Area of a Single Polar Loop

From Pole Crossings to One Loop

A polar curve can return to the pole at several parameter values. Those returns help identify where a loop may begin and end, but they do not by themselves tell us whether two parameter intervals trace different loops. To find the area enclosed by one loop, we need both a suitable interval and a check that the curve traces that boundary exactly once.

For a polar curve \(r=f(\theta)\), the curve reaches the pole when \(r=0\). Consecutive parameter values with \(r=0\) are natural candidates for the endpoints of a loop. On such an interval, the curve starts at the pole, leaves it, and returns. The key additional question is whether the resulting path is a single boundary traced once, rather than a boundary that crosses itself or is retraced.

Definition: A single polar loop is a closed curve traced exactly once over a parameter interval, with no repeated boundary points except the coincident start and end points. If the loop begins and ends at the pole, its interval endpoints satisfy \(r=0\).

Consecutive pole crossings are useful for locating a candidate loop interval, but they are not a complete test for a single loop. A curve can retrace the same geometric loop on a later interval. In that case, integrating over both intervals counts the same enclosed region more than once.

The Area Integral for One Loop

Suppose \(r=f(\theta)\) is continuous on \([a,b]\), \(f(a)=f(b)=0\), and the curve traces one loop exactly once as \(\theta\) runs from \(a\) to \(b\). The area \(A\) enclosed by that loop is

A=\frac12\int_a^b [f(\theta)]^2\,d\theta.

The square is important: the integrand is \(r^2\), not \(r\). If \(r\) is negative on part of the interval, the curve is represented in the opposite radial direction, but the area integrand remains \(r^2\).

Here is the reason for the factor \(\frac12\). A narrow sector of radius \(r\) and angle width \(\Delta\theta\) has area approximately \(\frac12r^2\Delta\theta\). Divide the interval into small pieces and add the areas of these narrow sectors. In the limit, the sum becomes the integral \(\frac12\int r^2\,d\theta\). This argument applies to a loop traced once; if a parameter interval retraces the boundary, the integral counts each traversal.

Find pole parameters.
Solve \(f(\theta)=0\) on a parameter range that covers the curve.

Select consecutive crossings.
Use consecutive roots as candidate endpoints \(a\) and \(b\).

Check the trace.
Confirm that the interval gives one closed boundary and does not retrace a loop already counted.

Integrate and verify.
Evaluate \(\frac12\int_a^b f(\theta)^2\,d\theta\), then check the antiderivative and the chosen interval.

Worked Examples

Worked Example: Area of a Cardioid Loop

Find the area enclosed by \(r=1+\cos\theta\). The pole crossings satisfy \(1+\cos\theta=0\), so \(\cos\theta=-1\), giving \(\theta=\pi+2k\pi\) for integers \(k\). Thus \(\pi\) and \(3\pi\) are consecutive pole crossings, and we use \([\pi,3\pi]\). On this interval \(r\geq 0\), and the curve makes one loop.

Apply the single-loop area formula and expand the square:

\begin{aligned} A &=\frac12\int_{\pi}^{3\pi}(1+\cos\theta)^2\,d\theta\\ &=\frac12\int_{\pi}^{3\pi}(1+2\cos\theta+\cos^2\theta)\,d\theta\\ &=\frac12\int_{\pi}^{3\pi}\left(\frac32+2\cos\theta+\frac12\cos(2\theta)\right)\,d\theta. \end{aligned}

Using \(\cos^2\theta=(1+\cos(2\theta))/2\), an antiderivative of the integrand before the outer factor \(\frac12\) is \(\frac32\theta+2\sin\theta+\frac14\sin(2\theta)\). Therefore,

\begin{aligned} A &=\frac12\left[\frac32\theta+2\sin\theta+\frac14\sin(2\theta)\right]_{\pi}^{3\pi}\\ &=\frac12\left(\frac92\pi- \frac32\pi\right)\\ &=\frac12(3\pi)=\frac{3\pi}{2}. \end{aligned}

Check the antiderivative: differentiating \(\frac32\theta+2\sin\theta+\frac14\sin(2\theta)\) gives \(\frac32+2\cos\theta+\frac12\cos(2\theta)=(1+\cos\theta)^2\). The endpoint evaluation is also consistent with the loop: \(r(\pi)=r(3\pi)=0\), while \(r(2\pi)=2\), so the curve leaves the pole and reaches a positive radius before returning.

Worked Example: A Loop With Its Pole Crossing at the Interval Endpoints

Find the area enclosed by \(r=2(1-\cos\theta)\). The pole crossings satisfy \(\cos\theta=1\), so consecutive crossings are \(0\) and \(2\pi\). The radius is nonnegative throughout that interval and is positive for \(0<\theta<2\pi\), so the interval traces one loop.

