The Area of a Polar Region

Swept from the pole

A polar curve $r=f\left(\theta \right)$r = f(θ) says how far the curve lies from the pole in each direction $\theta$θ. The region it encloses between two directions, $\theta =\alpha$θ = α and $\theta =\beta$θ = β, is everything a ray from the pole passes over as it turns from $\alpha$α to $\beta$β, out as far as the curve.

So the natural slice of that region is not a strip standing on the x-axis, as it is for the area under y = f(x). It is a thin wedge with its point at the pole, one for each small turn of the ray.

The cardioid $r=1+\cos \theta$r = 1 + cos θ, with the region swept as the ray turns from $\theta =0$θ = 0 to $\theta =\frac{\pi }{3}$θ = π/3 shaded. Every edge of the shaded piece except the curve itself runs to the pole: the initial line out to (2, 0), and the ray at $\frac{\pi }{3}$π/3 out to r = 1.5.

One thin sector

Take one wedge, of small angle $d\theta$dθ. Across so small a turn, r hardly changes, so the wedge is very nearly a sector of a circle of radius r. A full turn is $2\pi$2π, so the sector is the fraction $\frac{d\theta }{2\pi }$dθ/2π of the whole disc, and its area is $\left(\frac{d\theta }{2\pi }\right)\times \pi r^2=\frac{1}{2}{r}^{2}d\theta$(dθ/2π) × πr² = ½ r² dθ.

The same ½ appears if the wedge is treated as a thin triangle. Its two long sides are r, and the short side across the end is the arc $rd\theta$r dθ, so its area is half the base times the height: $\frac{1}{2}\times rd\theta \times r=\frac{1}{2}{r}^{2}d\theta$½ × r dθ × r = ½ r² dθ. Without the ½, the area measured is a rectangle r long and $rd\theta$r dθ wide, which is twice the wedge.

A sector of radius 3 and angle 30°, and the triangle on the same two radii. The sector is $\frac{1}{2}{r}^2\delta \theta =2.356$½r²δθ = 2.356 and the triangle $\frac{1}{2}{r}^2\sin \delta \theta =2.25$½r² sin δθ = 2.25, a ratio of 0.955. Drag the angle down to 10° and the two are 0.785 and 0.781, a ratio of 0.995. Switch to the rectangle, $r^2\delta \theta$r²δθ, and it is twice the sector at every angle.

Add the sectors

Adding the sectors from $\theta =\alpha$θ = α to $\theta =\beta$θ = β, and letting $d\theta$dθ shrink, turns the sum into an integral: $A=\frac{1}{2}\int r^{2}d\theta$A = ½∫ r² dθ, taken between the two bounding angles. The r is squared because a sector grows with the square of its radius: double r and the same wedge holds four times the area.

Since $r^2$r² is never negative, every sector adds area, whichever side of the pole it lies on. That is useful, and it is also the trap below: a stretch of $\theta$θ where r is negative still adds area, even when the curve there is one already drawn.

Check it on a quarter disc

Take r = 2 from $\theta =0$θ = 0 to $\theta =\frac{\pi }{2}$θ = π/2. Then $r^2=4$r² = 4 for every $\theta$θ, and the integral of 4 over a quarter turn is $4\times \frac{\pi }{2}=2\pi$4 × π/2 = 2π, so $A=\frac{1}{2}\times 2\pi =\pi$A = ½ × 2π = π. A quarter of the disc of radius 2 is a quarter of $4\pi$4π, which is $\pi$π as well.

Now the shaded piece of the cardioid. With $r=1+\cos \theta$r = 1 + cos θ, $r^2=1+2\cos \theta +{\cos }^2\theta$r² = 1 + 2 cos θ + cos² θ, and ${\cos }^2\theta =\frac{1}{2}\left(1+\cos 2\theta \right)$cos² θ = ½(1 + cos 2θ), so $r^2=\frac{3}{2}+2\cos \theta +\frac{1}{2}\cos 2\theta$r² = 3/2 + 2 cos θ + ½ cos 2θ. Its integral from 0 to $\frac{\pi }{3}$π/3 is $\left(\frac{3}{2}\right)\left(\frac{\pi }{3}\right)+2\sin \left(\frac{\pi }{3}\right)+\frac{1}{4}\sin \left(\frac{2\pi }{3}\right)=\frac{\pi }{2}+\sqrt{3}+\frac{\sqrt{3}}{8}$(3/2)(π/3) + 2 sin(π/3) + ¼ sin(2π/3) = π/2 + √3 + √3/8. Half of that is $\frac{\pi }{4}+\frac{9\sqrt{3}}{16}\approx 1.760$π/4 + 9√3/16 ≈ 1.760.

Over the full turn from 0 to $2\pi$2π, the cosine terms integrate to 0, leaving $\frac{1}{2}\times \frac{3}{2}\times 2\pi =\frac{3\pi }{2}\approx 4.712$½ × 3/2 × 2π = 3π/2 ≈ 4.712 for the whole cardioid.

