Core Definition
A polar rectangle R ⊂ R 2 R \subset \mathbb{R}^2 R ⊂ R 2
R = { ( r , θ ) ∈ R 2 ∣ a ≤ r ≤ b , α ≤ θ ≤ β } R = \{(r, \theta) \in \mathbb{R}^2 \mid a \le r \le b, \, \alpha \le \theta \le \beta\} R = {( r , θ ) ∈ R 2 ∣ a ≤ r ≤ b , α ≤ θ ≤ β }
where
0 ≤ a < b 0 \le a < b 0 ≤ a < b and
β − α ≤ 2 π \beta - \alpha \le 2\pi β − α ≤ 2 π .
The planar area of a polar rectangle is:
Area ( R ) = ∬ R 1 d A = ∫ α β ∫ a b r d r d θ = 1 2 ( b 2 − a 2 ) ( 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) Area ( R ) = ∬ R 1 d A = ∫ α β ∫ a b r d r d θ = 2 1 ( b 2 − a 2 ) ( β − α )
When
r r r is bounded by continuous angle-dependent functions
g 1 ( θ ) ≤ r ≤ g 2 ( θ ) g_1(\theta) \le r \le g_2(\theta) g 1 ( θ ) ≤ r ≤ g 2 ( θ ) , 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] [ a , b ] × [ c , d ]
In polar coordinates ( r , θ ) (r, \theta) ( r , θ ) r θ r\theta r θ annular sector in the physical x y xy x y
The inner and outer boundary curves r = a r = a r = a r = b r = b r = b
The angular edges θ = α \theta = \alpha θ = α θ = β \theta = \beta θ = β
Because the width of a polar sector grows proportionally with distance from the origin (s = r θ s = r\theta s = r θ
Δ A ≈ r ∗ Δ r Δ θ \Delta A \approx r^* \, \Delta r \, \Delta \theta Δ A ≈ r ∗ Δ r Δ θ 2. Common Geometries as Polar Rectangles
Full Circular Disk of Radius R R R a = 0 , b = R , α = 0 , β = 2 π a = 0, \, b = R, \, \alpha = 0, \, \beta = 2\pi a = 0 , b = R , α = 0 , β = 2 π Area = ∫ 0 2 π ∫ 0 R r d r d θ = ( 2 π ) [ 1 2 R 2 ] = π R 2 \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 Area = ∫ 0 2 π ∫ 0 R r d r d θ = ( 2 π ) [ 2 1 R 2 ] = π R 2 Annulus (Concentric Ring): a = R in , b = R out , α = 0 , β = 2 π a = R_{\text{in}}, \, b = R_{\text{out}}, \, \alpha = 0, \, \beta = 2\pi a = R in , b = R out , α = 0 , β = 2 π Area = π ( R out 2 − R in 2 ) \text{Area} = \pi(R_{\text{out}}^2 - R_{\text{in}}^2) Area = π ( R out 2 − R in 2 ) Annular Sector (Wedge): a = R in , b = R out , α ≤ θ ≤ β a = R_{\text{in}}, \, b = R_{\text{out}}, \, \alpha \le \theta \le \beta a = R in , b = R out , α ≤ θ ≤ β Area = 1 2 ( R out 2 − R in 2 ) ( β − α ) \text{Area} = \frac{1}{2}(R_{\text{out}}^2 - R_{\text{in}}^2)(\beta - \alpha) Area = 2 1 ( R out 2 − R in 2 ) ( β − α )
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) r = g ( θ )
