Unit 6, lesson 88

Areas Bounded by Polar Curves

Why the area swept by a polar curve is one half the integral of r squared, and how to set up areas of loops, petals and regions between two polar curves without counting anything twice.

  • Theory part 175
  • Practice part 176
  • Calculus of Polar Curves

Cutting a pizza gives slices, not strips. For a region that fans out from a center, slices are the natural pieces too. The area formula for polar curves comes from adding up very thin pizza slices, each one almost a sector of a circle.

Key terms

Sector approximation

Definition A circular sector of radius rr and central angle Δθ\Delta\theta in radians, 0<Δθ≤2π0 < \Delta\theta \le 2\pi, has area

12r2 Δθ.\frac{1}{2}r^2\,\Delta\theta.

Let ff be continuous and nonnegative on [a,b][a, b] with 0<b−a≤2π0 < b - a \le 2\pi, and let RR be the region bounded by r=f(θ)r = f(\theta) and the rays θ=a\theta = a and θ=b\theta = b. Divide [a,b][a, b] into nn subintervals of width Δθ\Delta\theta and choose θi∗\theta_i^* in the iith one. Then

Area(R)≈∑i=1n12f(θi∗)2 Δθ.\text{Area}(R) \approx \sum_{i=1}^{n} \frac{1}{2} f(\theta_i^*)^2 \,\Delta\theta.

In plain words On a thin wedge the curve is nearly at a constant distance from the pole, so the wedge is nearly a sector. Add up the sectors.

Example For r=θr = \theta on [0,π2][0, \frac{\pi}{2}] with two wedges of width π4\frac{\pi}{4} and midpoints π8\frac{\pi}{8} and 3π8\frac{3\pi}{8},

12⋅π4((π8)2+(3π8)2)=10π3512≈0.606.\frac{1}{2}\cdot\frac{\pi}{4}\left(\left(\frac{\pi}{8}\right)^2 + \left(\frac{3\pi}{8}\right)^2\right) = \frac{10\pi^3}{512} \approx 0.606.

The exact area, found below, is about 0.6460.646. More wedges bring the sum closer.

One half the integral of r squared

Definition With ff continuous and nonnegative on [a,b][a, b] and 0<b−a≤2π0 < b - a \le 2\pi, the area of the region bounded by r=f(θ)r = f(\theta), θ=a\theta = a and θ=b\theta = b is the limit of the sector sums,

A=∫ab12f(θ)2 dθ=12∫abr2 dθ.A = \int_a^b \frac{1}{2} f(\theta)^2 \, d\theta = \frac{1}{2}\int_a^b r^2 \, d\theta.

In plain words The integral replaces the sum, and the factor one half is the one from the sector formula. Do not forget it.

Example For r=θr = \theta on [0,π2][0, \frac{\pi}{2}] the area is

A=12∫0π/2θ2 dθ=16(π2)3=π348≈0.646.A = \frac{1}{2}\int_0^{\pi/2} \theta^2\, d\theta = \frac{1}{6}\left(\frac{\pi}{2}\right)^3 = \frac{\pi^3}{48} \approx 0.646.

For the whole cardioid r=1+cos⁡θr = 1 + \cos\theta, traced once on [0,2π][0, 2\pi],

A=12∫02π(1+2cos⁡θ+cos⁡2θ)dθ=12(2π+0+π)=3π2.A = \frac{1}{2}\int_0^{2\pi} \left(1 + 2\cos\theta + \cos^2\theta\right) d\theta = \frac{1}{2}(2\pi + 0 + \pi) = \frac{3\pi}{2}.

Radial intervals

Definition A region is described in polar form by radial intervals if it is the set of points with polar coordinates (r,θ)(r, \theta) such that

a≤θ≤b,g(θ)≤r≤f(θ),a \le \theta \le b, \qquad g(\theta) \le r \le f(\theta),

where ff and gg are continuous on [a,b][a, b], 0≤g(θ)≤f(θ)0 \le g(\theta) \le f(\theta) and 0<b−a≤2π0 < b - a \le 2\pi. Along each ray θ\theta the region is the segment from distance g(θ)g(\theta) to distance f(θ)f(\theta).