The circle r = 2, with the quarter swept from $\theta =0$θ = 0 to $\theta =\frac{\pi }{2}$θ = π/2 shaded: $\frac{1}{2}\times 4\times \frac{\pi }{2}=\pi$½ × 4 × π/2 = π, a quarter of the disc’s $4\pi$4π.

Where r is negative

The circle $r=2\cos \theta$r = 2 cos θ has radius 1 and passes through the pole. As $\theta$θ runs from $-\frac{\pi }{2}$−π/2 to $\frac{\pi }{2}$π/2, $\cos \theta$cos θ is positive and the circle is drawn once. The area is $\frac{1}{2}\int 4{\cos }^2\theta d\theta$½∫ 4 cos² θ dθ between those angles, and ${\cos }^2\theta$cos² θ averages ½ over that half turn, so $A=\frac{1}{2}\times 4\times \frac{1}{2}\times \pi =\pi$A = ½ × 4 × ½ × π = π, the area of a disc of radius 1.

From $\frac{\pi }{2}$π/2 to $\frac{3\pi }{2}$3π/2 the cosine is negative, so r is negative and each point is plotted on the opposite side of the pole, on the same circle again. Integrating over the full turn from 0 to $2\pi$2π counts every sector twice and gives $2\pi$2π.

So before choosing the limits, find where r = 0 and check which stretch of $\theta$θ draws the region exactly once. The rose in the first application below is the same trap with three petals.

$r=2\cos \theta$r = 2 cos θ for $\theta$θ from $-\frac{\pi }{2}$−π/2 to $\frac{\pi }{2}$π/2: the circle of radius 1 centered at (1, 0), drawn once and shaded once, with area $\pi$π. The next half turn would draw it again.

Between two curves

For the region between an outer curve and an inner one, sweep out to the outer curve, then take away the sweep out to the inner one. Both sweeps start at the pole, so the area is $\frac{1}{2}\int \left(r_{\text{out}}^2-r_{\text{in}}^2\right)d\theta$½∫ (r_out² − r_in²) dθ: the two squares subtract, not the two radii.

Between r = 2 and r = 1 over a quarter turn, $r_{\text{out}}^2-r_{\text{in}}^2=4-1=3$r_out² − r_in² = 4 − 1 = 3, so the area is $\frac{1}{2}\times 3\times \frac{\pi }{2}=\frac{3\pi }{4}\approx 2.356$½ × 3 × π/2 = 3π/4 ≈ 2.356. Squaring the gap instead, ${\left(2-1\right)}^2=1$(2 − 1)² = 1, gives $\frac{\pi }{4}\approx 0.785$π/4 ≈ 0.785, a third of the true area.

The limits come from where the curves meet. The cardioid $r=1+\cos \theta$r = 1 + cos θ is outside the circle r = 1 where $\cos \theta >0$cos θ > 0, from $-\frac{\pi }{2}$−π/2 to $\frac{\pi }{2}$π/2. There $r_{\text{out}}^2-r_{\text{in}}^2=2\cos \theta +{\cos }^2\theta$r_out² − r_in² = 2 cos θ + cos² θ, whose integral over that half turn is $4+\frac{\pi }{2}$4 + π/2, so the area is $2+\frac{\pi }{4}\approx 2.785$2 + π/4 ≈ 2.785.

The sweep out to r = 2 across a quarter turn, shaded, with the circle r = 1 dashed. The ring is the shaded quarter less the part inside the dashed circle: $\pi -\frac{\pi }{4}=\frac{3\pi }{4}$π − π/4 = 3π/4.

The usual mistakes

Using $\int ydx$∫ y dx. That builds the region from strips standing on the x-axis, and a polar equation gives r for each $\theta$θ, not y for each x.

Leaving r unsquared. $\int rd\theta$∫ r dθ adds up lengths; the area element is $\frac{1}{2}{r}^{2}d\theta$½ r² dθ.

Dropping the ½. For r = 2 over a quarter turn, $\int r^{2}d\theta$∫ r² dθ is $2\pi$2π, which is half the disc, not a quarter of it.

Squaring the difference. Between two curves the area is $\frac{1}{2}\int \left(r_{\text{out}}^2-r_{\text{in}}^2\right)d\theta$½∫ (r_out² − r_in²) dθ, not $\frac{1}{2}\int {\left(r_{\text{out}}-r_{\text{in}}\right)}^{2}d\theta$½∫ (r_out − r_in)² dθ.

Integrating round a full turn by habit. A curve that is complete before $\theta$θ reaches $2\pi$2π, or that runs through negative r, is then counted twice.

A flower bed, a garden path and an antenna

In the applications below, the petals of a rose are found from where r = 0, and one petal is integrated and multiplied. A spiral path is swept twice, and the strip of grass between its turns is the second sweep less the first. And a directional antenna is compared with a circle by subtracting squares on each side of the lines where the two edges meet.