The cardioid is traced once as varies from to

MATHEMATICS · Single correct · AP · Calculus BC · Parametric, Polar, and Vector-Valued Functions · Parametric, Polar, and Vector-Valued Functions

Problem

The cardioid r=1+cos⁡θr=1+\cos\theta is traced once as θ\theta varies from 00 to 2π2\pi. A student evaluates 12∫0π(1+cos⁡θ)2 dθ\dfrac12\int_{0}^{\pi}(1+\cos\theta)^2\,d\theta and then doubles the result to obtain the total area enclosed by the curve. Which of the following best justifies that this procedure yields the correct area? (a) Because r(θ)r(\theta) is periodic with period π\pi, integrating over 0≤θ≤π0\le \theta\le \pi gives the entire area, and doubling simply verifies the result. (b) For 0≤θ≤π0\le\theta\le\pi, the radius rr is positive while for π≤θ≤2π\pi\le\theta\le2\pi it is negative, so doubling compensates for these sign changes and yields the correct area. (c) The curve is symmetric with respect to the polar axis, so the interval 0≤θ≤π0\le\theta\le\pi covers exactly half of the region, and doubling produces the full area. (d) The cardioid is symmetric about the line θ=π2\theta=\dfrac{\pi}{2}, so 0≤θ≤π0\le\theta\le\pi contains one quarter of the region, and doubling gives the whole area.

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