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In order to find the arc length or area etc of a polar curve, you must integrate from $\theta_1$ to $\theta_2$. However, I'm having trouble finding the values of $\theta_1$ and $\theta_2$.

I know that they must constitute of one and only one full cycle of the curve. $r = \sin\theta$ completes its cycle when $\theta = 2\pi$ while $r = \cos\theta$ is done by $\pi$.

However, I'm confused as to where these numbers come from.

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I was told that I could find the values of $\theta$ by drawing the polar curves. In drawing $r = \cos\theta$, I found that once $\theta = \pi$, I had gotten my full circle. However, when drawing $r = \sin\theta$, I have the full circle by the time I get to $\theta = \pi$. Going to $2\pi$ only traces out the curve once more time. So why then is $2\pi$ considered the "full cycle" of sin?

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Also, how do the values of these full cycles change and what determines this change?

For example, I was trying to find the area of the region bounded by $r= 6\cos8\theta$ (this is the only information given. I originally tried $\int_0^\pi {(6\cos(6\theta))^2}d\theta $ but the right answer is found by taking the integral to $2\pi$ instead. But doesn't the cycle of cosine end at $\pi$?? When I found the area in $r=3\cos(5\theta)$, going to $\pi$ worked!

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  • $\begingroup$ Your intuition is correct. Who told you that you needed a full $2\pi$ to graph $r=\sin{\theta}$? $\endgroup$ – John Douma Apr 1 at 20:28
  • $\begingroup$ For finding the area of $r = 8sin(2\theta)$, I had to go to $2\pi$ for it to be correct. However, looking back now, I see that it must have something to do with the 2\theta part. $\endgroup$ – CodingMee Apr 2 at 0:48
  • $\begingroup$ If you multiply the angle by $2$ then you end up with different numbers of petals of the flower because the radius can go from $0$ to $r$ in an angle of $\frac{\pi}{4}$ instead of $\frac{\pi}{2}$. $\endgroup$ – John Douma Apr 2 at 1:32
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The correct way to do it would be to choose only the domain where $r\ge0$. That means for $r=\sin\theta$ you go from $0$ to $\pi$, while for $r=\cos\theta$ you choose $-\pi/2\le\theta\le\pi/2$.

For the other curves, you need to add together all pieces of the integral where $r>0$. For $r=6\cos(8\theta)$ you integrate $\theta$ between $-\frac{\pi}{16}$ and $\frac{\pi}{16}$, then from $\frac{3\pi}{16}$ to $\frac{5\pi}{16}$, and $\frac{7\pi}{16}$ to $\frac{9\pi}{16}$, and so on, until the upper limit is still a number less than $2\pi$. This way you have only one curve.

The reason your integration works from $0$ to $\pi$ and sometimes from $0$ to $2\pi$ is given by how many times you overlap the contour.

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  • $\begingroup$ I did not do the integration by hand. I used a calculator specifically for the probability of my flawed human self making some mistakes. Either way, thank you. I'll see if this works. $\endgroup$ – CodingMee Apr 1 at 20:33
  • $\begingroup$ I've made one small mistake, but I'm going to fix my answer. You need to add all regions where $r>0$, and I've picked only one. $\endgroup$ – Andrei Apr 1 at 20:36

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