Why Switching Curves Require a Split
To find the area between two polar curves, the integrand must use the curve farther from the pole as the outer radius and the nearer curve as the inner radius. Sometimes the same curve remains outer throughout the angle interval. In other problems, the curves intersect and exchange order. Then one formula with a fixed outer curve will not work over the entire interval.
The essential steps are to locate the intersection angles, test the radial order between consecutive intersections, and split the integral wherever that order changes. An intersection is a possible switching point, but it does not by itself prove that the order switches there. Testing values on either side settles the question.
Finding Where the Order Can Change
For two polar curves given by \(r=f(\theta)\) and \(r=g(\theta)\), their points on the same ray from the pole coincide when their radii are equal. Thus, solve \(f(\theta)=g(\theta)\) on the angle interval used to describe the region. These angles divide the interval into pieces on which the curves do not intersect.
On each piece, choose a test angle and compare the two radii. The larger nonnegative radius is farther from the pole and is the outer curve. If the radii are continuous, their difference cannot change sign on a piece without becoming zero somewhere on that piece. Such a zero would be another intersection. This is why one test angle determines the order throughout that piece.
In the examples below, both curves have positive radii on the intervals considered, so each angle describes a ray with an unambiguous inner and outer boundary. When working with curves that cross the pole or have negative radial values, first ensure that the angle interval traces the intended region correctly.
Set the two radius functions equal and solve within the chosen angle interval.
List the endpoints and intersection angles in increasing order.
Choose an angle inside each piece and compare the radii to identify the outer curve.
On each piece, integrate one-half of outer radius squared minus inner radius squared, then add the pieces.
Why the Area Integral Must Be Split
For a small angle change \(d\theta\), a sector with radius \(R\) has area approximately \(\frac12R^2d\theta\). The thin region between outer radius \(R(\theta)\) and inner radius \(r(\theta)\) therefore contributes approximately \(\frac12(R(\theta)^2-r(\theta)^2)d\theta\). Adding these contributions over an interval gives the polar area formula.
The subtraction must be outer squared minus inner squared so that each contribution is nonnegative. If the curves exchange order, keeping the same subtraction across the intersection makes the integrand negative on the piece where the chosen “outer” curve is actually inner. The correct approach is to reverse the subtraction there, which is exactly what splitting the integral accomplishes.
Here \(R(\theta)\) is the outer radius and \(r(\theta)\) is the inner radius at each angle. If their roles change at an angle \(c\), the area is written as the sum of one integral from \(a\) to \(c\) and another from \(c\) to \(b\), with the radii ordered correctly in each.
Worked Example: Two Curves Switch Order Once
Worked Example: Area Between Two Curves in the First Quadrant
Find the area between \(r_1=1+\cos\theta\) and \(r_2=1+\sin\theta\) for \(0\leq\theta\leq\frac{\pi}{2}\).
First find the intersection angles:
Test \(\theta=0\), which lies between \(0\) and \(\frac{\pi}{4}\):
Thus \(r_1\) is outer on \([0,\frac{\pi}{4}]\). Test \(\theta=\frac{\pi}{2}\), which lies between \(\frac{\pi}{4}\) and \(\frac{\pi}{2}\):
Thus \(r_2\) is outer on \([\frac{\pi}{4},\frac{\pi}{2}]\). The area is
The difference in the first integrand simplifies to
An antiderivative is \(F(\theta)=2(\sin\theta+\cos\theta)+\frac12\sin 2\theta\), since \(F'(\theta)=2(\cos\theta-\sin\theta)+\cos 2\theta\). Its values at the needed endpoints are
The first integral before the factor \(\frac12\) is \(F(\frac{\pi}{4})-F(0)=2\sqrt2-\frac32\). The second integral before the factor \(\frac12\) is the negative of the first difference over \([\frac{\pi}{4},\frac{\pi}{2}]\), so it equals \(F(\frac{\pi}{4})-F(\frac{\pi}{2})=2\sqrt2-\frac32\). Therefore,
Both contributions are positive, as they must be for areas. The result is approximately \(1.328\) square units. The intersection radii also agree: at \(\theta=\frac{\pi}{4}\), both equal \(1+\frac{\sqrt2}{2}\), confirming the shared boundary at the split.
