Why Inner-Loop Bounds Matter
A polar curve can cross the pole, the point whose distance from the origin is zero. Between two such crossings, the curve may trace a small loop inside a larger part of the graph. To find the area of that loop, the key task is to choose the angle interval that traces it exactly once.
The usual polar area formula applies, but the bounds must match the loop. Integrating over too large an interval can include other parts of the graph; integrating over too small an interval can leave out part of the loop. A negative value of \(r\) does not mean the area is negative: the distance from the pole is \(|r|\), and the area formula uses \(r^2\).
Finding the Interval for an Inner Loop
A pole crossing occurs at an angle \(\theta\) where \(r(\theta)=0\). For many inner-loop examples, the loop is traced between consecutive pole crossings across which \(r\) is negative. Solve \(r(\theta)=0\), determine the sign of \(r\) between the relevant roots, and then verify that the chosen interval traces the intended loop once.
When \(r<0\), the plotted point lies in the direction opposite the angle \(\theta\). This changes where the point appears, but it does not prevent the curve from being traced as \(\theta\) varies. In particular, a negative radius is not a reason to change the area formula or to take an absolute value inside it: squaring \(r\) already gives a nonnegative integrand.
Why the Polar Area Formula Works for a Loop
For a small angular change \(\Delta\theta\), a sector with radius \(r\) has area approximately \(\frac12r^2\Delta\theta\). Partition an interval \([\alpha,\beta]\) into small angular pieces. Adding these sector areas gives an approximation to the area swept out; as the partition becomes finer, the sum approaches the integral of \(\frac12r^2\).
This reasoning concerns distance from the pole, which is \(|r|\). Since \((|r|)^2=r^2\), the same expression works when the radius is negative. The crucial condition is that the interval covers the loop once. If the curve traces a portion more than once, the integral counts the swept area with that repetition.
Worked Examples
Worked Example: A Limaçon with an Inner Loop
Find the area of the inner loop of \(r=1+2\cos\theta\).
First locate the pole crossings:
On \(0\leq\theta\leq2\pi\), the crossings are \(\theta=\frac{2\pi}{3}\) and \(\theta=\frac{4\pi}{3}\). At \(\theta=\pi\), \(r=1+2\cos\pi=-1<0\), so \(r\) is negative between these roots. This interval traces the inner loop once.
Apply the formula and simplify the square:
Evaluate each difference. The angle term is \(\frac32(4\pi/3-2\pi/3)=\pi\). The sine terms are \(2(-\sqrt3/2-\sqrt3/2)=-2\sqrt3\) and \(\frac12(\sqrt3/2-(-\sqrt3/2))=\frac{\sqrt3}{2}\). Therefore,
Check: Since \(\pi\approx3.142\) and \(\frac{3\sqrt3}{2}\approx2.598\), the area is approximately \(0.543\), which is positive. The expression is the integral over the interval between the two pole crossings, not over the entire curve.
Worked Example: Bounds Involving an Inverse Trigonometric Function
Find the area of the inner loop of \(r=2+3\cos\theta\) over one revolution.
Set \(r=0\) to find the crossings. Let \(\alpha=\arccos(-2/3)\), where \(0<\alpha<\pi\). The crossings on \(0\leq\theta\leq2\pi\) are \(\alpha\) and \(2\pi-\alpha\). At \(\theta=\pi\), \(r=2+3(-1)=-1<0\), so the radius is negative between these crossings. This is the interval tracing the inner loop once.
Square the radius and use \(\cos^2\theta=(1+\cos2\theta)/2\):
Because \(\sin(2\pi-\alpha)=-\sin\alpha\), the sine difference in the middle term is \(-2\sin\alpha\). Also, \(\sin\alpha=\sqrt5/3\), so that contribution is \(6(-2\sqrt5/3)=-4\sqrt5\). For the last term, \(\sin(4\pi-2\alpha)=-\sin2\alpha\), giving a difference of \(-2\sin2\alpha\). Since \(\sin2\alpha=2\sin\alpha\cos\alpha=-4\sqrt5/9\), this contribution is \(\frac98(8\sqrt5/9)=\sqrt5\). The angle contribution is \(\frac{17}{4}(2\pi-2\alpha)=\frac{17}{2}(\pi-\alpha)\). Thus,
Check: Using \(\alpha\approx2.301\), the angle term is about \(7.149\) and \(3\sqrt5\approx6.708\); their difference is about \(0.441>0\). The crossing angles are symmetric about \(\pi\), as expected for this cosine equation.
