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Area of a
Polar Region

Learn to compute the area enclosed by polar curves using the integral formula \( A = \frac{1}{2}\int_{\alpha}^{\beta} [r(\theta)]^2\, d\theta \).

AP Calculus BC

What Are Polar Coordinates?

In Cartesian coordinates we locate a point with \( (x, y) \). In polar coordinates, we use a distance \( r \) from the origin and an angle \( \theta \) from the positive \( x \)-axis. The conversion is:

x y \(P(r, \theta)\) \(\theta\) \(r\) O

Why Area in Polar Is Different

In Cartesian coordinates, we sum thin rectangular strips of width \( dx \). In polar coordinates, we sweep through angles: each thin slice is a circular sector, not a rectangle.

Cartesian
Vertical rectangles: \( dA = f(x)\,dx \)
Polar
Circular sectors: \( dA = \tfrac{1}{2}r^2\,d\theta \)

Building the Area Integral

\(d\theta\) \(r(\theta)\) O \(dA\) Each thin sector: \(dA = \tfrac{1}{2}r^2\,d\theta\)

Partition the angle from \( \alpha \) to \( \beta \) into \( n \) sub-angles of width \( \Delta\theta \). Each sector has approximate area \( \frac{1}{2}[r(\theta_k)]^2 \Delta\theta \). Taking the limit as \( n \to \infty \) gives the Riemann sum turning into a definite integral.

The Polar Area Formula

1
\( r(\theta) \) is the polar function giving the distance from the origin.
2
\( \alpha \) and \( \beta \) are the angles bounding the region.
3
The \( \frac{1}{2} \) comes from the sector area formula \( \frac{1}{2}r^2\theta \).

Sectors vs. Rectangles

Cartesian Polar
Slice shape Thin rectangle Thin sector
Width element \( dx \) or \( dy \) \( d\theta \)
Height/radius \( f(x) \) \( r(\theta) \)
Area element \( f(x)\,dx \) \( \tfrac{1}{2}r^2\,d\theta \)
Total area \( \int_a^b f(x)\,dx \) \( \frac{1}{2}\int_\alpha^\beta r^2\,d\theta \)

How to Find Polar Area

1
Sketch the curve (or use key values of \( r(\theta) \)) to identify the region.
2
Find the bounds \( \alpha \) and \( \beta \). Often set \( r = 0 \) to find where the curve returns to the origin.
3
Set up the integral: \( A = \frac{1}{2}\int_\alpha^\beta [r(\theta)]^2\,d\theta \).
4
Expand and simplify the integrand. Use identities like \( \cos^2\theta = \frac{1+\cos 2\theta}{2} \).
5
Evaluate the definite integral and confirm the answer is positive.

Useful Trig Identities

When you square polar functions you often get \( \cos^2\theta \) or \( \sin^2\theta \). These power-reduction identities are essential:

\( \cos^2\theta \)
\( \dfrac{1 + \cos 2\theta}{2} \)
\( \sin^2\theta \)
\( \dfrac{1 - \cos 2\theta}{2} \)

Example 1: Circle \( r = 3 \)

\(r = 3\)

Example 2: Cardioid \( r = 1 + \cos\theta \)

O \(r = 2\) \(r = 0\) at \(\theta = \pi\)

Cardioid Solution

1
Expand: \( (1+\cos\theta)^2 = 1 + 2\cos\theta + \cos^2\theta \).
2
Apply identity: Replace \( \cos^2\theta \) with \( \frac{1+\cos 2\theta}{2} \). Integrand becomes \( \frac{3}{2} + 2\cos\theta + \frac{\cos 2\theta}{2} \).
3
Integrate: \( \int_0^{2\pi}\!\bigl(\frac{3}{2} + 2\cos\theta + \frac{\cos 2\theta}{2}\bigr)\,d\theta = \frac{3}{2}(2\pi) + 0 + 0 = 3\pi \).
4
Multiply by \( \frac{1}{2} \): \( A = \frac{1}{2}(3\pi) = \) \( \frac{3\pi}{2} \).

Example 3: One Petal of \( r = \cos 2\theta \)

O \(\theta = \pi/4\) \(\theta = -\pi/4\)

Using Symmetry

Many polar curves are symmetric about the polar axis, the line \( \theta = \frac{\pi}{2} \), or the origin. Exploit this to halve your work:

Polar axis
If \( r(\theta) = r(-\theta) \), integrate \( 0 \) to \( \pi \) and double.
Line \( \theta = \frac{\pi}{2} \)
If \( r(\pi - \theta) = r(\theta) \), integrate over half and double.
Origin
If \( r(\theta + \pi) = r(\theta) \), integrate one period and double.

