Split an overlap by its radial boundary

Try

Two polar curves can bound a common region without one being the outer curve throughout. Consider A: r=2 cos θ and B: r=2 sin θ. Convert the first equation by multiplying by r: x²+y²=2x, hence (x−1)²+y²=1. The second similarly gives x²+(y−1)²=1. They are unit circles centered at (1,0) and (0,1).

Draw both disks and mark their shared interior. The entire overlap lies in the first quadrant: the first disk has x≥0 and the second has y≥0. This observation lets us use rays with 0≤θ≤π/2 and nonnegative radial distances. We are finding the area inside both disks, not the area inside one and outside the other.

Predict

On a ray close to the positive x-axis, which circle is reached first? Predict which formula bounds the shared part of that ray. Repeat near the positive y-axis. If the answer changes, one integral using a single radial boundary cannot describe the entire lens. Predict the switching angle before opening the calculation.

Also predict whether solving 2 cos θ=2 sin θ finds every Cartesian intersection. Remember that the origin has more than one polar representation.

Experiment

Which circle limits the overlap?

Circle A: r=2 cos θ, center (1,0), radius 1. Circle B: r=2 sin θ, center (0,1), radius 1. The shaded lens is inside both disks. The probe uses 0≤θ≤π/2, where both radii are nonnegative.

Probe angle in radians 01π/162π/163π/16π/45π/166π/167π/16π/2

Predict whether A or B supplies the shorter radial boundary, then inspect the labeled endpoints. The dashed ray is a direction, not another boundary of the full lens.

Text equivalent: the circle-marked A endpoint is (1.70711, 0.70711); the square-marked B endpoint is (0.70711, 0.29289). The thick radial segment ends at (0.70711, 0.29289), inside both disks. Coordinates are rounded for display; geometry uses unrounded values.

Reveal boundary and areaReset

At θ=0, B reaches the origin while A reaches (2,0). At θ=π/2, A reaches the origin while B reaches (0,2). Thus the origin is a shared Cartesian point represented at different angles. Solving only rA(θ)=rB(θ) on this interval finds (1,1), not every intersection.

Temporary exploration only; record your split-integral argument in the written practice. Changing the angle hides the previous calculation.

Move the probe through π/8, π/4 and 3π/8. Read both endpoint coordinates and identify the shorter segment from the origin. The overlap extends from radial distance zero to the smaller of the two radii. It is not the radial strip between the curves.

Angular interval	Smaller nonnegative radius	Radial overlap
0 to π/4	2 sin θ from B	0≤r≤2 sin θ
π/4 to π/2	2 cos θ from A	0≤r≤2 cos θ

At π/4 both radii are √2, giving Cartesian point (1,1). At θ=0, curve B reaches the origin but A reaches (2,0). At θ=π/2, curve A reaches the origin but B reaches (0,2). Thus the origin belongs to both circles even though their radius formulas are unequal at either of these angles.

Check that these are all intersections independently: subtract the two Cartesian circle equations to get x=y. Substituting gives 2x²=2x, hence x=0 or 1. The only common points are (0,0) and (1,1). This check prevents a missed intersection from silently producing wrong bounds.

Understand

A thin polar sector with radius R and angular width dθ contributes approximately R²dθ/2. Integration gives the exact area for this continuously varying radial boundary. Splitting where the smaller radius changes, the lens area is

one half the integral of (2 sin θ)² from 0 to π/4, plus one half the integral of (2 cos θ)² from π/4 to π/2.

Use 2 sin²θ=1−cos 2θ and 2 cos²θ=1+cos 2θ. The first integral becomes [θ−sin(2θ)/2] from 0 to π/4, which is π/4−1/2. The second becomes [θ+sin(2θ)/2] from π/4 to π/2 and has the same value. The total is π/2−1 square units, approximately 0.5708. This is positive and smaller than either unit disk’s area π, useful checks on the scale.

