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

Question

Quizplus · Accepted Answer

We know that the area enclosed within a polar curve can be found using the formula $\frac{1}{2}\int_{\alpha}^{\beta}r^2\,d\theta$, where $\alpha$ and $\beta$ are the angles that correspond to the boundaries of the region enclosed by one loop of the curve.

For the given polar curve $r=7\cos(8\theta)$, we need to find these boundaries. This can be done by finding the values of $\theta$ for which $r=0$ and for which $r$ changes signs, which occur at the following values of $\theta$: $\frac{\pi}{16}$, $\frac{5\pi}{16}$, $\frac{9\pi}{16}$, $\frac{13\pi}{16}$, $\frac{17\pi}{16}$, $\frac{21\pi}{16}$, $\frac{25\pi}{16}$, and $\frac{29\pi}{16}$.

We are interested in the region enclosed by one loop of the curve, which corresponds to the interval $\frac{5\pi}{16}\leq\theta\leq\frac{13\pi}{16}$. Therefore, we have

$$\begin{aligned} A &= \frac{1}{2}\int_{\frac{5\pi}{16}}^{\frac{13\pi}{16}}(7\cos(8\theta))^2\,d\theta \ &= \frac{1}{2}\int_{\frac{5\pi}{16}}^{\frac{13\pi}{16}}49\cos^2(8\theta)\,d\theta \ &= \frac{49}{2}\cdot\frac{1}{2}\int_{\frac{5\pi}{16}}^{\frac{13\pi}{16}}(1+\cos(16\theta))\,d\theta \ &= \frac{49}{4}\left[\theta+\frac{1}{16}\sin(16\theta)\right]_{\frac{5\pi}{16}}^{\frac{13\pi}{16}} \ &= \frac{49}{4}\cdot\frac{8\pi}{16} \ &= \boxed{\frac{49\pi}{32}}. \end{aligned}$$

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 $r = 7 \cos 8 \theta$