Solving for the curves and gives

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

Problem

Solving r1(θ)=r2(θ)r_1(\theta)=r_2(\theta) for the curves r1(θ)=3sin⁡θr_1(\theta)=3\sin\theta and r2(θ)=3(1−sin⁡θ)r_2(\theta)=3(1-\sin\theta) gives sin⁡θ=12\sin\theta=\tfrac12, whose solutions in [0,π][0,\pi] are θ=π6\theta=\tfrac{\pi}{6} and θ=5π6\theta=\tfrac{5\pi}{6}. When finding the area of the region enclosed by both curves, only θ=π6\theta=\tfrac{\pi}{6} is used as an integration bound. Which of the following best explains why θ=5π6\theta=\tfrac{5\pi}{6} is omitted? (a) Beyond θ=π2\theta=\tfrac{\pi}{2} the two curves never intersect again, so θ=5π6\theta=\tfrac{5\pi}{6} is not a necessary bound. (b) For θ=5π6\theta=\tfrac{5\pi}{6} the cardioid already lies completely inside the circle, so that angle does not mark a change in which curve forms the outer boundary. (c) Because the graphs are symmetric about the yy-axis, integrating to θ=5π6\theta=\tfrac{5\pi}{6} would simply duplicate the area already accounted for on [ −π6, π6][\,-\tfrac{\pi}{6},\,\tfrac{\pi}{6}]. (d) At θ=5π6\theta=\tfrac{5\pi}{6} the computed radii are positive, but the rays point to the left of the yy-axis, outside the sector being used for the integral.

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