Measure a region by small sectors

Try

In polar coordinates, a point is described by an angle θ and a directed radius r. Convert (2,π) and (−2,0) using x=r cos θ and y=r sin θ. Both give the Cartesian point (−2,0). Predict why a plot of r against θ is not itself the Cartesian shape of the curve.

Predict

A sector of a circle with radius r and radian angle Δθ has area r²Δθ/2. If r changes with angle, predict the integral obtained by adding many thin sectors. Then ask whether integrating over two full traversals should count the same region once or twice.

Experiment

For r=2cos θ, restrict θ to [−π/2,π/2], where r is nonnegative. Convert selected points before sketching.

θ	r	Cartesian point
−π/2	0	origin
0	2	(2,0)
π/2	0	origin

The identity r²=2r cos θ becomes x²+y²=2x, or (x−1)²+y²=1. Thus this interval traces a circle of radius 1 once. Its polar area is (1/2)∫₋π/2^π/2 4cos²θdθ=π, agreeing with the geometric circle area. The unit-circle investigation can refresh the coordinate conversion, but it does not determine the tracing interval for an unfamiliar polar curve.

Understand

For a nonnegative radial boundary r(θ) over an interval that describes the intended sector region without overlap, area is (1/2)∫r²dθ. The derivation adds small sector areas as the angular partition is refined. Angles are measured in radians; using degree numbers directly would introduce an incorrect scale factor.

For a region between nonnegative outer and inner radial boundaries on the same angular interval, subtract their squared radii: area is (1/2)∫(R²−r²)dθ. Between radii 2 and 1 over [0,π/3], the area is π/2. Squaring the difference of radii instead measures a different quantity, just as the analogous washer error did in volume problems.

Polar curves can use negative r, which places the point on the opposite ray. The area formula may still compute a correctly interpreted loop when its parameter interval is analyzed carefully, but r² erases sign information and cannot detect repeated coverage by itself. First locate zeros, intersections and repeated points; use symmetry only after identifying which pieces are congruent and nonoverlapping.

For r=cos(2θ), the right petal runs from −π/4 to π/4. Its area is (1/2)∫cos²(2θ)dθ=π/8. The curve has four petals; an interval covering the full rose is not an interval for just one petal. A numerical integral can be perfectly calculated and still answer the wrong region question if the bounds trace extra pieces.

When two polar curves intersect, solving equal radius expressions at the same angle can miss intersections represented by different angle-radius pairs or the origin. Convert or inspect zero-radius conditions where needed. A sketch should label the actual region, not merely overlay two plots and assume every enclosed-looking patch belongs to the requested area.

Apply

For each practice region, give its angle interval, sign of radius and traversal count before integrating. Use the Cartesian circle conversion as an independent check. In the rose problem, identify consecutive zeros enclosing one petal and justify the symmetry or power-reduction step used to evaluate it.

Continue

Next compute the length of a polar curve. Changing radius and changing direction both contribute to its velocity, so area and length use different integrands.