Why overlap area requires a piecewise setup
When two polar curves bound regions that both extend from the pole outward, their common region follows whichever boundary is nearer to the pole at each angle. That boundary can change as the angle changes. The central task in an overlap-area problem is therefore not just to find intersections: it is to find every angular interval on which the same curve supplies the nearer boundary.
A circle centered at the pole has constant radius \(r=R\), where \(R>0\). A cardioid in standard orientation can be written \(r=a(1+\cos\theta)\), where \(a>0\). Its radius is nonnegative for every angle. We will use these two curves to practise solving intersections, dividing the angular domain, and evaluating the resulting integrals.
A reliable workflow
The polar area formula assigns area \(\frac12\int r^2\,d\theta\) to a region traced once with nonnegative radius over its angular interval. For an overlap, use that formula on the boundary of the common region, not automatically on either entire original region. For the centered circle and standard cardioid here, both regions reach outward from the pole at every angle, so comparing their radii gives the overlap boundary directly.
For a full circle and cardioid of this form, \(0\leq\theta\leq2\pi\) traces each region once.
Set the radii equal and solve within the chosen domain. Check whether an equality gives a genuine boundary intersection or only the pole.
Test one angle between consecutive intersection angles. The smaller nonnegative radius bounds the common region there.
Use one integral for each interval and add the results, including the factor \(\frac12\).
Solving the equality locates where the ordering might change, but does not by itself tell us which curve is nearer on either side. A test angle supplies that information. If two radii are continuous and are unequal on an interval, their ordering cannot switch inside that interval without another equality; this is why testing one point per interval is enough.
Worked Example: A circle and a cardioid with quadrantal intersections
Worked Example: A circle and a cardioid with quadrantal intersections
Find the common area of the circle \(r=2\) and the cardioid \(r=2(1+\cos\theta)\) for \(0\leq\theta\leq2\pi\).
At an intersection, the radii are equal: $$ 2=2(1+\cos\theta) \quad\Longrightarrow\quad \cos\theta=0. $$ Thus the intersection angles in this domain are \(\theta=\frac{\pi}{2}\) and \(\theta=\frac{3\pi}{2}\). Substituting either angle into the cardioid gives \(r=2\), confirming both intersection points.
Test \(\theta=0\), which lies in the interval from \(0\) to \(\frac{\pi}{2}\). The cardioid radius is \(2(1+\cos 0)=4\), which exceeds the circle radius \(2\), so the circle is nearer to the pole. On the middle interval, test \(\theta=\pi\): the cardioid radius is \(2(1+\cos\pi)=0\), so the cardioid is nearer. On the final interval, test \(\theta=2\pi\): the cardioid radius is \(4\), so the circle is nearer again. Therefore $$ A=\frac12\left[ \int_0^{\pi/2}4\,d\theta +\int_{\pi/2}^{3\pi/2}4(1+\cos\theta)^2\,d\theta +\int_{3\pi/2}^{2\pi}4\,d\theta \right]. $$
An antiderivative needed for the cardioid integral is $$ F(\theta)=\frac{3\theta}{2}+2\sin\theta+\frac{\sin 2\theta}{4}, $$ because \(F'(\theta)=\frac32+2\cos\theta+\frac12\cos2\theta=(1+\cos\theta)^2\). The two circle integrals together contribute \(\frac12(4\pi)=2\pi\). For the middle interval, $$ \int_{\pi/2}^{3\pi/2}(1+\cos\theta)^2\,d\theta =F(3\pi/2)-F(\pi/2) =\left(\frac{9\pi}{4}-2\right)-\left(\frac{3\pi}{4}+2\right) =\frac{3\pi}{2}-4. $$ Including the factor \(4\) in the squared cardioid radius and the area factor \(\frac12\), this interval contributes \(2(\frac{3\pi}{2}-4)=3\pi-8\). Hence $$ A=2\pi+(3\pi-8)=5\pi-8. $$
The boundary check gives \(r=2\) for both curves at each reported intersection, and differentiating \(F\) recovers the cardioid’s squared-radius integrand. The area is positive, as expected for a nonempty overlap.
Why the minimum-radius rule works
At a fixed angle \(\theta\), suppose the circle occupies radial distances from \(0\) to \(r_1(\theta)\), and the cardioid occupies distances from \(0\) to \(r_2(\theta)\). A point lies in both regions precisely when its radius \(r\) satisfies both \(0\leq r\leq r_1(\theta)\) and \(0\leq r\leq r_2(\theta)\). Together, those inequalities say $$ 0\leq r\leq\min(r_1(\theta),r_2(\theta)). $$ Thus the common region has the smaller radius as its boundary on that ray. Applying the polar area formula to this boundary gives the stated integral. The proof relies on the regions extending continuously outward from the pole on the rays under consideration; if a region has a radial gap or does not exist on some angles, its radial description must be adjusted instead of applying the rule blindly.
The intersection angles divide the domain into intervals on which the order can be tested. Between consecutive equality angles, continuous radii cannot exchange order without becoming equal somewhere. This justifies the piecewise method, provided all relevant intersections have been found and the radii are compared on each resulting interval.
Worked Example: A non-quadrantal intersection
Worked Example: A non-quadrantal intersection
Find the common area of the circle \(r=2\) and the cardioid \(r=\frac32(1+\cos\theta)\) over \(0\leq\theta\leq2\pi\).
