Polar coordinates and area factors

Try

A disk of radius 2 is described in Cartesian coordinates by x²+y²≤4. Its circular boundary makes rectangular slices involve square roots. Polar coordinates instead record distance r≥0 from the origin and angle θ in radians: x=r cos θ and y=r sin θ. The disk becomes 0≤r≤2, 0≤θ≤2π. A coordinate change should simplify the region, the integrand, or preferably both.

Predict

Predict whether a small parameter rectangle of widths Δr and Δθ represents area ΔrΔθ in the plane. At larger radius, the same angular interval spans a longer arc. A small sector near radius r has transverse width approximately rΔθ, so its area is approximately rΔrΔθ. The missing radial factor is why treating dr dθ as ordinary planar area produces incorrect disk formulas.

Experiment

Compute the area using ∫₀²π∫₀²r dr dθ=4π. Omitting r would give 4π too for this particular radius, an accidental coincidence that makes this a poor isolated test. Change the radius to 3: the correct area is 9π, while the incorrect expression gives 6π. This comparison illustrates why verification needs cases that distinguish competing formulas.

Now integrate the density ρ=x²+y² on the radius-3 disk. It becomes r², and the area factor contributes another r. The integral is ∫₀²π∫₀³r³ dr dθ=81π/2, not an integral of r² alone.

Understand

The coordinate map has derivative columns (cos θ,sin θ) and (−r sin θ,r cos θ). Their determinant is r. An absolute determinant gives local unsigned area scaling, so dA=r dr dθ on the usual domain r≥0. The origin and the angular seam are coordinate exceptions of area zero; the disk’s interior is otherwise covered once if a single full angular interval is used.

For a sector α≤θ≤β and radial bounds a(θ)≤r≤b(θ), build the integral with those bounds and the radial factor. Radii must remain nonnegative under this convention. If a proposed upper radius is negative, that angular range does not describe the intended ordinary sector and must be reconsidered.

Apply

Complete the square in x²+y²≤4y: the region is the disk centered at (0,2) with radius 2. In polar form r²≤4r sin θ. The convenient description is 0≤θ≤π and 0≤r≤4sin θ. Its area is ∫₀π8sin²θ dθ=4π, consistent with the Cartesian center-radius description.

Repair “a circle always requires θ from 0 to 2π.” The shifted disk above lies in the upper half-plane, and rays from the origin meet it only for angles between 0 and π. A full angular interval with the same radial bound would introduce negative radii or duplicate a different description. Bounds depend on the chosen origin and the region.

Continue

As optional depth, explain why allowing negative radii can be convenient for drawing polar curves but complicates integration multiplicity. A plotting convention and a one-to-one change-of-variables domain serve different purposes. Likewise, a curve equation r=f(θ) is only a boundary until an interior radial interval is specified.

Record a disk, an annulus and a sector with their bounds, then test each setup using the integrand 1. For every nonconstant density, separately transform the function and multiply by the area factor. The next lesson extends this bookkeeping to solids, where cylindrical and spherical coordinates add another dimension and another scale factor.