Worked Example: A Sine Equation with a Negative-Radius Interval
Find the area of the inner loop of \(r=2+4\sin\theta\) over \(0\leq\theta\leq2\pi\).
The pole crossings satisfy \(2+4\sin\theta=0\), so \(\sin\theta=-1/2\). Thus \(\theta=7\pi/6\) and \(\theta=11\pi/6\). At \(\theta=3\pi/2\), \(r=2+4(-1)=-2<0\), confirming that the loop is traced between these crossings.
The angle contribution is \(6(11\pi/6-7\pi/6)=4\pi\). The cosine contribution is \(-8(\sqrt3/2-(-\sqrt3/2))=-8\sqrt3\). For the last contribution, \(\sin(11\pi/3)=-\sqrt3/2\) and \(\sin(7\pi/3)=\sqrt3/2\), so it is \(-2(-\sqrt3)=2\sqrt3\). Therefore,
Check: The exact area is positive because \(4\pi\approx12.566\) and \(6\sqrt3\approx10.392\). The loop is centered in the interval where \(r<0\), and both endpoints are pole crossings.
A Common Mistake: Integrating the Radius Instead of Its Square
The polar area formula is not \(\int r\,d\theta\), nor is it \(\frac12\int r\,d\theta\). It is \(\frac12\int r^2\,d\theta\). A negative result from integrating \(r\) over an inner-loop interval does not represent a negative area; it signals that the wrong integrand was used.
For example, with \(r=2+4\cos\theta\), the crossings are \(2\pi/3\) and \(4\pi/3\), and the radius is negative between them. The incorrect unsquared integral evaluates as follows:
The endpoint evaluation matters: the sine contribution is \(-4\sqrt3\), not \(-8\sqrt3\). But even with that evaluation corrected, this is still not an area, because the radius was not squared. The correct setup is \(\frac12\int_{2\pi/3}^{4\pi/3}(2+4\cos\theta)^2\,d\theta\), which has a nonnegative integrand.
A Reliable Setup
Before evaluating an integral, confirm that its bounds and integrand match the loop. The following sequence keeps the geometry and calculation connected.
Solve \(r(\theta)=0\) in the angular range being considered.
Use the relevant consecutive crossings, check the sign of \(r\) between them, and confirm the curve traces the intended loop once.
Use \(\frac12\int_{\alpha}^{\beta}r^2\,d\theta\), with the selected crossing angles as bounds.
Show the endpoint substitutions carefully. The final area should be nonnegative and should correspond to the chosen loop alone.
Why the Interval Must Match the Loop
The factor \(\frac12r^2\) measures the area swept per unit angle, so the bounds determine exactly which angular sweep is counted. Integrating through pole crossings can join separate pieces of the curve in the parameter even when they belong to different parts of the graph. Conversely, selecting only part of the interval that traces a loop gives only part of its area.
For this reason, solving \(r=0\) is necessary but not always sufficient. A curve may have several pole crossings, or a chosen interval may retrace part of the graph. Use the equation, a sign check, and the curve’s tracing behavior together. For a standard inner-loop example, two consecutive crossings surrounding an interval where \(r<0\) identify the intended bounds; still check that no additional tracing occurs on that interval.
Check Your Understanding
For each question, identify the appropriate bounds or area setup before evaluating.
- For \(r=1+2\cos\theta\), what are the pole-crossing angles on \(0\leq\theta\leq2\pi\), and which interval traces the inner loop?
- Why does a negative value of \(r\) not make the polar area integral negative?
- Write the area integral for the inner loop of \(r=2+3\cos\theta\), using \(\alpha=\arccos(-2/3)\).
- For \(r=2+4\sin\theta\), verify that \(\theta=3\pi/2\) lies between its two pole crossings and that \(r\) is negative there.
- Explain why integrating \(\frac12r^2\) over an interval that traces the loop more than once would not give the area of a single tracing.
