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Of the area inside the smaller loop of the equation $r = 1-2sin\theta$

Here's my attempt at a solution:

The shape has an inner and an outer loop, both of which will terminate at the origin. Therefore, I want to find the range of the equation which only draws the inner loop. I would assume I would select a range for $\theta$ which would set $r=0$, however, I'm honestly not sure how I would go about doing that. I can't just take all the points where $r = 0$, because then the outer loop will be included as well. How do I set this integral up?

Note: I've tried to find answers online and stumbled onto this, but cannot make heads or tails of what is being described. The document claims that $ θ = ±π/4$ and $θ = ±3π/4$ are the candidates, but when I use the equation $sin\theta = 1/2$, I end up with π/6.

http://jacobi.math.wvu.edu/~hjlai/Teaching/Tip-Pdf/Tip3-34.pdf (Example 3)

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The bounds $\frac{\pi}{4}$ and $\frac{3\pi}{4}$ in the document linked to are wrong. Probably a typo. You are right about $\frac{\pi}{6}$.

We go from $\frac{\pi}{6}$ to $\pi-\frac{\pi}{6}$. The convention used is that where $r$ is negative we reflect the result for $|r|$ across the origin, or equivalently rotate through b$180^\circ$.

So we integrate $\frac{1}{2}(1-2\sin\theta)^2$ from $\frac{\pi}{6}$ to $\frac{5\pi}{6}$ or equivalently, taking advantage of symmetry, we calculate $\int_{\pi/6}^{\pi/2}(1-2\sin\theta)^2\,d\theta$.

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  • $\begingroup$ Okay, this makes sense, thank you. My only real confusion is how we know we haven't "overshot" the small loop and ended up in the large one as well. Is it just because those are the first two values which set r = 0 ? Also, why are you integrating only half of the function? EDIT: Ah, I wasn't clear on your formatting, I think I understand. Thank you. $\endgroup$ – Mock Dec 8 '14 at 3:49
  • $\begingroup$ You are welcome. The only way I feel secure is by making a sketch. From $\pi/6$ to $5\pi/6$, $r$ grows from $0$ to $1$ and then shrinks to $0$. That's the inner loop. In the outer loop $r$ grows to $3$. $\endgroup$ – André Nicolas Dec 8 '14 at 3:54
  • $\begingroup$ I am integrating over half the interval and multiplying the result by $2$, that's why the $\frac{1}{2}$ has disappeared. For me taking advantage of symmetry is a reflex. $\endgroup$ – André Nicolas Dec 8 '14 at 3:55

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