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Polar Rectangles and Annular Sectors

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Core Definition

A polar rectangle R⊂R2R \subset \mathbb{R}^2 is a planar region bounded by concentric circular arcs of constant radii and radial rays of constant angles:

R={(r,θ)∈R2∣a≤r≤b, α≤θ≤β}R = \{(r, \theta) \in \mathbb{R}^2 \mid a \le r \le b, \, \alpha \le \theta \le \beta\}
where 0≤a<b0 \le a < b and β−α≤2π\beta - \alpha \le 2\pi.

The planar area of a polar rectangle is:

Area(R)=∬R1 dA=∫αβ∫abr dr dθ=12(b2−a2)(β−α)\text{Area}(R) = \iint_R 1 \, dA = \int_{\alpha}^{\beta} \int_{a}^{b} r \, dr \, d\theta = \frac{1}{2}(b^2 - a^2)(\beta - \alpha)
When rr is bounded by continuous angle-dependent functions g1(θ)≤r≤g2(θ)g_1(\theta) \le r \le g_2(\theta), the domain is classified as a general polar region.

Concept Explanation

1. Geometry of a Polar Rectangle

In Cartesian coordinates, a rectangle [a,b]×[c,d][a, b] \times [c, d] has straight, orthogonal boundaries parallel to the coordinate axes.

In polar coordinates (r,θ)(r, \theta), a "rectangle" in the rθr\theta-plane corresponds to an annular sector in the physical xyxy-plane:

  • The inner and outer boundary curves r=ar = a and r=br = b are concentric circles centered at the origin.
  • The angular edges θ=α\theta = \alpha and θ=β\theta = \beta are straight radial rays emanating outward from the origin.

Because the width of a polar sector grows proportionally with distance from the origin (s=rθs = r\theta), the area element expands linearly with radius:

ΔA≈r∗ Δr Δθ\Delta A \approx r^* \, \Delta r \, \Delta \theta

2. Common Geometries as Polar Rectangles

  • Full Circular Disk of Radius RR: a=0, b=R, α=0, β=2πa = 0, \, b = R, \, \alpha = 0, \, \beta = 2\pi.
    Area=∫02π∫0Rr dr dθ=(2π)[12R2]=πR2\text{Area} = \int_{0}^{2\pi} \int_{0}^{R} r \, dr \, d\theta = (2\pi)\left[\frac{1}{2}R^2\right] = \pi R^2
  • Annulus (Concentric Ring): a=Rin, b=Rout, α=0, β=2πa = R_{\text{in}}, \, b = R_{\text{out}}, \, \alpha = 0, \, \beta = 2\pi.
    Area=π(Rout2−Rin2)\text{Area} = \pi(R_{\text{out}}^2 - R_{\text{in}}^2)
  • Annular Sector (Wedge): a=Rin, b=Rout, α≤θ≤βa = R_{\text{in}}, \, b = R_{\text{out}}, \, \alpha \le \theta \le \beta.
    Area=12(Rout2−Rin2)(β−α)\text{Area} = \frac{1}{2}(R_{\text{out}}^2 - R_{\text{in}}^2)(\beta - \alpha)

3. General Polar Regions: Radial Functional Bounds

When a region's perimeter cannot be bounded by constant radii, its boundary is defined by a polar curve r=g(θ)r = g(\theta):

  • Rays as Outer Bounds: For an angular sweep θ∈[α,β]\theta \in [\alpha, \beta], a radial probe starting at the origin extends through the domain, entering at r=g1(θ)r = g_1(\theta) and exiting at r=g2(θ)r = g_2(\theta).
  • Offset Circles Passing Through the Origin: A circle centered at (a,0)(a, 0) with radius aa has Cartesian equation:
    (x−a)2+y2=a2  ⟹  x2−2ax+a2+y2=a2  ⟹  x2+y2=2ax(x - a)^2 + y^2 = a^2 \implies x^2 - 2ax + a^2 + y^2 = a^2 \implies x^2 + y^2 = 2ax
    Substituting x2+y2=r2x^2 + y^2 = r^2 and x=rcos⁡θx = r\cos\theta:
    r2=2arcos⁡θ  ⟹  r=2acos⁡θr^2 = 2ar\cos\theta \implies r = 2a\cos\theta
    Because r≥0r \ge 0, the circle lies entirely in the right half-plane (x≥0x \ge 0), meaning the angular sweep is:
    −π2≤θ≤π2-\frac{\pi}{2} \le \theta \le \frac{\pi}{2}

Engineering Applications

  • Automotive & Powertrain Engineering: Sizing friction lining contact patches on clutch plates and disc brake calipers, where pads occupy annular sectors to balance contact pressure and thermal venting.
  • Aerospace & Defense: Defining radar and sonar coverage sectors where surveillance sweeps cover specific angular azimuths between minimum and maximum operational detection ranges.
  • Civil & Municipal Engineering: Designing circular sedimentation clarification tanks with central baffle wells and concentric peripheral effluent launders.
  • Structural Engineering: Evaluating cross-sectional properties of hollow structural steel (HSS) circular tubes and tunnel liner casings.

Example Problems

Example 1: Area of an Annular Sector

Find the area of the region RR in the first quadrant bounded between the circles x2+y2=4x^2 + y^2 = 4 and x2+y2=16x^2 + y^2 = 16, and between the lines y=13xy = \frac{1}{\sqrt{3}}x and y=xy = x.

Solution

Step 1: Determine Radial Limits
State the Circle Polar Radius Identity:

x2+y2=r2x^2 + y^2 = r^2
For x2+y2=4  ⟹  r1=2x^2 + y^2 = 4 \implies r_1 = 2.
For x2+y2=16  ⟹  r2=4x^2 + y^2 = 16 \implies r_2 = 4.
The radial limits are:
2≤r≤42 \le r \le 4

Step 2: Determine Angular Limits
State the Polar Angle Tangent Formula:

tan⁡θ=yx\tan\theta = \frac{y}{x}
For the lower boundary line y=13xy = \frac{1}{\sqrt{3}}x:
tan⁡θ1=13  ⟹  θ1=π6\tan\theta_1 = \frac{1}{\sqrt{3}} \implies \theta_1 = \frac{\pi}{6}
For the upper boundary line y=xy = x:
tan⁡θ2=1  ⟹  θ2=π4\tan\theta_2 = 1 \implies \theta_2 = \frac{\pi}{4}
Thus, the angular sweep is:
π6≤θ≤π4\frac{\pi}{6} \le \theta \le \frac{\pi}{4}

Step 3: Formulate and Evaluate the Area Integral
State the Polar Rectangle Area Formula:

Area(R)=∫θ1θ2∫r1r2r dr dθ\text{Area}(R) = \int_{\theta_1}^{\theta_2} \int_{r_1}^{r_2} r \, dr \, d\theta
Substitute the bounds:
Area(R)=∫π6π4∫24r dr dθ\text{Area}(R) = \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \int_{2}^{4} r \, dr \, d\theta
Evaluate the inner integral:
∫24r dr=[r22]24=162−42=8−2=6\int_{2}^{4} r \, dr = \left[ \frac{r^2}{2} \right]_{2}^{4} = \frac{16}{2} - \frac{4}{2} = 8 - 2 = 6
Evaluate the outer integral:
Area(R)=∫π6π46 dθ=6(π4−π6)=6(3π−2π12)=6(π12)=π2\text{Area}(R) = \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} 6 \, d\theta = 6 \left( \frac{\pi}{4} - \frac{\pi}{6} \right) = 6 \left( \frac{3\pi - 2\pi}{12} \right) = 6 \left(\frac{\pi}{12}\right) = \frac{\pi}{2}

Example 2: Area of an Offset Circle via Polar Integration

Compute the area of the disk (x−1)2+y2≤1(x - 1)^2 + y^2 \le 1 using polar coordinates.

Solution

Step 1: Convert the Cartesian Boundary to Polar Form
Expand the circle equation:

x2−2x+1+y2=1  ⟹  x2+y2=2xx^2 - 2x + 1 + y^2 = 1 \implies x^2 + y^2 = 2x
Substitute x2+y2=r2x^2 + y^2 = r^2 and x=rcos⁡θx = r\cos\theta:
r2=2rcos⁡θ  ⟹  r=2cos⁡θr^2 = 2r\cos\theta \implies r = 2\cos\theta

Step 2: Determine Angular and Radial Bounds
Since r≥0r \ge 0, we require cos⁡θ≥0\cos\theta \ge 0, which corresponds to the first and fourth quadrants:

−π2≤θ≤π2-\frac{\pi}{2} \le \theta \le \frac{\pi}{2}
For each θ\theta, the radius extends from the origin (r=0r = 0) to the perimeter curve:
0≤r≤2cos⁡θ0 \le r \le 2\cos\theta

Step 3: Formulate and Evaluate the Integral
State the General Polar Area Integral:

Area=∫−π2π2∫02cos⁡θr dr dθ\text{Area} = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_{0}^{2\cos\theta} r \, dr \, d\theta
Evaluate the inner integral:
∫02cos⁡θr dr=[r22]02cos⁡θ=4cos⁡2θ2=2cos⁡2θ\int_{0}^{2\cos\theta} r \, dr = \left[ \frac{r^2}{2} \right]_{0}^{2\cos\theta} = \frac{4\cos^2\theta}{2} = 2\cos^2\theta
Substitute the half-angle identity cos⁡2θ=1+cos⁡(2θ)2\cos^2\theta = \frac{1 + \cos(2\theta)}{2}:
Area=∫−π2π22(1+cos⁡(2θ)2)dθ=∫−π2π2(1+cos⁡(2θ)) dθ\text{Area} = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 2 \left( \frac{1 + \cos(2\theta)}{2} \right) d\theta = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (1 + \cos(2\theta)) \, d\theta
Evaluate across the symmetric limits:
Area=[θ+12sin⁡(2θ)]−π2π2=(π2+0)−(−π2+0)=π\text{Area} = \left[ \theta + \frac{1}{2}\sin(2\theta) \right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = \left(\frac{\pi}{2} + 0\right) - \left(-\frac{\pi}{2} + 0\right) = \pi
(This confirms the geometric formula πR2=π(1)2=π\pi R^2 = \pi(1)^2 = \pi.)

Connections & References