrdθ
Evaluate the inner integral with respect to rrr:
∫01+cos⁡θr dr=[r22]01+cos⁡θ=12(1+cos⁡θ)2\int_{0}^{1+\cos\theta} r \, dr = \left[ \frac{r^2}{2} \right]_{0}^{1+\cos\theta} = \frac{1}{2}(1 + \cos\theta)^2∫01+cosθ​rdr=[2r2​]01+cosθ​=21​(1+cosθ)2

Step 3: Expand and Apply Half-Angle Identity
Expand the integrand:

12(1+2cos⁡θ+cos⁡2θ)\frac{1}{2}(1 + 2\cos\theta + \cos^2\theta)21​(1+2cosθ+cos2θ)
Substitute the half-angle identity cos⁡2θ=1+cos⁡(2θ)2\cos^2\theta = \frac{1 + \cos(2\theta)}{2}cos2θ=21+cos(2θ)​:
=12(1+2cos⁡θ+12+12cos⁡(2θ))=12(32+2cos⁡θ+12cos⁡(2θ))= \frac{1}{2}\left(1 + 2\cos\theta + \frac{1}{2} + \frac{1}{2}\cos(2\theta)\right) = \frac{1}{2}\left(\frac{3}{2} + 2\cos\theta + \frac{1}{2}\cos(2\theta)\right)=21​(1+2cosθ+21​+21​cos(2θ))=21​(23​+2cosθ+21​cos(2θ))

Step 4: Evaluate the Outer Integral (dθd\thetadθ)

Area=12[32θ+2sin⁡θ+14sin⁡(2θ)]02π\text{Area} = \frac{1}{2} \left[ \frac{3}{2}\theta + 2\sin\theta + \frac{1}{4}\sin(2\theta) \right]_{0}^{2\pi}Area=21​[23​θ+2sinθ+41​sin(2θ)]02π​
Evaluate at the boundaries:
Area=12[(32(2π)+2sin⁡(2π)+14sin⁡(4π))−0]=12(3π+0+0)=3π2\text{Area} = \frac{1}{2} \left[ \left(\frac{3}{2}(2\pi) + 2\sin(2\pi) + \frac{1}{4}\sin(4\pi)\right) - 0 \right] = \frac{1}{2}(3\pi + 0 + 0) = \frac{3\pi}{2}Area=21​[(23​(2π)+2sin(2π)+41​sin(4π))−0]=21​(3π+0+0)=23π​

Connections & References

  • Parent Concepts: Double Integrals in Polar Coordinates, Double Integrals over a Rectangular Region
  • Related Notes: Polar Rectangles and Annular Sectors, Volume and Surface Integration in Polar Coordinates
  • Course Sessions: Session 15 - Double Integrals in Polar Coordinates