AP Calculus BC / Parametric Equations, Polar Coordinates, and Vector-Valued Functions / Question No. 27

27. Which of the following gives the area of the region inside the polar curve \( r = 1 + \cos\theta \) and outside the polar curve \( r = 2\cos\theta \)?

\( \frac{1}{2} \int_{0}^{2\pi} (1 - \cos\theta)^2 \, d\theta \)

\( \int_{0}^{2\pi} (1 + 2\cos\theta)^2 \, d\theta \)

\( \int_{0}^{2\pi} (1 + \cos\theta)^2 \, d\theta - \pi \)

\( \frac{1}{2} \int_{0}^{2\pi} (1 + \cos\theta)^2 \, d\theta - \pi \)

Answer is: option4

\( \frac{1}{2} \int_{0}^{2\pi} (1 + \cos\theta)^2 \, d\theta - \pi \)

Solution:

\( \frac{1}{2} \int_{0}^{2\pi} (1 + \cos\theta)^2 \, d\theta - \pi \)

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Home Practice Questions AP Calculus BC Parametric Equations, Polar Coordinates, and Vector-Valued Functions Question No. 27

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