Squaring the radius and integrating gives

\begin{aligned} A &=\frac12\int_0^{2\pi}[2(1-\cos\theta)]^2\,d\theta\\ &=2\int_0^{2\pi}(1-2\cos\theta+\cos^2\theta)\,d\theta\\ &=2\int_0^{2\pi}\left(\frac32-2\cos\theta+\frac12\cos(2\theta)\right)\,d\theta\\ &=2\left[\frac32\theta-2\sin\theta+\frac14\sin(2\theta)\right]_0^{2\pi}\\ &=2(3\pi)=6\pi. \end{aligned}

To check the calculation, differentiate the bracketed expression: its derivative is \(\frac32-2\cos\theta+\frac12\cos(2\theta)=(1-\cos\theta)^2\), the expanded squared radius divided by \(4\). Including the factor \(2\) outside the integral therefore gives the required integrand \(2(1-\cos\theta)^2\). At both endpoints \(r=0\), as required for this pole-to-pole loop.

Worked Example: Consecutive Pole Crossings Can Bound a Retraced Loop

Consider \(r=3\sin\theta\). Its pole crossings occur at integer multiples of \(\pi\), so \([0,\pi]\) and \([\pi,2\pi]\) are consecutive intervals between crossings. We first find the area on \([0,\pi]\):

\begin{aligned} A_{[0,\pi]} &=\frac12\int_0^\pi(3\sin\theta)^2\,d\theta\\ &=\frac92\int_0^\pi\sin^2\theta\,d\theta\\ &=\frac92\int_0^\pi\frac{1-\cos(2\theta)}{2}\,d\theta\\ &=\frac92\left[\frac{\theta}{2}-\frac{\sin(2\theta)}{4}\right]_0^\pi\\ &=\frac92\left(\frac{\pi}{2}\right)=\frac{9\pi}{4}. \end{aligned}

The antiderivative check is direct: differentiating \(\frac{\theta}{2}-\frac{\sin(2\theta)}{4}\) gives \(\frac12-\frac12\cos(2\theta)=\sin^2\theta\). To see what the curve traces, write \(x=r\cos\theta\) and \(y=r\sin\theta\). Since \(r=3\sin\theta\),

x^2+\left(y-\frac32\right)^2 =9\sin^2\theta\cos^2\theta+ \left(3\sin^2\theta-\frac32\right)^2 =\frac94.

Thus the curve lies on the circle with centre \((0,\frac32)\) and radius \(\frac32\). More specifically, for \(0\leq\theta\leq\pi\), the points at \(\theta\) and \(\theta+\pi\) are identical: \(r(\theta+\pi)=-r(\theta)\), and the reversed radial direction represents the same point. Therefore \([0,\pi]\) traces the circular loop once, and \([\pi,2\pi]\) retraces that same loop. The enclosed area is \(\frac{9\pi}{4}\), not twice that value. Integrating over \([0,2\pi]\) would give \(\frac{9\pi}{2}\), because it counts the same circular loop twice.

A Common Counting Mistake

A frequent error is to assume that every interval between consecutive pole crossings creates a distinct loop. The example \(r=3\sin\theta\) shows why that assumption fails. The equation \(r=0\) gives crossings at \(0,\pi,2\pi\), but the first interval and the second interval trace the same circle, not two separate circles.

Pitfall: Consecutive pole crossings identify candidate intervals, not necessarily distinct enclosed regions. Before adding areas or reporting a loop area, check whether the corresponding parameter intervals trace different geometric boundaries or repeat the same one.

For the mistaken calculation, one might evaluate \(\frac12\int_0^{2\pi}9\sin^2\theta\,d\theta=\frac{9\pi}{2}\) and call that the area of the circle. That is the area counted with multiplicity two: \([0,\pi]\) traces the circle once, then \([\pi,2\pi]\) retraces it. The correction is to use one interval that traces the loop once, such as \([0,\pi]\), giving \(\frac12\int_0^\pi9\sin^2\theta\,d\theta=\frac{9\pi}{4}\).

Why the Interval Check Matters

The integral measures area accumulated along the parameterized trace. It does not automatically detect that the same boundary has been traversed before. Consequently, a correct integral with an interval that covers a repeated trace can produce a number larger than the geometric area of the region.

For a single-loop calculation, confirm these points before integrating:

The endpoints are the intended pole crossings, when the loop begins and ends at the pole.
The curve does not reach the pole at an intervening parameter value.
The chosen interval traces the loop boundary once rather than repeating an earlier trace.
The integrand is the square of the full radius function, including any coefficient or parentheses.

These checks keep the geometric question and the parameter calculation aligned. Once the interval is established, the remaining work is an ordinary definite integral; the essential geometric judgment is deciding which traversal represents one loop.

Check Your Understanding

Use the definition of a single loop and the area formula to answer these questions.

What equation identifies parameter values at which a polar curve reaches the pole?
Why do consecutive pole crossings give candidate loop endpoints rather than a guarantee of a distinct loop?
For \(r=1+\cos\theta\), what are a pair of consecutive pole crossings, and what is the area on that interval?
For \(r=3\sin\theta\), why does integrating over \([0,2\pi]\) count the circular region twice?
What should be checked about the trace before using \(\frac12\int_a^b r^2\,d\theta\) as the area of one loop?

What You'll Learn