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Some curves are far simpler in polar form, where a point is given by its distance r from the pole and the angle θ of its direction. A cardioid, a spiral or a rose is one short equation in r and θ. This lesson sketches polar curves from their equations, finds the area swept by the radius using thin sectors, avoids counting a region twice when a curve is traced more than once, and finds arc length in polar form.
A point from a distance and an angle
In polar coordinates a point is given by its distance r from the pole O and the angle θ that OP makes with the initial line, measured anticlockwise. The Cartesian coordinates are x = r cos θ, y = r sin θ, so r² = x² + y².
A polar curve r = f(θ) is traced as θ runs through a stated interval, here 0 ≤ θ < 2π, -π < θ ≤ π or part of one, with r ≥ 0 throughout. To sketch one:
- tabulate r at the quarter turns and any angle where r = 0 or r is largest;
- use symmetry: if f(-θ) = f(θ) the curve is symmetric about the initial line;
- at a θ where r = 0, the curve reaches the pole in the direction θ.
For the cardioid r = 1 + cos θ, 0 ≤ θ < 2π: r = 2 at θ = 0, r = 1 at θ = π/2 and 3π/2, and r = 0 at θ = π. Since cos(-θ) = cos θ, it is symmetric about the initial line, and r ≥ 0 everywhere.
The cardioid r = 1 + cos θ
A heart-shaped curve with its cusp at the pole.
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Figure data
This table is included in print while it is open.
| θ | r |
|---|---|
| 0 | 2 |
| π/2 | 1 |
| 2π/3 | 1/2 |
| π | 0 |
| 3π/2 | 1 |
Area sweeps out thin sectors
A thin sector with angle δθ and radius r has area about (1/2)r² δθ. Adding these and letting δθ → 0 gives the area swept out by the radius as θ goes from α to β:
The formula measures area swept by the moving radius, so each part of the region must be swept exactly once.
Worked example 1
The area inside a cardioid
Worked example 1
The area inside a cardioid
Find the area enclosed by r = a(1 + cos θ), 0 ≤ θ < 2π, where a > 0.
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Square r and rewrite cos²θ
Method
Expand r² and use cos² θ = (1/2)(1 + cos 2θ).Reason
Each resulting term has a standard integral.Working
Answer
r² = a²(1 + 2 cos θ + cos² θ) = a²(3/2 + 2 cos θ + 1/2 cos 2θ).Integrate once round the curve
Method
Integrate from 0 to 2π, which traces the curve exactly once.Reason
Over a full turn, cos θ and cos 2θ integrate to zero.Working
Answer
A = 1/2∫₀^2πa²(3/2 + 2 cos θ + 1/2 cos 2θ)dθ = (1/2)a² · 3/2 · 2π = (3π a²)/2.
Quick check
Using symmetry, 2 × 1/2∫₀^π r² dθ gives the same result. The cardioid fits inside the circle r = 2a of area 4π a², and (3/2)π a² is less than that, as it must be.
Spot the mistake 1
A circle counted twice
Spot the mistake 1
A circle counted twice
A student finds the area enclosed by r = cos θ as 1/2∫₀^2π cos² θ dθ = π/2. The curve is a circle of diameter 1. Explain the error and give the correct area.
Write your answer, then compare it with the solution
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Find where r is not negative
Method
cos θ ≥ 0 only for -π/2 ≤ θ ≤ π/2.Reason
This syllabus uses curves with r ≥ 0; over a full turn, cos² θ sweeps the same circle a second time.Working
Answer
A = 1/2∫_(-π/2)^(π/2) cos² θ dθ = 1/2 · π/2 = π/4.
Common mistake
A circle of diameter 1 has area π(1/2)² = π/4, which the student’s π/2 doubles. Before integrating, state the interval on which r ≥ 0 and the curve is traced once.
Check your understanding 1
Area swept by a spiral
Check your understanding 1
Area swept by a spiral
Find the exact area swept out by the radius of the spiral r = θ as θ increases from 0 to π.
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The spiral never retraces itself on this interval, because r increases with θ, so no part of the region is counted twice.
Arc length in polar form
With x = r cos θ and y = r sin θ, the arc-length formula for a parametric curve simplifies, because (dx/(dθ))² + (dy/(dθ))² = r² + (dr/(dθ))²:
Common mistakes
- Using ∫ r dθ or forgetting the factor 1/2 in A = 1/2∫ r² dθ.
- Integrating over an interval in which the curve is traced twice, so that the area is doubled.
- Ignoring values of θ where r would be negative, which this syllabus excludes from a curve’s domain.
- Writing the polar arc length as ∫ square root of (1 + (dr/(dθ))²) dθ; it is ∫ square root of (r² + (dr/(dθ))²) dθ.
- Using degrees instead of radians for θ in an integral.
Before you move on
Try it yourself 1
The perimeter of a cardioid
Try it yourself 1
The perimeter of a cardioid
Show that the total length of the cardioid r = a(1 + cos θ), 0 ≤ θ < 2π, is 8a.
Write your answer, then compare it with the solution
Hint 1
Use a half-angle identity
1 + cos θ = 2 cos² (θ/2). After taking the square root, check the sign of cos(θ/2) on your interval.
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Simplify the integrand
Method
dr/dθ = -a sin θ.Reason
A perfect square under the root makes the integral elementary.Working
Answer
r² + (dr/(dθ))² = a²(1 + 2 cos θ + cos² θ + sin² θ) = 2a²(1 + cos θ) = 4a² cos² θ/2.Keep the absolute value and use symmetry
Method
square root of (4a² cos² (θ/2)) = 2a|cos(θ/2)|.Reason
cos(θ/2) is negative for π < θ < 2π, so integrating 2a cos(θ/2) from 0 to 2π would give 0.Working
Answer
Use the symmetric upper half:
s = 2∫₀^π2a cos θ/2 dθ = 2[4a sin θ/2]₀^π = 8a.
Quick check
A length of 0 is a clear warning that a square root has lost its sign. With a = 1, a calculator’s numerical integral of square root of (2 + 2 cos θ) from 0 to 2π also gives 8. Compare your reasoning with the solution; this answer is self-checked, not marked.
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