Rays as Outer Bounds: For an angular sweep θ ∈ [ α , β ] \theta \in [\alpha, \beta] θ ∈ [ α , β ] r = g 1 ( θ ) r = g_1(\theta) r = g 1 ( θ ) r = g 2 ( θ ) r = g_2(\theta) r = g 2 ( θ ) Offset Circles Passing Through the Origin: A circle centered at ( a , 0 ) (a, 0) ( a , 0 ) a a a ( x − a ) 2 + y 2 = a 2 ⟹ x 2 − 2 a x + a 2 + y 2 = a 2 ⟹ x 2 + y 2 = 2 a x (x - a)^2 + y^2 = a^2 \implies x^2 - 2ax + a^2 + y^2 = a^2 \implies x^2 + y^2 = 2ax ( x − a ) 2 + y 2 = a 2 ⟹ x 2 − 2 a x + a 2 + y 2 = a 2 ⟹ x 2 + y 2 = 2 a x x 2 + y 2 = r 2 x^2 + y^2 = r^2 x 2 + y 2 = r 2 x = r cos θ x = r\cos\theta x = r cos θ r 2 = 2 a r cos θ ⟹ r = 2 a cos θ r^2 = 2ar\cos\theta \implies r = 2a\cos\theta r 2 = 2 a r cos θ ⟹ r = 2 a cos θ r ≥ 0 r \ge 0 r ≥ 0 x ≥ 0 x \ge 0 x ≥ 0 − π 2 ≤ θ ≤ π 2 -\frac{\pi}{2} \le \theta \le \frac{\pi}{2} − 2 π ≤ θ ≤ 2 π
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 R R R x 2 + y 2 = 4 x^2 + y^2 = 4 x 2 + y 2 = 4 x 2 + y 2 = 16 x^2 + y^2 = 16 x 2 + y 2 = 16 y = 1 3 x y = \frac{1}{\sqrt{3}}x y = 3 1 x y = x y = x y = x
Solution
Step 1: Determine Radial Limits Circle Polar Radius Identity :
x 2 + y 2 = r 2 x^2 + y^2 = r^2 x 2 + y 2 = r 2
For
x 2 + y 2 = 4 ⟹ r 1 = 2 x^2 + y^2 = 4 \implies r_1 = 2 x 2 + y 2 = 4 ⟹ r 1 = 2 .
For
x 2 + y 2 = 16 ⟹ r 2 = 4 x^2 + y^2 = 16 \implies r_2 = 4 x 2 + y 2 = 16 ⟹ r 2 = 4 .
The radial limits are:
Step 2: Determine Angular Limits Polar Angle Tangent Formula :
tan θ = y x \tan\theta = \frac{y}{x} tan θ = x y
For the lower boundary line
y = 1 3 x y = \frac{1}{\sqrt{3}}x y = 3 1 x :
tan θ 1 = 1 3 ⟹ θ 1 = π 6 \tan\theta_1 = \frac{1}{\sqrt{3}} \implies \theta_1 = \frac{\pi}{6} tan θ 1 = 3 1 ⟹ θ 1 = 6 π
For the upper boundary line
y = x y = x y = x :
tan θ 2 = 1 ⟹ θ 2 = π 4 \tan\theta_2 = 1 \implies \theta_2 = \frac{\pi}{4} tan θ 2 = 1 ⟹ θ 2 = 4 π
Thus, the angular sweep is:
π 6 ≤ θ ≤ π 4 \frac{\pi}{6} \le \theta \le \frac{\pi}{4} 6 π ≤ θ ≤ 4 π Step 3: Formulate and Evaluate the Area Integral Polar Rectangle Area Formula :
Area ( R ) = ∫ θ 1 θ 2 ∫ r 1 r 2 r d r d θ \text{Area}(R) = \int_{\theta_1}^{\theta_2} \int_{r_1}^{r_2} r \, dr \, d\theta Area ( R ) = ∫ θ 1 θ 2 ∫ r 1 r 2 r d r d θ
Substitute the bounds:
Area ( R ) = ∫ π 6 π 4 ∫ 2 4 r d r d θ \text{Area}(R) = \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \int_{2}^{4} r \, dr \, d\theta Area ( R ) = ∫ 6 π 4 π ∫ 2 4 r d r d θ
Evaluate the inner integral:
∫ 2 4 r d r = [ r 2 2 ] 2 4 = 16 2 − 4 2 = 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 ∫ 2 4 r d r = [ 2 r 2 ] 2 4 = 2 16 − 2 4 = 8 − 2 = 6
