AP Calculus BC 9.9 Finding the Area of the Region Bounded by Two Polar Curves - MCQs - Exam Style Questions

Intersection with \(r=1\): \(4\sin^{2}(2\theta)=1\Rightarrow \sin(2\theta)=\tfrac12\) (since \(0\le2\theta\le\pi\)).

Thus \(\theta=\tfrac{\pi}{12}\) and \(\theta=\tfrac{5\pi}{12}\). On this interval \(r_{\text{outer}}=4\sin^{2}(2\theta)\ge1=r_{\text{inner}}\).

Area between polar curves:
\(\displaystyle A=\int_{\pi/12}^{5\pi/12}\frac12\big(r_{\text{outer}}^{2}-r_{\text{inner}}^{2}\big)\,d\theta =\int_{\pi/12}^{5\pi/12}\!\Big(8\sin^{4}(2\theta)-\tfrac12\Big)d\theta.\)

Use \(\sin^{4}x=\tfrac{3}{8}-\tfrac12\cos(2x)+\tfrac18\cos(4x)\) with \(x=2\theta\):
\(\displaystyle \int \sin^{4}(2\theta)d\theta =\tfrac{3}{8}\theta-\tfrac{1}{8}\sin(4\theta)+\tfrac{1}{64}\sin(8\theta).\)

So
\(\displaystyle A=\Big[3\theta-\sin(4\theta)+\tfrac18\sin(8\theta)\Big]_{\pi/12}^{5\pi/12}-\tfrac12\Big(\tfrac{5\pi}{12}-\tfrac{\pi}{12}\Big).\)

Values: \(\sin(4\cdot\tfrac{\pi}{12})=\sin(\tfrac{\pi}{3})=\tfrac{\sqrt3}{2}\),
\(\sin(4\cdot\tfrac{5\pi}{12})=\sin(\tfrac{5\pi}{3})=-\tfrac{\sqrt3}{2}\),
\(\sin(8\cdot\tfrac{\pi}{12})=\sin(\tfrac{2\pi}{3})=\tfrac{\sqrt3}{2}\),
\(\sin(8\cdot\tfrac{5\pi}{12})=\sin(\tfrac{4\pi}{3})=-\tfrac{\sqrt3}{2}\).

Hence \(A=\tfrac{5\pi}{6}+\tfrac{7\sqrt3}{8}\approx 4.134\).

✅ Answer: (B)