The curves and enclose a region for

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

Problem

The curves

r=3+2\cos\theta

and

r=4

enclose a region for

0\le\theta\le 2\pi

. They meet at

\theta=\pi/3

and

\theta=5\pi/3

. Which of the following expressions equals the exact area of the region bounded by the two curves? (a)

\dfrac12\left[\displaystyle\int_{0}^{\pi/3}\!\big((3+2\cos\theta)-4\big)\,d\theta+\int_{\pi/3}^{5\pi/3}\!\big(4-(3+2\cos\theta)\big)\,d\theta+\int_{5\pi/3}^{2\pi}\!\big((3+2\cos\theta)-4\big)\,d\theta\right]

(b)

\dfrac12\left[\displaystyle\int_{0}^{\pi/3}\!\big((3+2\cos\theta)^2-4^2\big)\,d\theta+\int_{\pi/3}^{5\pi/3}\!\big(4^2-(3+2\cos\theta)^2\big)\,d\theta+\int_{5\pi/3}^{2\pi}\!\big((3+2\cos\theta)^2-4^2\big)\,d\theta\right]

(c)

\dfrac12\left[\displaystyle\int_{0}^{\pi/3}\!\big(4^2-(3+2\cos\theta)^2\big)\,d\theta+\int_{\pi/3}^{5\pi/3}\!\big((3+2\cos\theta)^2-4^2\big)\,d\theta+\int_{5\pi/3}^{2\pi}\!\big(4^2-(3+2\cos\theta)^2\big)\,d\theta\right]

(d)

\dfrac12\displaystyle\int_{0}^{2\pi}\!\big((3+2\cos\theta)^2-4^2\big)\,d\theta