## Calculus 8th Edition

$$A=(\frac{\pi}{3}+\sqrt 3)$$
Area can be calculated as: $$A=\frac{1}{2}\int(1+2cos3\theta)^{2}d \theta$$ or, $$A=\frac{1}{2}\int(1+4cos3\theta+4cos^{2}3\theta)d \theta$$ or, $$A=\frac{1}{2}[3\theta+\frac{4}{3}sin3\theta+\frac{1}{6}sin6\theta]$$ In order to find the area we will take the help of limits and then we will subtract the small loop from the big loop. For the big loop , use the interval of the half of the loop and take double of it. Thus, $$A=2.\frac{1}{2}[3\theta+\frac{4}{3}sin3\theta+\frac{1}{6}sin6\theta]_{0}^{2\pi/9}-\frac{1}{2}[3\theta+\frac{4}{3}sin3\theta+\frac{1}{6}sin6\theta]_{2\pi/9}^{4\pi/9}$$ Hence, $$A=(\frac{\pi}{3}+\sqrt 3)$$