Find the area of the region enclosed by one loop of the curve.
$$r = 7 \cos 8 \theta$$
A) $\frac { 49 \pi } { 11 }$
B) $\frac { 49 \pi } { 32 }$
C) $\pi$
D) $49 \pi$
E) $\frac { \pi } { 8 }$

Question

Quizplus · Accepted Answer

The area enclosed by one loop of the polar curve $r = 7 \cos 8 \theta$ is given by $\frac{1}{2} \int_{\alpha}^{\beta} r^2 \, d\theta$, where $\alpha$ and $\beta$ are the limits for one loop. For $r = 7 \cos 8 \theta$, one loop occurs when $\theta$ ranges from $0$ to $\frac{\pi}{8}$. The area is $\frac{1}{2} \int_{0}^{\frac{\pi}{8}} (7 \cos 8\theta)^2 \, d\theta = \frac{49}{2} \int_{0}^{\frac{\pi}{8}} \cos^2 8\theta \, d\theta$. Using the identity $\cos^2 x = \frac{1 + \cos 2x}{2}$, the integral becomes $\frac{49}{4} \int_{0}^{\frac{\pi}{8}} (1 + \cos 16\theta) \, d\theta$. Evaluating this gives $\frac{49 \pi}{32}$.

Find the Area of the Region Enclosed by One Loop r=7cos⁡8θr = 7 \cos 8 \thetar=7cos8θ

Find the Area of the Region Enclosed by One Loop $r = 7 \cos 8 \theta$