# Find the Area Using Polar Coordinates and a Double Integral

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.

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 $\int_{\pi/6}^{\pi/2}(1-2\sin\theta)^2\,d\theta$.
• 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 $3$. – André Nicolas Dec 8 '14 at 3:54
• 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. – André Nicolas Dec 8 '14 at 3:55