7.4 Area and Arc Length in Polar Coordinates

Big idea

Area under a graph is built from rectangles because a rectangle is the shape a thin vertical strip of the region approximates. In polar coordinates the region is not swept by vertical strips. It is swept by a ray pivoting about the pole, and the thin piece cut off between two nearby directions is not a rectangle but a narrow wedge of a circle. Rebuilding the integral from the right elementary shape is the whole content of the area formula, and it is why a factor of one half appears where no factor appeared before.

A circular sector of radius $r$ spanning angle $\Delta\theta$ has area $\frac12 r^2 \Delta\theta$, since it is the fraction $\Delta\theta / (2\pi)$ of a full disc of area $\pi r^2$. Add those wedges over a range of directions and pass to the limit, and you get an integral of $\frac12 r^2$ with respect to $\theta$. The area between two polar curves is the same idea with the inner wedge subtracted from the outer one before integrating.

The hard part is not the formula but the limits. A polar region is bounded by directions, so finding the right two means solving for where the boundary curve reaches the pole or where two curves cross. Worse, a polar curve can trace itself more than once as $\theta$ runs through a full turn, and integrating over the full turn then counts some of the area twice. This is a genuine trap, it catches careful students, and the defence is to sketch first and integrate second.

Two more computations round out the toolkit. Arc length in polar form comes from the parametrization $x = r\cos\theta$, $y = r\sin\theta$, where the algebra collapses into a clean integrand involving $r$ and its derivative. And the slope of a polar curve is not $dr/d\theta$, which measures something else entirely; it comes from differentiating the rectangular coordinates with respect to $\theta$ and dividing.

Decoder

The area swept out by a polar curve between two directions is the integral of one half the square of the radius with respect to the angle.

The factor of one half never appears in the area between two rectangular graphs, so it has to be remembered, and the surest way to keep it is to remember where it came from: a wedge of angle $\Delta\theta$ is $\Delta\theta/(2\pi)$ of a whole disc, and $\pi r^2$ times that fraction is $\frac12 r^2 \Delta\theta$.

The phrase between two directions is where the difficulty actually sits. For a single loop that returns to the pole, the limits are consecutive angles where $r = 0$. For a region bounded by two curves, they are angles where the curves meet, found by solving an equation. For a curve that closes up before $\theta$ has advanced a full turn, they stop where the tracing stops.

The classic mistake is integrating from $0$ to $2\pi$ by reflex. For a four-petal rose or a three-petal rose this overcounts, because the full turn traces the curve twice. It also fails for any region that is not the whole interior of a closed curve. Before writing limits, sketch the curve, mark the angles where $r$ is zero, and decide which interval of angles sweeps your region exactly once.

Definitions and results

Area of a polar region. If $r = f(\theta)$ is continuous and nonnegative on $[\alpha,\beta]$ with $\beta - \alpha \le 2\pi$, the region bounded by the curve and the rays $\theta = \alpha$ and $\theta = \beta$ has area

$$ A = \frac{1}{2}\int_{\alpha}^{\beta} \big[ f(\theta) \big]^2 \, d\theta $$

Area between two curves. If $f(\theta) \ge g(\theta) \ge 0$ on $[\alpha,\beta]$, the region between them is

$$ A = \frac{1}{2}\int_{\alpha}^{\beta} \Big( \big[ f(\theta) \big]^2 - \big[ g(\theta) \big]^2 \Big) \, d\theta $$

The squares are subtracted inside the integral. Subtracting the radii first and then squaring is wrong.

Limits from the pole. A loop that begins and ends at the pole is swept between two consecutive solutions of $f(\theta) = 0$, and those angles are its limits.

Limits from an intersection. Two curves may cross at a point where their $\theta$ values differ, because polar names are not unique, so solving $f(\theta) = g(\theta)$ can miss intersections. The pole is the usual offender: check separately whether each curve passes through it.

Double tracing. A rose $r = a\cos(n\theta)$ with $n$ odd is traced twice as $\theta$ runs from $0$ to $2\pi$, so integrating over the full turn returns twice the enclosed area. With $n$ even it is traced once.

Arc length in polar form. For a smooth curve $r = f(\theta)$ traced once on $[\alpha,\beta]$,

$$ L = \int_{\alpha}^{\beta} \sqrt{\big[ f(\theta) \big]^2 + \big[ f'(\theta) \big]^2}\; d\theta $$

This follows from the parametric arc length integral, since $\left(\frac{dx}{d\theta}\right)^2 + \left(\frac{dy}{d\theta}\right)^2$ simplifies to $r^2 + (r')^2$ after expanding and using the Pythagorean identity.

Slope of a polar curve. With $x = f(\theta)\cos\theta$ and $y = f(\theta)\sin\theta$,

$$ \frac{dy}{dx} = \frac{f'(\theta)\sin\theta + f(\theta)\cos\theta}{f'(\theta)\cos\theta - f(\theta)\sin\theta} $$

whenever the denominator is nonzero. The quantity $dr/d\theta$ by itself measures how fast the distance from the pole changes, which is not the slope of the tangent line.

Tangents at the pole. If $f(\theta_0) = 0$ and $f'(\theta_0) \neq 0$, the formula collapses to $\tan\theta_0$, so the curve leaves the pole along the line $\theta = \theta_0$.

