Why Circle–Cardioid Areas Need Care
A circle and a cardioid may intersect at several points, and their relative distance from the pole can change as the angle changes. To find the area between them, it is not enough to find intersection angles: on each interval, we also need to know which curve has the larger radius.
We will study the circle \(r=2\cos\theta\) and the cardioid \(r=2(1-\cos\theta)\). They are symmetric about the polar axis. The bounded region of interest lies between their nonzero intersection points, with the circle arc passing through its rightmost point and the cardioid arc passing through the pole.
Find the Intersections and the Radial Order
For nonzero intersections represented at the same angle, set the radii equal:
Dividing by \(2\) and solving gives \(\cos\theta=1/2\), so the nonzero intersections occur at \(\theta=\pm\pi/3\) on the relevant circle arc. Their common radius is \(1\).
There is also an intersection at the pole. The cardioid reaches the pole at \(\theta=0\), while the circle reaches it when \(\cos\theta=0\). Since the pole can be represented by different angles in polar coordinates, equating the radii at the same angle does not reveal every intersection. For this region, the two nonzero intersections give the angle bounds; the pole lies on the cardioid boundary between them.
Now compare the radii inside those bounds:
For \(-\pi/3<\theta<\pi/3\), \(\cos\theta>1/2\), so \(4\cos\theta-2>0\). The circle is therefore outer throughout the interval, and the cardioid is inner. At either endpoint the radii are equal.
Why the Difference of Squared Radii Works
The polar area formula gives the area swept by a radius \(r(\theta)\) over an angle interval as \(\frac12\int r(\theta)^2\,d\theta\). At each angle, the region between an outer radius \(R(\theta)\) and an inner radius \(r(\theta)\) is the outer swept area minus the inner swept area. Thus, provided the radii are nonnegative and the radial order is fixed on the interval,
If the order changes, the difference inside the integral would change sign. Splitting at the angle where the radii are equal keeps the outer radius first on every piece and ensures each area contribution is counted positively.
Worked Example: The Symmetric Region Between the Circle and Cardioid
Worked Example: The Symmetric Region Between the Circle and Cardioid
Find the area between \(r=2\cos\theta\) and \(r=2(1-\cos\theta)\) from their upper to lower nonzero intersections.
Step 1: Find the angle bounds. Equating the radii gives \(2\cos\theta=2(1-\cos\theta)\), so \(4\cos\theta=2\), or \(\cos\theta=1/2\). The bounds are \(-\pi/3\) and \(\pi/3\).
Step 2: Check which radius is larger. At \(\theta=0\), the circle has radius \(2\) and the cardioid has radius \(0\). More generally, on this interval \(4\cos\theta-2>0\), so the circle is outer.
Step 3: Set up and simplify the area.
Step 4: Evaluate and check. An antiderivative is \(4\sin\theta-2\theta\), since its derivative is \(4\cos\theta-2\). Therefore,
This is positive, as required for an area. Numerically it is about \(2.74\) square units. As a further check, the integrand \(4\cos\theta-2\) is positive between the bounds and zero at both endpoints, matching the radial-order check.
Worked Example: The Upper Half of the Same Region
Worked Example: The Upper Half of the Same Region
Find the area in the preceding region above the polar axis.
Step 1: Set the bounds and order. The upper half corresponds to \(0\leq\theta\leq\pi/3\). The circle remains outer because \(4\cos\theta-2\geq0\) on this interval.
Step 2: Use the squared-radius difference.
The antiderivative checks because \(\frac{d}{d\theta}(4\sin\theta-2\theta)=4\cos\theta-2\). Doubling this result gives \(4\sqrt3-\frac{4\pi}{3}\), exactly the symmetric area from the first example. That agrees with the symmetry of both curves about the polar axis.
Worked Example: A Circle and a Smaller Cardioid
Worked Example: A Circle and a Smaller Cardioid
Find the area between the circle \(r=2\cos\theta\) and the cardioid \(r=1-\cos\theta\), between their nonzero intersections.
Step 1: Find the bounds. Equating the radii gives \(2\cos\theta=1-\cos\theta\), so \(\cos\theta=1/3\). Let \(\alpha=\arccos(1/3)\). The relevant bounds are \(-\alpha\) and \(\alpha\).
Step 2: Confirm the order. The difference between the circle radius and cardioid radius is \(2\cos\theta-(1-\cos\theta)=3\cos\theta-1\). For \(-\alpha<\theta<\alpha\), \(\cos\theta>1/3\), so the circle is outer. At \(\theta=0\), their radii are \(2\) and \(0\), which also confirms this order.
Step 3: Integrate.
An antiderivative of the integrand, including the factor \(1/2\), is \(F(\theta)=\frac{\theta}{4}+\frac{3}{8}\sin(2\theta)+\sin\theta\). Indeed,
Since \(\cos\alpha=1/3\), we have \(\sin\alpha=2\sqrt2/3\) and \(\sin(2\alpha)=4\sqrt2/9\). The integrand is even, so
The formula is positive, and the antiderivative differentiates to the required half-integrand. The bounds are also consistent with the circle being outer inside the interval and the radii agreeing at its endpoints.
A Common Error: Subtracting Radii Instead of Their Squares
A tempting but incorrect setup is to integrate one-half the difference of the radii. For the first example, that mistake would give
This is only half the correct area. The area swept by a polar radius depends on its square, not on the radius alone. The correct setup is \(\frac12\int(R^2-r^2)\,d\theta\), which gives \(4\sqrt3-\frac{4\pi}{3}\) for the full region. A useful check is that the integrand should be one-half the difference of squared radii and should be nonnegative when the outer curve is listed first.
A Reliable Setup for These Problems
For a circle-and-cardioid area problem, organize the work in this order:
Equate the radii to find nonzero intersections, and consider separately whether the pole is also on both curves.
Use the intersections that bound the particular region being measured; do not assume every intersection bounds the same region.
Compare the radii on each interval, using a test angle or the sign of their difference.
Use one-half the outer squared radius minus the inner squared radius, splitting wherever the order changes.
Confirm the integrand is nonnegative on each piece, verify the antiderivative, and check that the final area is positive.
The main challenge is usually not the integration. It is correctly identifying which bounded region is intended and maintaining the correct outer-minus-inner order. The intersection angles, radial comparison, and squared-radius formula together make the setup precise.
Check Your Understanding
Use the circle \(r=2\cos\theta\) and cardioid \(r=2(1-\cos\theta)\) unless a question specifies otherwise.
- What are the nonzero intersection angles, and what is the common radius there?
- Why does equating the radii at the same angle fail to identify the pole as an intersection?
- Which curve is outer for \(-\pi/3<\theta<\pi/3\), and what inequality establishes this?
- Write the area integral for the upper half of the bounded region without evaluating it.
- Why must the area integrand involve the difference of squared radii rather than the difference of radii?