Evaluate the outer integral:
Area ( R ) = ∫ π 6 π 4 6 d θ = 6 ( π 4 − π 6 ) = 6 ( 3 π − 2 π 12 ) = 6 ( π 12 ) = π 2 \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} Area ( R ) = ∫ 6 π 4 π 6 d θ = 6 ( 4 π − 6 π ) = 6 ( 12 3 π − 2 π ) = 6 ( 12 π ) = 2 π
Example 2: Area of an Offset Circle via Polar Integration
Compute the area of the disk ( x − 1 ) 2 + y 2 ≤ 1 (x - 1)^2 + y^2 \le 1 ( x − 1 ) 2 + y 2 ≤ 1
Solution
Step 1: Convert the Cartesian Boundary to Polar Form
x 2 − 2 x + 1 + y 2 = 1 ⟹ x 2 + y 2 = 2 x x^2 - 2x + 1 + y^2 = 1 \implies x^2 + y^2 = 2x x 2 − 2 x + 1 + y 2 = 1 ⟹ x 2 + y 2 = 2 x
Substitute
x 2 + y 2 = r 2 x^2 + y^2 = r^2 x 2 + y 2 = r 2 and
x = r cos θ x = r\cos\theta x = r cos θ :
r 2 = 2 r cos θ ⟹ r = 2 cos θ r^2 = 2r\cos\theta \implies r = 2\cos\theta r 2 = 2 r cos θ ⟹ r = 2 cos θ Step 2: Determine Angular and Radial Bounds r ≥ 0 r \ge 0 r ≥ 0 cos θ ≥ 0 \cos\theta \ge 0 cos θ ≥ 0
− π 2 ≤ θ ≤ π 2 -\frac{\pi}{2} \le \theta \le \frac{\pi}{2} − 2 π ≤ θ ≤ 2 π
For each
θ \theta θ , the radius extends from the origin (
r = 0 r = 0 r = 0 ) to the perimeter curve:
0 ≤ r ≤ 2 cos θ 0 \le r \le 2\cos\theta 0 ≤ r ≤ 2 cos θ Step 3: Formulate and Evaluate the Integral General Polar Area Integral :
Area = ∫ − π 2 π 2 ∫ 0 2 cos θ r d r d θ \text{Area} = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_{0}^{2\cos\theta} r \, dr \, d\theta Area = ∫ − 2 π 2 π ∫ 0 2 c o s θ r d r d θ
Evaluate the inner integral:
∫ 0 2 cos θ r d r = [ r 2 2 ] 0 2 cos θ = 4 cos 2 θ 2 = 2 cos 2 θ \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 ∫ 0 2 c o s θ r d r = [ 2 r 2 ] 0 2 c o s θ = 2 4 cos 2 θ = 2 cos 2 θ
Substitute the half-angle identity
cos 2 θ = 1 + cos ( 2 θ ) 2 \cos^2\theta = \frac{1 + \cos(2\theta)}{2} cos 2 θ = 2 1 + c o s ( 2 θ ) :
Area = ∫ − π 2 π 2 2 ( 1 + cos ( 2 θ ) 2 ) d θ = ∫ − π 2 π 2 ( 1 + cos ( 2 θ ) ) d θ \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 Area = ∫ − 2 π 2 π 2 ( 2 1 + cos ( 2 θ ) ) d θ = ∫ − 2 π 2 π ( 1 + cos ( 2 θ )) d θ
Evaluate across the symmetric limits:
Area = [ θ + 1 2 sin ( 2 θ ) ] − π 2 π 2 = ( π 2 + 0 ) − ( − π 2 + 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 Area = [ θ + 2 1 sin ( 2 θ ) ] − 2 π 2 π = ( 2 π + 0 ) − ( − 2 π + 0 ) = π (This confirms the geometric formula π R 2 = π ( 1 ) 2 = π \pi R^2 = \pi(1)^2 = \pi π R 2 = π ( 1 ) 2 = π
Connections & References