In plain words Stand at the pole and look along one ray. The region should start at one curve and end at the other. If that is true for every ray in the range, the region has a clean polar description.

Example The region inside r=3sin⁡θr = 3\sin\theta and outside r=1+sin⁡θr = 1 + \sin\theta is

π6≤θ≤5π6,1+sin⁡θ≤r≤3sin⁡θ.\frac{\pi}{6} \le \theta \le \frac{5\pi}{6}, \qquad 1 + \sin\theta \le r \le 3\sin\theta.

Along the ray θ=π2\theta = \frac{\pi}{2} it runs from r=2r = 2 to r=3r = 3.

Area between curves

Definition For a region described by radial intervals as above,

A=12∫ab(f(θ)2−g(θ)2)dθ.A = \frac{1}{2}\int_a^b \left( f(\theta)^2 - g(\theta)^2 \right) d\theta.

In plain words Take the area out to the outer curve and subtract the area out to the inner curve. Note that you subtract the squares, not the square of the difference.

Example For the region inside r=3sin⁡θr = 3\sin\theta and outside r=1+sin⁡θr = 1 + \sin\theta, the outer radius squared minus the inner one is 9sin⁡2θ−(1+sin⁡θ)29\sin^2\theta - (1 + \sin\theta)^2, so

A=12∫π/65π/6(8sin⁡2θ−1−2sin⁡θ)dθ.A = \frac{1}{2}\int_{\pi/6}^{5\pi/6} \left(8\sin^2\theta - 1 - 2\sin\theta\right) d\theta.

Using 8sin⁡2θ=4−4cos⁡2θ8\sin^2\theta = 4 - 4\cos 2\theta, the antiderivative is 3θ−2sin⁡2θ+2cos⁡θ3\theta - 2\sin 2\theta + 2\cos\theta. Between the limits the term 3θ3\theta contributes 2π2\pi, the term −2sin⁡2θ-2\sin 2\theta contributes

−2(−32−32)=23,-2\left(-\frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2}\right) = 2\sqrt{3},

and the term 2cos⁡θ2\cos\theta contributes

2(−32−32)=−23.2\left(-\frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2}\right) = -2\sqrt{3}.

The last two cancel, so A=12(2π)=πA = \frac{1}{2}(2\pi) = \pi.

Intersections

Definition The limits of integration for a region between r=f(θ)r = f(\theta) and r=g(θ)r = g(\theta) are angles at which the curves meet. Solve f(θ)=g(θ)f(\theta) = g(\theta), and also check whether the pole lies on both curves, since the curves may reach the pole at different angles. A sketch is needed to decide which curve is outer on each interval.

In plain words The algebra finds most crossing points, the pole has to be checked on its own, and the picture tells you which piece of which curve bounds the region.

Example The region inside both r=sin⁡θr = \sin\theta and r=cos⁡θr = \cos\theta is bounded by the arc of r=sin⁡θr = \sin\theta for 0≤θ≤π40 \le \theta \le \frac{\pi}{4} and the arc of r=cos⁡θr = \cos\theta for π4≤θ≤π2\frac{\pi}{4} \le \theta \le \frac{\pi}{2}. Solving sin⁡θ=cos⁡θ\sin\theta = \cos\theta gives θ=π4\theta = \frac{\pi}{4}, and the other corner is the pole. By symmetry about the line y=xy = x,

A=2⋅12∫0π/4sin⁡2θ dθ=[θ2−sin⁡2θ4]0π/4=π8−14.A = 2 \cdot \frac{1}{2}\int_0^{\pi/4} \sin^2\theta\, d\theta = \left[\frac{\theta}{2} - \frac{\sin 2\theta}{4}\right]_0^{\pi/4} = \frac{\pi}{8} - \frac{1}{4}.

Loops

Definition If θ1<θ2\theta_1 < \theta_2 are consecutive zeros of a continuous ff, with θ2−θ1≤2π\theta_2 - \theta_1 \le 2\pi and ff of one sign on (θ1,θ2)(\theta_1, \theta_2), the arc for θ1≤θ≤θ2\theta_1 \le \theta \le \theta_2 is a loop through the pole, traced once. The area it encloses is

A=12∫θ1θ2f(θ)2 dθ,A = \frac{1}{2}\int_{\theta_1}^{\theta_2} f(\theta)^2\, d\theta,

which is valid even when f<0f < 0 on the interval, because only f2f^2 enters.