Total Area: Rose Curves

For \( r = \cos n\theta \) or \( r = \sin n\theta \):

\( n \) odd
\( n \) petals. Total area: \( n \times \) (one petal area)
\( n \) even
\( 2n \) petals. Total area: \( 2n \times \) (one petal area)

Common Mistakes

✕ Forgetting \( \frac{1}{2} \)
The formula is \( \frac{1}{2}\int r^2\,d\theta \), not \( \int r^2\,d\theta \). Missing the half doubles your answer.
✕ Wrong bounds
Integrating \( 0 \) to \( 2\pi \) for a rose petal gives the total area. One petal needs a smaller interval.
✕ Not squaring \( r \)
You integrate \( [r(\theta)]^2 \), not \( r(\theta) \). Forgetting to square changes the problem entirely.

Quick Reference Table

Curve Bounds Area
\( r = a \) \( 0 \) to \( 2\pi \) \( \pi a^2 \)
\( r = a\cos\theta \) \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \) \( \frac{\pi a^2}{4} \)
\( r = 1 + \cos\theta \) \( 0 \) to \( 2\pi \) \( \frac{3\pi}{2} \)
\( r = \cos 2\theta \) (1 petal) \( -\frac{\pi}{4} \) to \( \frac{\pi}{4} \) \( \frac{\pi}{8} \)
\( r = \sin 3\theta \) (1 petal) \( 0 \) to \( \frac{\pi}{3} \) \( \frac{\pi}{12} \)

Worked practice: Practice Problems, Problem 4 — applied

Work through this original lesson problem, then reveal the solution and illustrations.

A microphone has a polar pickup pattern modeled by r = 3 + 3 sin θ, where r represents the sensitivity (in arbitrary units) in the direction θ. (a) Sketch the curve and identify it by name. (b) Set up, but do not evaluate, an integral expression for the total area enclosed by this pickup pattern. (c) Evaluate the integral from part (b) to find the total enclosed area. (d) The manufacturer claims the microphone covers an area of 45 square units. Is this claim correct? Justify your answer.

Show answer and explanation

Rubric (5 points): Point 1 (Part a): The curve r = 3 + 3 sin θ is a cardioid (since the constant equals the coefficient of sin θ). It is oriented vertically with the cusp at the origin when θ = 3π/2 (where sin θ = −1 gives r = 0). Award 1 point for correctly identifying the curve as a cardioid and providing a reasonable sketch showing symmetry about the line θ = π/2. Point 2 (Part b): The full cardioid is traced for θ ∈ [0, 2π]. The area integral is A = ½ ∫₀^(2π) [3 + 3 sin θ]² dθ = ½ ∫₀^(2π) 9(1 + sin θ)² dθ. Award 1 point for the correct integral with correct bounds and the ½ factor. Point 3 (Part c — expansion): Expand (1 + sin θ)² = 1 + 2 sin θ + sin² θ = 1 + 2 sin θ + (1 − cos 2θ)/2 = 3/2 + 2 sin θ − ½ cos 2θ. Award 1 point for correct algebraic expansion and use of the power-reduction identity. Point 4 (Part c — evaluation): ∫₀^(2π) [3/2 + 2 sin θ − ½ cos 2θ] dθ = 3π + 0 + 0 = 3π. Therefore A = (9/2)(3π) = 27π/2. Award 1 point for correct antidifferentiation and final answer. Point 5 (Part d): 27π/2 ≈ 42.41 square units, which is less than 45. The manufacturer's claim is incorrect. Award 1 point for a correct numerical comparison with a clear justification statement.

Upward cardioid response r=3+3sinθ: maximum 6, side response 3, rear null zero, and polar area 27π/2 in squared response units.
The rotated cardioid has maximum microphone response upward and a rear null; radial values are sensitivity units rather than distance.
Microphone context photograph with an orientation cue; it does not encode a quantitative response curve.
Microphone context photograph with an orientation cue; it does not encode a quantitative response curve.

Question 1 of 5

What is the area enclosed by the circle \( r = 4 \)?

Question 2 of 5

Which integral gives the area of one petal of \( r = \sin 2\theta \)?

Question 3 of 5

The area enclosed by the cardioid \( r = 2(1 + \sin\theta) \) is:

Question 4 of 5

How many petals does \( r = \cos 3\theta \) have, and what is the total area?

Question 5 of 5

In the polar area formula \( A = \frac{1}{2}\int_\alpha^\beta [r(\theta)]^2\,d\theta \), the factor \( \frac{1}{2} \) comes from: