AP Calculus BC 9.8 Find the Area of a Polar Region - MCQs -Exam Style Questions
No-CalcQuestion
What is the area of the region bounded by the polar curve \( r \;=\; 0.45\theta \;+\; 0.2\cos(0.5\theta) \) for \( 0 \le \theta \le \pi \) and the \(x\)-axis?
(B) \(1.310\)
(C) \(2.567\)
(D) \(2.621\)
▶️ Answer/Explanation
Correct answer: (A) \(1.283\)
Area in polar coordinates: \[ A \;=\; \frac{1}{2}\int_{0}^{\pi} r^{2}\,d\theta \;=\; \frac{1}{2}\int_{0}^{\pi} \!\big(0.45\theta + 0.2\cos(0.5\theta)\big)^{2}\,d\theta. \] Numerical evaluation gives \( A \approx 1.283 \).
No-Calc Question
Let \(R\) be the region in the first quadrant bounded by the polar curve \(r=\sqrt{6\sin(3\theta)}\). What is the area of \(R\)?
(B) \(2\)
(C) \(4\)
(D) \(6\)
▶️ Answer/Explanation
Area in polar form:
\(\displaystyle A=\tfrac12\int_{\alpha}^{\beta} r^2\,d\theta\).
Here \(r^2=6\sin(3\theta)\). The first-quadrant loop occurs for \(0\le \theta \le \tfrac{\pi}{3}\).
\(\displaystyle A=\tfrac12\int_{0}^{\pi/3} 6\sin(3\theta)\,d\theta =3\int_{0}^{\pi/3}\sin(3\theta)\,d\theta =3\Big[-\tfrac13\cos(3\theta)\Big]_{0}^{\pi/3}\).
\(\displaystyle A= -\big(\cos\pi-\cos0\big)= -(-1-1)=2\).
✅ Answer: (B)