Worked Example: Two Switches Over a Full Turn
Worked Example: Avoiding Signed Cancellation
Find the area between \(r_1=2+\cos\theta\) and \(r_2=2+\sin\theta\) for \(0\leq\theta\leq2\pi\). Both radii are positive throughout this interval. Intersections satisfy \(\cos\theta=\sin\theta\), giving \(\theta=\frac{\pi}{4}\) and \(\theta=\frac{5\pi}{4}\) within the interval.
At \(\theta=0\), \(r_1=3\) and \(r_2=2\), so \(r_1\) is outer on \([0,\frac{\pi}{4}]\). At \(\theta=\frac{\pi}{2}\), \(r_1=2\) and \(r_2=3\), so \(r_2\) is outer on \([\frac{\pi}{4},\frac{5\pi}{4}]\). At \(\theta=\frac{3\pi}{2}\), \(r_1=2\) and \(r_2=1\), so \(r_1\) is outer on \([\frac{5\pi}{4},2\pi]\).
The squared-radius difference, taken in the order \(r_1^2-r_2^2\), is
An antiderivative is \(F(\theta)=4(\sin\theta+\cos\theta)+\frac12\sin 2\theta\); differentiating gives \(F'(\theta)=4(\cos\theta-\sin\theta)+\cos 2\theta\). The area contributions are then:
On the middle interval, \(r_2\) is outer, which explains the reversed order of the \(F\)-values in \(A_2\). On the first and last intervals, \(r_1\) is outer. Their contributions are not equal: the first is \(2\sqrt2-\frac74\), and the last is \(2\sqrt2+\frac74\). Their sum is \(4\sqrt2\), and the total area is
The unequal first and last contributions arise from the unequal interval lengths and the changing values of the integrand. No symmetry claim is needed to add them. As a check, each contribution is positive, and the two curves have equal radii at both split angles.
Worked Example: One Switch on a Restricted Interval
Worked Example: Area on a Half-Turn
Find the area between \(r_1=2+\cos\theta\) and \(r_2=2\) for \(0\leq\theta\leq\pi\). Their radii are equal when \(2+\cos\theta=2\), so \(\cos\theta=0\), giving the one interior intersection \(\theta=\frac{\pi}{2}\).
At \(\theta=0\), \(r_1=3>2=r_2\), so \(r_1\) is outer on \([0,\frac{\pi}{2}]\). At \(\theta=\pi\), \(r_1=1<2=r_2\), so \(r_2\) is outer on \([\frac{\pi}{2},\pi]\). The difference in the first order is
An antiderivative is \(G(\theta)=4\sin\theta+\frac{\theta}{2}+\frac14\sin2\theta\). Indeed, \(G'(\theta)=4\cos\theta+\frac12+\frac12\cos2\theta=4\cos\theta+\cos^2\theta\). Its endpoint values are \(G(0)=0\), \(G(\frac{\pi}{2})=4+\frac{\pi}{4}\), and \(G(\pi)=\frac{\pi}{2}\).
The reversed endpoint order in the second contribution accounts for \(r_2\) becoming outer. The two contributions are positive and sum to \(4\) square units.
A Common Mistake: Integrating the Signed Difference
A tempting shortcut is to integrate \(r_1^2-r_2^2\) across the whole interval and take one-half of the result, without checking whether that difference changes sign. For the full-turn example, that shortcut gives
because each sine or cosine term integrates to zero over a full turn. The resulting “area” would be zero, even though the curves enclose a positive area between them.
The mistake is that signed contributions on intervals with opposite radial order cancel. Area does not cancel in this way. The correction is to split at the intersection angles, put the outer radius first on every interval, and add the positive contributions. Equivalently, for nonnegative radii the integrand can be written as one-half the absolute value of the squared-radius difference, but splitting explicitly makes the outer and inner curves clear.
Check Your Understanding
For each question, identify the angle intervals and radial order before setting up the area.
- For \(r_1=3+\cos\theta\) and \(r_2=3+\sin\theta\) on \(0\leq\theta\leq\frac{\pi}{2}\), find the intersection angle and state which curve is outer on each side.
- Why is it enough to test one angle between consecutive intersections when the radius functions are continuous?
- For \(r_1=2+\cos\theta\) and \(r_2=2\) on \(0\leq\theta\leq\pi\), write the split area integral without evaluating it.
- In the full-turn example, why does integrating the signed squared-radius difference over the entire interval fail to give the area?
- Does every intersection necessarily mark a change in radial order? Explain what additional check is needed.