In plain words To find the area of a petal or a loop, integrate between the two angles where it leaves and returns to the pole.

Example The inner loop of r=1+2cos⁡θr = 1 + 2\cos\theta runs from θ=2π3\theta = \frac{2\pi}{3} to 4π3\frac{4\pi}{3}. Its area is

12∫2π/34π/3(1+4cos⁡θ+4cos⁡2θ)dθ=12[3θ+4sin⁡θ+sin⁡2θ]2π/34π/3=π−332≈0.544.\frac{1}{2}\int_{2\pi/3}^{4\pi/3} \left(1 + 4\cos\theta + 4\cos^2\theta\right) d\theta = \frac{1}{2}\Big[3\theta + 4\sin\theta + \sin 2\theta\Big]_{2\pi/3}^{4\pi/3} = \pi - \frac{3\sqrt{3}}{2} \approx 0.544.

Geometric area and tracing multiplicity

Definition Let ff be continuous on [a,b][a, b]. The integral

12∫abf(θ)2 dθ\frac{1}{2}\int_a^b f(\theta)^2\, d\theta

counts each part of a region as many times as the interval [a,b][a, b] sweeps over it. It gives the geometric area only when every point of the region, apart from a set of zero area, is swept exactly once.

In plain words The integral cannot see that a petal was already drawn. Choose the angles so that every piece is covered once.

Example Over [0,2π][0, 2\pi] the rose r=sin⁡3θr = \sin 3\theta gives

12∫02πsin⁡23θ dθ=π2.\frac{1}{2}\int_0^{2\pi}\sin^2 3\theta\, d\theta = \frac{\pi}{2}.

But r=sin⁡3θr = \sin 3\theta traces its three petals twice on [0,2π][0, 2\pi], so the area of the rose is π4\frac{\pi}{4}. For r=cos⁡2θr = \cos 2\theta the same integral over [0,2π][0, 2\pi] also gives π2\frac{\pi}{2}, and here it is right, since the four petals are each traced once.

Check yourself

  1. Find the area inside r=2sin⁡θr = 2\sin\theta. What goes wrong if you integrate over [0,2π][0, 2\pi]?
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The circle is traced once on [0,π][0, \pi], so its area is

12∫0π4sin⁡2θ dθ=π,\frac{1}{2}\int_0^{\pi} 4\sin^2\theta\, d\theta = \pi,

the area of a circle of radius 11. Over [0,2π][0, 2\pi] the circle is traced twice and you get 2π2\pi.

  1. Find the area of one petal of r=cos⁡3θr = \cos 3\theta.
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One petal is −π6≤θ≤π6-\frac{\pi}{6} \le \theta \le \frac{\pi}{6}, and its area is

12∫−π/6π/6cos⁡23θ dθ=12⋅π6=π12.\frac{1}{2}\int_{-\pi/6}^{\pi/6} \cos^2 3\theta\, d\theta = \frac{1}{2}\cdot\frac{\pi}{6} = \frac{\pi}{12}.
  1. Find the area inside the circle r=1r = 1 and outside the cardioid r=1−cos⁡θr = 1 - \cos\theta.
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The curves meet where cos⁡θ=0\cos\theta = 0, and the circle is outside for −π2≤θ≤π2-\frac{\pi}{2} \le \theta \le \frac{\pi}{2}. Since 1−(1−cos⁡θ)2=2cos⁡θ−cos⁡2θ1 - (1 - \cos\theta)^2 = 2\cos\theta - \cos^2\theta, the area is

A=12∫−π/2π/2(2cos⁡θ−cos⁡2θ)dθ=12(4−π2)=2−π4.A = \frac{1}{2}\int_{-\pi/2}^{\pi/2} \left(2\cos\theta - \cos^2\theta\right) d\theta = \frac{1}{2}\left(4 - \frac{\pi}{2}\right) = 2 - \frac{\pi}{4}.

How this lesson connects