Given the polar curve $r=4\sin \left(5\theta \right)$, what is the area enclosed by one loop of the curve?

• A. $\frac{4\pi }{5}$
• B. $\frac{16\pi }{5}$
• C. $\frac{8\pi }{25}$
• D. $\frac{8\pi }{5}$

Answer: (D)

Verified step by step guidance

1. Recall that the area enclosed by one loop of a polar curve given by \(r = f(\theta)\) is calculated using the formula: \(\text{Area} = \frac{1}{2} \int_{\alpha}^{\beta} r^2 \, d\theta\) where \(\alpha\) and \(\beta\) are the limits of integration corresponding to one loop of the curve.
2. Identify the function and the limits for one loop. Here, \(r = 4 \sin(5\theta)\). Since the sine function completes one full period over \$2\pi\(, and the argument is \)5\theta\(, one loop corresponds to the interval where \)\sin(5\theta)\( goes from 0 to its next zero, i.e., from \)\theta = 0\( to \)\theta = \frac{\pi}{5}$.
3. Set up the integral for the area of one loop: \(\text{Area} = \frac{1}{2} \int_0^{\frac{\pi}{5}} (4 \sin(5\theta))^2 \, d\theta = \frac{1}{2} \int_0^{\frac{\pi}{5}} 16 \sin^2(5\theta) \, d\theta\).
4. Use the trigonometric identity to simplify the integrand: \(\sin^2 x = \frac{1 - \cos(2x)}{2}\), so \$16 \sin^2(5\theta) = 16 \times \frac{1 - \cos(10\theta)}{2} = 8 (1 - \cos(10\theta))$.
5. Rewrite the integral and integrate term-by-term: \(\text{Area} = \frac{1}{2} \int_0^{\frac{\pi}{5}} 8 (1 - \cos(10\theta)) \, d\theta = 4 \int_0^{\frac{\pi}{5}} (1 - \cos(10\theta)) \, d\theta\). Then compute the integral of \$1\( and \)\cos(10\theta)$ over the interval, and simplify to find the area enclosed by one loop.