There is a second geometric verification. Each lens half is a circular sector of angle π/2 minus a right triangle with legs 1. Each sector has area π/4 and each triangle area 1/2; doubling their difference reproduces π/2−1. The integral and geometric dissection use the same boundary but compute its area in different ways.

Why not integrate half the difference of the squared radii? That construction measures a region between an inner and an outer boundary along a ray. Here the common region starts at the origin, so its inner radius is zero. Why not use the smaller formula for a full turn? Negative radii and repeated tracing would complicate the interpretation; the first-quadrant argument already provides a single coverage of this region.

The widget draws the full circles on equally scaled Cartesian axes. Its shaded boundary is sampled for rendering, but its area is computed from the exact antiderivatives above. A dense drawing does not prove the integral, and the rounded displayed coordinates should not replace exact intersection calculations.

Apply

Describe the area swept from angle zero to a probe angle α. For 0≤α≤π/4 it is α−sin(2α)/2. For π/4≤α≤π/2 it is α+sin(2α)/2−1. Check that both formulas agree at the switch and that the second gives the total at π/2.

Write a correction to “the only intersection is (1,1), because the radii agree only at π/4.” Include the two different angle representations of the origin. Then explain how uniformly multiplying every radial length by a positive factor k changes the area by k² while preserving the switching angle. Your written practice uses a changed scale so that the live answer alone is insufficient.

A signed-radius and retracing transfer

The single curve r=2cos θ provides a useful warning before integrating a new overlap. Multiplying by r and converting to Cartesian coordinates gives (x−1)²+y²=1. It is a radius-one circle. Over [0,2π], the integral of r²/2 is 2π, yet the enclosed disk area is only π. The mismatch is repeated tracing, not an integration error.

To prove the repetition, increase θ by π. The radius changes sign, and so do cos θ and sin θ. Both products r cos θ and r sin θ stay the same, so the Cartesian point is repeated. Negative radius is not a negative physical distance along the original ray; it represents a point in the opposite direction. Squaring a signed radius does not automatically correct multiplicity.

A single trace is obtained on [−π/2,π/2], where r is nonnegative. Interior ray directions give distinct non-origin boundary points; both endpoints give the origin. The sector integral over that interval is π. An endpoint repeated at one point contributes no extra area. By contrast [0,π/2] sweeps only the upper half-disk.

Move the angle control and compare 60° with 240°, then 90° with 270°. Explain the coordinates, not just the sign of r. Before computing an unfamiliar common region, record the requested region, every Cartesian intersection, the chosen angle interval and a single-coverage argument. This retry isolates a tracing criterion; it does not claim to solve arbitrary polar intersections automatically.

Signed radii and repeated tracing

Review criterion: Translate to Cartesian points and establish single coverage before interpreting a polar area integral.

Predict, then change one input

Angle θ (degrees):

60

Signed radius r

1

Cartesian x=r cos θ

0.5

Cartesian y=r sin θ

0.866025

Compare θ with θ+180° where the control permits it. The signed radius and direction both reverse, so the Cartesian point repeats. The circle has radius 1, not 2.

Explain the changed case: A repeated circle

For r=2cos θ, what does (1/2)∫₀²π r² dθ=2π represent?

• The curve is (x−1)²+y²=1.
• θ and θ+π produce the same Cartesian point.

Write a reason before selecting a claim. Feedback is public practice, not a proof grader.

Twice the area of the unit disk, because the curve is traced twice.The disk area is 2π because negative radii add another half.Zero, because positive and negative radii cancel.

Switch to changed retryReset this practice

Inputs and choices are temporary. Nothing is saved, no lesson completion or score is changed, and repeating a public case is not unseen assessment.

Continue

For each new pair of curves, identify the requested region, check Cartesian intersections and tracing, choose the radial boundary on each interval, and only then integrate. Revisit polar area for sector-area derivation. Matching equal radii is a useful equation, but it is not a general theorem that all intersections share the same angle representation.