Find the area of the specified region.
-Inside the graph of $\mathrm { r } = 2 + \cos 3 \theta$
A) $\frac { 9 \pi } { 2 } + \frac { 17 } { 24 }$
B) $9 \pi$
C) $\frac { 9 \pi } { 2 }$
D) $9 \pi + \frac { 17 } { 12 }$

Question

Quizplus · Accepted Answer

The graph of $\mathrm {r} = 2+\cos 3	heta$ is a rose curve with three petals (similar to $\mathrm{r}=2\cos 3	heta$) but with a minimum value of $1$ instead of $-2$. Therefore, we can find the area of one petal and multiply it by $3$.

To find the area of one petal, we need to evaluate the integral $\frac{1}{2}\int_{\alpha}^{\beta} r^2 \ d	heta$, where $\alpha$ and $\beta$ are the boundaries of one petal. To find the boundaries, we need to solve the equation $2+\cos 3	heta=0$:

\begin{align*}
\cos 3	heta & = -2 \
3	heta &= \pi + 2n\pi, \quad 	ext{or} \quad 3	heta = -\pi + 2n\pi \quad 	ext{where} \quad n \in \mathbb{Z} \
	heta & = \frac{\pi}{3} + \frac{2n\pi}{3}, \quad 	ext{or} \quad 	heta = -\frac{\pi}{9} + \frac{2n\pi}{3}, \quad 	ext{where} \quad n \in \mathbb{Z}
\end{align*}

The boundaries of one petal are $	heta = \frac{\pi}{3}$ and $	heta = -\frac{\pi}{9}$, so we have:

\begin{align*}
\frac{1}{2}\int_{\pi/3}^{-\pi/9} (2+\cos 3	heta)^2 \ d	heta &= \frac{1}{2}\int_{\pi/3}^{-\pi/9} (5+4\cos 3	heta+\cos^2 3	heta) \ d	heta \
&= \frac{1}{2}\left[5	heta+2\sin 3	heta + \frac{1}{9}\sin^3 3	hetaight]_{\pi/3}^{-\pi/9} \
&= \frac{9\pi}{4} + \frac{17}{48}
\end{align*}

Therefore, the area of the entire graph is $3\left(\frac{9\pi}{4} + \frac{17}{48}ight) = \boxed{	extbf{(C) } \frac{9\pi}{2}}$.

Find the Area of the Specified Region r=2+cos⁡3θ\mathrm { r } = 2 + \cos 3 \thetar=2+cos3θ

Find the Area of the Specified Region $\mathrm { r } = 2 + \cos 3 \theta$