# Find the area bounded by $r^{2} = 9 cos2\Theta$

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?

• what are you using $\cos \theta$ or $\cos 2\theta$? Jan 23, 2020 at 14:37
• Use $\cos x$ for $\cos x$. Jan 23, 2020 at 14:38
• I am sorry. I edited the title. I am dealing with a lemniscate. Jan 23, 2020 at 14:43

$$Area =4\times\frac{1}{2} \int_{0}^{\pi/4} 9 \cos 2\theta d \theta =9 \sin 2\theta |_{0}^{\pi/4}=9.$$
• A typo. The integral is $9\sin2\theta$. But that still gives 9. Jan 23, 2020 at 15:21