For this question, I first made a graph for the polar curve (lemniscate):
The lower bound is obviously 0 (r = 3 is at $\Theta = 0$). So, I solved for the theta at the pole by letting r be equal to 0.
$0 = 9cos(2\Theta )$
$0 = cos(2\Theta )$
$\frac{cos^{-1}(0)}{2}=\Theta $
$\frac{\pi }{4} = \Theta $
Finally, I inputted these values into my calculator to find the area. I multiplied it to four because I believe that I am only getting the area of each half of the curve.
$4(\frac{1}{2})\int_{0}^{\frac{\pi }{4}} (9cos\Theta )d\Theta =9units^{2}$
Are my solution and answer correct?
Best Answer
$$Area =4\times\frac{1}{2} \int_{0}^{\pi/4} 9 \cos 2\theta d \theta =9 \sin 2\theta |_{0}^{\pi/4}=9.$$
Yest you are right.