Find the Area of a Polar Region or the Area Bounded by a Single Polar Curve BC only

A thin sector of radius $r$ and angle $d\theta$ has area $\frac{1}{2}r^{2}\,d\theta$, so $A = \frac{1}{2}\int_{\alpha}^{\beta} r^{2}\,d\theta$. The one-half and the square go missing independently of each other and of the limits, and the squaring belongs inside the integral as it has since 8.9; comparing an answer against a circle whose area is known settles all three at once, since $r = 2\cos\theta$ must give $\pi$. The limits of one petal are consecutive solutions of $r = 0$: for $r = 2\sin 3\theta$ those are $\theta = 0$ and $\frac{\pi}{3}$, giving a petal of area $\frac{\pi}{3}$ and a whole rose of $\pi$.

Integrating that rose over $[0, 2\pi]$ returns $2\pi$, exactly double, because the three petals are completely traced by $[0, \pi]$ and the rest of the revolution redraws them. The rectangular instinct that going all the way round encloses the region fails here, since a full turn of the angle may be more than the curve needs. In general $r = a\sin(n\theta)$ has $n$ petals over $[0, \pi]$ for odd $n$ and $2n$ over $[0, 2\pi]$ for even $n$. A negative radius plots backwards along the ray and contributes area normally, since $r^{2}$ ignores the sign, while a limaçon's inner loop needs its own interval and is subtracted rather than added.