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!
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.