Worked examples

One petal of a rose

Find the area enclosed by one petal of $r = 2\cos(3\theta)$.

First the limits. The petal begins and ends where the curve is at the pole, so solve $\cos(3\theta) = 0$. The solutions nearest zero are $3\theta = -\pi/2$ and $3\theta = \pi/2$, giving $\theta = -\pi/6$ and $\theta = \pi/6$. That pair of directions sweeps one petal exactly once.

Now the integral, using $\cos^2 u = \frac{1 + \cos 2u}{2}$:

$$ A = \frac{1}{2}\int_{-\pi/6}^{\pi/6} 4\cos^2(3\theta)\,d\theta = 2\int_{-\pi/6}^{\pi/6} \frac{1 + \cos(6\theta)}{2}\,d\theta = \int_{-\pi/6}^{\pi/6} \big( 1 + \cos(6\theta) \big)\,d\theta $$

The antiderivative is $\theta + \frac{\sin(6\theta)}{6}$. At $\theta = \pi/6$ this is $\frac{\pi}{6} + \frac{\sin\pi}{6} = \frac{\pi}{6}$, and at $\theta = -\pi/6$ it is $-\frac{\pi}{6}$. The difference is $\frac{\pi}{3}$, so one petal has area $\frac{\pi}{3}$.

Now the warning. This rose has three petals, so the total enclosed area is $\pi$. Integrating $\frac12 r^2$ from $0$ to $2\pi$ returns $2\pi$, exactly double, because an odd-numbered rose is traced twice over a full turn. The formula was not wrong; the limits were.

Between a circle and a smaller circle

Find the area inside $r = 2\sin\theta$ and outside $r = 1$.

The curve $r = 2\sin\theta$ is a circle of radius $1$ sitting above the pole, and $r = 1$ is the unit circle centred at the pole. They meet where $2\sin\theta = 1$, that is $\theta = \pi/6$ and $\theta = 5\pi/6$. Between those directions $2\sin\theta$ is the larger radius, so those are the limits.

$$ A = \frac{1}{2}\int_{\pi/6}^{5\pi/6} \big( 4\sin^2\theta - 1 \big)\,d\theta $$

Replace $4\sin^2\theta$ by $2 - 2\cos 2\theta$, so the integrand becomes $1 - 2\cos 2\theta$ and the antiderivative is $\theta - \sin 2\theta$. At $\theta = 5\pi/6$: $\frac{5\pi}{6} - \sin\frac{5\pi}{3} = \frac{5\pi}{6} + \frac{\sqrt3}{2}$. At $\theta = \pi/6$: $\frac{\pi}{6} - \sin\frac{\pi}{3} = \frac{\pi}{6} - \frac{\sqrt3}{2}$. Subtracting gives $\frac{2\pi}{3} + \sqrt3$, and halving:

$$ A = \frac{\pi}{3} + \frac{\sqrt3}{2} \approx 1.913 $$

Sanity check the size. The disc bounded by $r = 2\sin\theta$ has radius $1$ and area $\pi \approx 3.14$, and the region computed is the part of it outside the unit circle, which the sketch shows is a bit more than half.

Length of a cardioid

Find the total length of $r = 1 + \cos\theta$.

Here $f'(\theta) = -\sin\theta$, so the quantity under the radical is

$$ (1+\cos\theta)^2 + \sin^2\theta = 1 + 2\cos\theta + \cos^2\theta + \sin^2\theta = 2 + 2\cos\theta $$

Use the half-angle identity $1 + \cos\theta = 2\cos^2(\theta/2)$ to write this as $4\cos^2(\theta/2)$, whose square root is $2\left| \cos(\theta/2) \right|$. On $0 \le \theta \le \pi$ the half-angle stays in the first quadrant, so the absolute value can be dropped, and the curve is symmetric about the polar axis. Therefore

$$ L = 2\int_0^{\pi} 2\cos\!\left( \frac{\theta}{2} \right) d\theta = 2\Big[ 4\sin\!\left( \frac{\theta}{2} \right) \Big]_0^{\pi} = 2(4 - 0) = 8 $$

The cardioid has length $8$. As a rough check, it stretches from the pole out to $r = 2$, so it should be longer than a circle of radius $1$, whose circumference is about $6.28$.

A tangent line in polar form

Find the slope of the tangent to $r = 1 + \cos\theta$ at $\theta = \pi/2$.

At that angle $f(\pi/2) = 1 + 0 = 1$ and $f'(\pi/2) = -\sin(\pi/2) = -1$. Substitute into the slope formula, with $\cos(\pi/2) = 0$ and $\sin(\pi/2) = 1$:

$$ \frac{dy}{dx} = \frac{(-1)(1) + (1)(0)}{(-1)(0) - (1)(1)} = \frac{-1}{-1} = 1 $$

The point itself is $(0,1)$, so the tangent line is $y = x + 1$. Note that $dr/d\theta = -1$ there, which is not the slope and does not resemble it: the distance from the pole is shrinking while the tangent line rises.

Practice

Start with area. The integrand is short, so put your effort into the limits: sketch the curve, locate the angles where it reaches the pole or meets the other curve, and confirm the interval sweeps your region exactly once.

Then tangents. These ask for $dy/dx$ at a given angle, for horizontal and vertical tangents, and for the directions in which a curve leaves the pole.

Quiz

Five items on polar area, regions between curves, arc length and tangent slopes.