Polar curves, area and arc length

Sketch polar curves with non-negative r, then find the area they sweep out and their arc length.

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Enlarged diagram

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:

  1. tabulate r at the quarter turns and any angle where r = 0 or r is largest;
  2. use symmetry: if f(-θ) = f(θ) the curve is symmetric about the initial line;
  3. 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.

As θ increases from 0 to π, r falls from 2 to 0 and the upper half is traced; from π to 2π the lower half is traced by symmetry.
Figure data

This table is included in print while it is open.

Values of r = 1 + cos θ
θr
02
π/21
2π/31/2
π0
3π/21

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:

A = 1/2∫α^β r² dθ.

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

Find the area enclosed by r = a(1 + cos θ), 0 ≤ θ < 2π, where a > 0.

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  1. 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).
  2. 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.

Spot the mistake 1

A circle counted twice

About 5 min

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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  1. 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.

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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A = 1/2∫₀^πθ² dθ = 1/2 · π³/3 = π³/6 ≈ 5.17.

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:

s = ∫α^β square root of (r² + (dr/(dθ))²) dθ.

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

Minimal support

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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  1. 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.
  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.

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