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For this question, I first made a graph for the polar curve (lemniscate):enter image description here

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?

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

1 Answer 1

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$$Area =4\times\frac{1}{2} \int_{0}^{\pi/4} 9 \cos 2\theta d \theta =9 \sin 2\theta |_{0}^{\pi/4}=9.$$

Yest you are right.

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  • $\begingroup$ A typo. The integral is $9\sin2\theta$. But that still gives 9. $\endgroup$
    – almagest
    Jan 23, 2020 at 15:21

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