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Calculate the area bounded by $r = 1 - 2 \sin\theta$ and the axes of Cartesian plane.

I have the solution to the problem. But I was wondering about how we find the $\theta$ boundaries of the area. Finding the $\theta$ intercepts doesn't seem to help much. Is it by trial and error? I am just trying to see what the author of my textbook did to come up with the required $\theta$ intervals.

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  • $\begingroup$ Are you looking for areas of the six individual non-overlapping regions? $\endgroup$ Dec 12 '16 at 20:23
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We know that on one cycle from $0$ to $2\pi$, the curve passes through the origin when $\theta=\frac{\pi}{6}$ and when $\theta=\frac{5\pi}{6}$ by setting $1-2\sin\theta=0$.

It crosses an axis when $\theta=0,\frac{\pi}{2},\pi$,and $\frac{3\pi}{2}$.

So the first interval is $0\le\theta\le\frac{\pi}{6}$, the second is $\frac{\pi}{6}\le\theta\le\frac{\pi}{2}$, we can skip the next two regions as "repeats." The next region having distinct area is $\frac{5\pi}{6}\le\theta\le\frac{3\pi}{2}$.

graph of y=1-2sin(x)

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  • $\begingroup$ @ John Wayland Bales, very nice. Thank you. $\endgroup$ Dec 13 '16 at 0:46

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