Equating radii gives $$ 2=\frac32(1+\cos\theta) \quad\Longrightarrow\quad \cos\theta=\frac13. $$ Let \(\alpha=\arccos(\frac13)\), so \(0<\alpha<\frac{\pi}{2}\). The intersection angles are \(\alpha\) and \(2\pi-\alpha\). Substitution gives cardioid radius \(\frac32(1+\frac13)=2\) at both angles.
At \(\theta=0\), the cardioid radius is \(3\), so the circle is nearer on \([0,\alpha]\). At \(\theta=\pi\), the cardioid radius is \(0\), so the cardioid is nearer on \([\alpha,2\pi-\alpha]\). At \(\theta=2\pi\), the cardioid radius is again \(3\), so the circle is nearer on \([2\pi-\alpha,2\pi]\). Since \(\sin\alpha=\frac{2\sqrt2}{3}\), the area is $$ A=\frac12\left[ \int_0^\alpha 4\,d\theta +\int_\alpha^{2\pi-\alpha}\frac94(1+\cos\theta)^2\,d\theta +\int_{2\pi-\alpha}^{2\pi}4\,d\theta \right]. $$
The circle pieces together contribute \(4\alpha\) after applying the area factor. Using the same antiderivative \(F\) as above, the cardioid integral without its factor \(\frac94\) is $$ F(2\pi-\alpha)-F(\alpha) =3\pi-3\alpha-4\sin\alpha-\frac12\sin2\alpha. $$ Here \(\sin2\alpha=2\sin\alpha\cos\alpha=\frac{4\sqrt2}{9}\), so this difference is \(3\pi-3\alpha-\frac{26\sqrt2}{9}\). Therefore $$ A=4\alpha+\frac98\left(3\pi-3\alpha-\frac{26\sqrt2}{9}\right) =\frac{27\pi}{8}+\frac{5\alpha}{8}-\frac{13\sqrt2}{4}, \qquad \alpha=\arccos\!\left(\frac13\right). $$
The radius equality checks at both endpoints, and \(F'(\theta)=(1+\cos\theta)^2\) verifies the integrated factor. The cardioid lies inside the circle on most angles in this example, while the circle itself limits the overlap near \(\theta=0\) and \(2\pi\); the piecewise integral reflects both parts.
Worked Example: The circle is the nearer boundary on most angles
Worked Example: The circle is the nearer boundary on most angles
Find the common area of \(r=1\) and \(r=2(1+\cos\theta)\) over one full turn.
The intersections satisfy $$ 1=2(1+\cos\theta) \quad\Longrightarrow\quad \cos\theta=-\frac12. $$ Thus \(\theta=\frac{2\pi}{3}\) and \(\theta=\frac{4\pi}{3}\). At both angles, the cardioid radius is \(1\), so these are intersections. At \(\theta=0\), the cardioid radius is \(4\), making the circle nearer. At \(\theta=\pi\), the cardioid radius is \(0\), making the cardioid nearer. The circle bounds the overlap on the remaining angular interval, whose total length is \(\frac{4\pi}{3}\). Hence its contribution is \(\frac12\cdot\frac{4\pi}{3}=\frac{2\pi}{3}\).
For the cardioid interval, the antiderivative \(F\) gives $$ \int_{2\pi/3}^{4\pi/3}(1+\cos\theta)^2\,d\theta =F(4\pi/3)-F(2\pi/3) =\pi-\frac{7\sqrt3}{4}. $$ The squared cardioid radius is \(4(1+\cos\theta)^2\), so this part contributes \(2(\pi-\frac{7\sqrt3}{4})\) to the area. The total is $$ A=\frac{2\pi}{3}+2\left(\pi-\frac{7\sqrt3}{4}\right) =\frac{8\pi}{3}-\frac{7\sqrt3}{2}. $$
Substitution confirms radius \(1\) at both intersection angles, and the antiderivative check \(F'=(1+\cos\theta)^2\) applies here as well. The result is positive, and it is smaller than the area \(\pi\) of the circle, as any overlap with that circle must be.
A common mistake: integrating the outer curve
In the first example, the cardioid is farther from the pole than the circle near \(\theta=0\). A tempting but incorrect setup is to integrate the cardioid’s squared radius over the entire turn and call that the overlap. That computes the area of the whole cardioid, not its common area with the circle. For example, at \(\theta=0\), the cardioid reaches \(r=4\), but points with \(2<r\leq4\) are outside the circle and therefore outside the overlap.
The correction is to use the smaller radius on each interval: the circle on \([0,\frac{\pi}{2}]\) and \([\frac{3\pi}{2},2\pi]\), and the cardioid on \([\frac{\pi}{2},\frac{3\pi}{2}]\). Another related error is to subtract one curve’s area from the other’s without establishing that the desired region is a difference. Common area is bounded radially by the nearer curve; it is not generally the difference of the two full areas.
Check Your Understanding
For each question, identify the angular intervals before evaluating or interpreting the integral.
- For \(r=1\) and \(r=2(1+\cos\theta)\), what are the intersection angles in \(0\leq\theta\leq2\pi\)?
- In the first worked example, why does the cardioid bound the overlap near \(\theta=\pi\), while the circle bounds it near \(\theta=0\)?
- For two nonnegative radial boundaries, explain why the overlap radius is their minimum rather than their maximum.
- What must be checked after solving an equation for intersection angles?
- In what kind of radial region would the simple minimum-radius rule need modification?
