- #1

archaic

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- Homework Statement
- Find the area of the region inside the circle ##r=-2\sin\theta## and outside the circle ##r=1##.

- Relevant Equations
- $$A=\frac 12\int_{\theta_1}^{\theta_2}(r^2-r^2_0)d\theta$$

$$-2\sin\theta=1\Leftrightarrow\theta=-\frac{\pi}{6},\,-\frac{5\pi}{6}\\

\begin{align*}

\int_{-\frac{\pi}{6}}^{-\frac{5\pi}{6}}\frac 12\left(4\sin^2\theta-1\right)d\theta

&=\int_{-\frac{\pi}{6}}^{-\frac{5\pi}{6}}\frac 12\left(1-2\cos2\theta\right)d\theta\\

&=\int_{-\frac{\pi}{6}}^{-\frac{5\pi}{6}}\frac 12d\theta-\int_{-\frac{\pi}{6}}^{-\frac{5\pi}{6}}\cos2\theta d\theta\\

&=\frac 12\left[\theta-\sin2\theta\right]_{-\frac{\pi}{6}}^{-\frac{5\pi}{6}}\\

&=-\frac{5\pi}{12}-\frac{\sqrt 3}{4}+\frac{\pi}{12}-\frac{\sqrt 3}{4}\\

&=-\frac{\pi}{3}-\frac{\sqrt 3}{2}

\end{align*}\\

A=\frac{\pi}{3}+\frac{\sqrt 3}{2}$$

\begin{align*}

\int_{-\frac{\pi}{6}}^{-\frac{5\pi}{6}}\frac 12\left(4\sin^2\theta-1\right)d\theta

&=\int_{-\frac{\pi}{6}}^{-\frac{5\pi}{6}}\frac 12\left(1-2\cos2\theta\right)d\theta\\

&=\int_{-\frac{\pi}{6}}^{-\frac{5\pi}{6}}\frac 12d\theta-\int_{-\frac{\pi}{6}}^{-\frac{5\pi}{6}}\cos2\theta d\theta\\

&=\frac 12\left[\theta-\sin2\theta\right]_{-\frac{\pi}{6}}^{-\frac{5\pi}{6}}\\

&=-\frac{5\pi}{12}-\frac{\sqrt 3}{4}+\frac{\pi}{12}-\frac{\sqrt 3}{4}\\

&=-\frac{\pi}{3}-\frac{\sqrt 3}{2}

\end{align*}\\

A=\frac{\pi}{3}+\frac{\sqrt 3}{2}$$

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