Separability in Polar Integrals | Tony's Intertwingled Memex

Core Definition

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

f(r, \theta) = g(r) \cdot h(\theta)

then by Fubini's Theorem, the double integral factors into the direct algebraic product of two independent single-variable definite integrals:

\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)

(Note: The Jacobian factor $r$ is always grouped with the radial function $g(r)$ ).

Concept Explanation

1. The Two Non-Negotiable Conditions for Separability

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

Constant Polar Boundaries (Polar Rectangle): All four limits of integration ( $r \in [a, b]$ and $\theta \in [\alpha, \beta]$ ) must be strictly numeric constants. If the radial bound depends on the angle (e.g., $0 \le r \le 1 + \cos\theta$ or $r \le 2\sin\theta$ ), the inner variable cannot be isolated from the outer boundary, and the shortcut cannot be used.
Multiplicative Integrand Structure: The function $f(r, \theta)$ must be a pure product of radial and angular terms. If the integrand contains un-factorable sums (such as $r^2 + \sin\theta$ or $\sqrt{r + \cos\theta}$ ), 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 $\sin\theta$ or $\cos\theta$ over an entire symmetric period $[0, 2\pi]$ or $[-\pi/2, \pi/2]$ ), 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	$\int_{0}^{2\pi} \int_{1}^{3} r^3 \cos^2\theta \, dr \, d\theta$	Product: $\left(\int_0^{2\pi} \cos^2\theta \, d\theta\right) \left(\int_1^3 r^3 \, dr\right)$
Additive Integrand	$\int_{0}^{\pi} \int_{0}^{2} (r^2 + \sin\theta) \, r \, dr \, d\theta$	Must evaluate nested inside-out, or split by linearity into sum of two integrals
Variable Radial Bound	$\int_{0}^{2\pi} \int_{0}^{1+\cos\theta} r \, dr \, d\theta$	Strictly sequential: Inner $r$ -evaluation produces functions of $\theta$ that feed the outer integral

Engineering Applications

Structural Mechanics: Calculating centroid coordinates $(\bar{x}, \bar{y})$ and second moments of area $I_x = \iint y^2 \, dA = \iint r^3 \sin^2\theta \, dr \, d\theta$ for symmetric annular circular columns.
Aerodynamics & Turbomachinery: Evaluating axisymmetric velocity distributions $v(r, \theta) = v(r)$ over circular rotor disks, where the angular integral trivially reduces to a factor of $2\pi$ .
Electromagnetics: Computing electric flux $\Phi_E = \iint_D E(r) \cos\theta \, dA$ across aperture antennas and microwave waveguide cross-sections.

Example Problems

Example 1: Evaluating via the Product Separability Shortcut

Evaluate the double integral:

I = \iint_R r^2 \sin\theta \, dA

over the semicircular region

R = \{(r, \theta) \mid 0 \le r \le 2, \, 0 \le \theta \le \pi\}

Solution

Step 1: Check Separability Conditions

Bounds are constants: $r \in [0, 2]$ and $\theta \in [0, \pi]$ .
Full integrand with the Jacobian $r$ : $f(r, \theta) \cdot r = (r^2 \sin\theta) \cdot r = r^3 \sin\theta$ This is purely multiplicative: $g(r) = r^3$ and $h(\theta) = \sin\theta$ .

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

I = \left( \int_{0}^{\pi} \sin\theta \, d\theta \right) \left( \int_{0}^{2} r^3 \, dr \right)

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

\int_{0}^{\pi} \sin\theta \, d\theta = [-\cos\theta]_0^\pi = (-\cos\pi) - (-\cos 0) = -(-1) - (-1) = 1 + 1 = 2

Evaluate the radial integral:

\int_{0}^{2} r^3 \, dr = \left[ \frac{r^4}{4} \right]_0^2 = \frac{16}{4} - 0 = 4

Step 4: Compute the Final Product

I = 2 \times 4 = 8

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

Find the area of the region enclosed by the cardioid $r = 1 + \cos\theta$ for $0 \le \theta \le 2\pi$ .

Solution

Step 1: Formulate the Region Bounds
The upper radial limit depends on $\theta$ :

0 \le r \le 1 + \cos\theta, \quad 0 \le \theta \le 2\pi

Because the upper bound is a function of

\theta

, the integral is non-separable and must be integrated sequentially from the inside out.

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

\text{Area} = \int_{0}^{2\pi} \int_{0}^{1+\cos\theta} r \, dr \, d\theta

Evaluate the inner integral with respect to

r

\int_{0}^{1+\cos\theta} r \, dr = \left[ \frac{r^2}{2} \right]_{0}^{1+\cos\theta} = \frac{1}{2}(1 + \cos\theta)^2

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

\frac{1}{2}(1 + 2\cos\theta + \cos^2\theta)

Substitute the half-angle identity

\cos^2\theta = \frac{1 + \cos(2\theta)}{2}

= \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)

Step 4: Evaluate the Outer Integral ( $d\theta$ )

\text{Area} = \frac{1}{2} \left[ \frac{3}{2}\theta + 2\sin\theta + \frac{1}{4}\sin(2\theta) \right]_{0}^{2\pi}

Evaluate at the boundaries:

\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}

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