Given the polar curve $r=4+3-\sin \left(\theta \right)$, what is the area enclosed by one loop of the curve?

• A. $\frac{25\pi }{2}$
• B. $\frac{49\pi }{2}$
• C. $14\pi$
• D. $21\pi$

Answer: (B)

Verified step by step guidance

1. Recall that the area enclosed by a polar curve \( r = f(\theta) \) between \( \theta = a \) and \( \theta = b \) is given by the integral formula: \[\text{Area} = \frac{1}{2} \int_{a}^{b} r^2 \, d\theta\]
2. Identify the given polar curve: \[r = 4 + 3 - \sin(\theta) = 7 - \sin(\theta)\] (Note: The problem states \( r = 4 + 3 - \sin(\theta) \), so simplify the expression first.)
3. Determine the interval \( [a, b] \) for one complete loop of the curve. Since the curve is defined for \( \theta \) from 0 to \( 2\pi \), and the function is continuous and periodic, one loop corresponds to \( \theta \) going from 0 to \( 2\pi \).
4. Set up the integral for the area enclosed by one loop: \[\text{Area} = \frac{1}{2} \int_{0}^{2\pi} (7 - \sin(\theta))^2 \, d\theta\]
5. Expand the square inside the integral and use trigonometric identities to simplify the integral before integrating term-by-term: \[ (7 - \sin(\theta))^2 = 49 - 14 \sin(\theta) + \sin^2(\theta) \] Recall that \( \sin^2(\theta) = \frac{1 - \cos(2\theta)}{2} \) to help with integration.