Area of an Inner Loop

Why an Inner Loop Needs Its Own Bounds

A limacon with an inner loop passes through the pole and traces a smaller loop before continuing around its larger outer portion. Finding the area of that inner loop is not simply a matter of choosing a familiar full-turn interval: the bounds must begin and end at pole crossings and must trace the small loop exactly once.

The key steps are to find where the radius is zero, determine where it is negative, and use that interval in the polar area formula. A negative polar radius does not mean a negative area. It means that the plotted point lies in the direction opposite to the angle, and the squared radius in the area formula is still nonnegative.

Finding the Pole Crossings

Consider the standard cosine limacon

The curve passes through the pole when \(r=0\). Thus its pole-crossing angles satisfy \(\cos\theta=-a/b\). There are two distinct crossings in a turn precisely when \(b>a\), because then \(-a/b\) lies strictly between \(-1\) and \(1\). This is the case that produces an inner loop.

Define \(\alpha=\arccos(-a/b)\), taking \(\alpha\) between \(0\) and \(\pi\). Since cosine has the same value at \(\alpha\) and \(2\pi-\alpha\), the two crossings in \([0,2\pi]\) are \(\alpha\) and \(2\pi-\alpha\). Between them, cosine is less than \(-a/b\), so \(a+b\cos\theta<0\).

Why does this interval trace the inner loop, rather than just part of it? As \(\theta\) runs from one zero of \(r\) to the next, the plotted point starts at the pole, moves along the negative-radius portion of the limacon, and returns to the pole. The point at angle \(\theta\) with negative radius is plotted in the direction \(\theta+\pi\); that reversal is what places this part of the curve inside the larger loop. The parameter passes through the interval once, so the pole-to-pole arc closes the inner loop once.

The condition \(b>a\) matters. If \(b=a\), the radius reaches zero at a single angle where cosine is \(-1\), rather than becoming negative over an interval; there is no inner loop of positive area. If \(b<a\), the radius never reaches zero and never becomes negative, so this form of limacon has no inner loop.

Deriving the Inner-Loop Area Formula

The polar area formula states that a curve \(r=r(\theta)\) traced once over \(\theta\in[u,v]\) encloses area \(\frac12\int_u^v r(\theta)^2\,d\theta\), provided the interval traces the boundary of the region once. It follows by adding the areas of thin sectors: a sector with radius \(r\) and angular width \(d\theta\) has area approximately \(\frac12r^2\,d\theta\). Integrating gives the area of the region. The earlier material on deriving the polar area formula establishes this result; here the important task is choosing the interval that encloses this particular loop.

For the cosine limacon, substitute the two pole-crossing angles as the limits. The integral can also be evaluated in a reusable form.

To evaluate the integral, use \(\int\cos\theta\,d\theta=\sin\theta\) and \(\cos^2\theta=(1+\cos2\theta)/2\). An antiderivative of \(r^2\) is therefore

To see that the stated interval gives the inner-loop area, first note that its endpoints satisfy \(r=0\). As just described, the curve starts and ends at the pole on this interval. Moreover, the radius is strictly negative between the endpoints, so this is exactly the part of the parameter range that traces the inner loop. Consequently, the polar area formula applies to this single closed trace. In particular, there is no need to insert an extra minus sign when \(r<0\): the area integrand is \(r^2\).

Evaluating the antiderivative at the endpoints gives a general formula. Write \(s=\sin\alpha=\sqrt{1-a^2/b^2}\); since \(\alpha\in(0,\pi)\), \(s>0\), and \(\sin2\alpha=-2(a/b)s\). Also, \(\sin(2\pi-\alpha)=-\sin\alpha\) and \(\sin(4\pi-2\alpha)=-\sin2\alpha\). Thus

For the final equality, substituting \(\sin2\alpha=-2(a/b)\sin\alpha\) changes the last term in the first line to \(+ab\sin\alpha\); together with \(-4ab\sin\alpha\), this gives \(-3ab\sin\alpha\). Therefore the inner-loop area is

Worked Examples

Common Mistake: Choosing an Open Arc

A frequent error is to use the interval from \(0\) to the first pole crossing, as if reaching the pole once were enough to enclose the inner loop. For \(r=2+4\cos\theta\), that mistaken interval is \([0,2\pi/3]\). Its endpoints are not both at the pole: at \(\theta=0\), \(r=6\), while at \(\theta=2\pi/3\), \(r=0\). It traces an arc from a point on the outer part of the curve to the pole, not a closed inner-loop boundary.

This is not the inner-loop area; the interval does not trace a closed loop. The correction is to locate both pole crossings and use \([2\pi/3,4\pi/3]\), which returns to the pole after tracing the inner loop. A useful check before integrating is therefore: do both endpoints satisfy \(r=0\), and does the interval trace the small loop once?

Why the Method Matters

For a limacon with an inner loop, the polar radius changes sign, but the curve remains an ordinary geometric boundary. Understanding how the negative-radius interval corresponds to the inner trace prevents two distinct errors: using bounds that do not close the loop, and treating a negative radius as a negative contribution to area. The area formula squares the radius, while the interval identifies which region is being measured.

For \(r=a+b\cos\theta\), the procedure is systematic: verify \(0<a<b\), solve \(r=0\), choose the consecutive crossings enclosing the negative-radius interval, and integrate \(r^2/2\) over those bounds. For a sine limacon, solve its own crossing equation and select the interval where its radius is negative. In either case, checking the endpoints and the sign of the radius in the interval is part of setting up the calculation correctly.