A student plans to find the area of the region that lies inside and outside by evaluating

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

Problem

A student plans to find the area of the region that lies inside r=4sin⁡θr=4\sin\theta and outside r=2sin⁡θr=2\sin\theta by evaluating 12∫0π ⁣((4sin⁡θ)2−(2sin⁡θ)2) dθ\displaystyle \tfrac12\int_{0}^{\pi}\!\bigl((4\sin\theta)^2-(2\sin\theta)^2\bigr)\,d\theta. Which of the following best justifies whether the limits of integration 0≤θ≤π0\le\theta\le\pi are appropriate? (a) They are inappropriate because the outer curve changes at θ=π2\theta=\tfrac{\pi}{2}, so the integral must be split at that point. (b) They are appropriate because 4sin⁡θ≥2sin⁡θ≥04\sin\theta\ge2\sin\theta\ge0 for every θ\theta in [0,π][0,\pi], so the curves never switch roles on that interval. (c) They are inappropriate because the region is symmetric about the polar axis, so one should integrate from 00 to π2\tfrac{\pi}{2} and double instead. (d) They are appropriate only if the integrand is replaced by ∣ (4sin⁡θ)2−(2sin⁡θ)2∣|\,(4\sin\theta)^2-(2\sin\theta)^2| throughout [0,π][0,\pi].

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