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Find the area of the region $R$ given by two curves.

So the region $R$ describes the area that is common between the two curves: $\begin{align*} \text{Function 1: } r&= 2\sin(\theta)\\ \text{Function 2: }r&= \frac{3}{2} - \sin(\theta). \end{align*}$

I have to find the area that is common between the two graphs..

What I did was just take the integral of function 1 (from 0 to $2\pi$) then subtract it from the integral of function 2 (from 0 to $2\pi$)... which gave me ($2-3\pi$) as my final answer... is this correct?

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    Per chance, after careful analysis of the regions, one could also avail of the change of variables?2012-06-10

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The graph of $r=2\sin(\theta)$ is a circle with radius $1$ centered at $(0,1)$. It lies entirely in qu The graph of $r = \frac{3}{2}-\sin(\theta)$ is a cardiod; call it $\mathcal{C}_2$

From $\theta=0$ to $\theta=\frac{\pi}{6}$, $\mathcal{C}_2$ is "outside" of $\mathcal{C}_1$. They intersect at $\theta=\frac{\pi}{6}$.

From $\theta=\frac{\pi}{6}$ to $\theta=\frac{5\pi}{6}$, $\mathcal{C}_1$ is "outside" $\mathcal{C}_2$. They intersect when $\theta=\frac{5\pi}{6}$.

From $\theta=\frac{5\pi}{6}$ to $\theta=\pi$, $\mathcal{C}_2$ is "outside $\mathcal{C}_1$.

From $\theta=\pi$ to $\theta=2\pi$, $\mathcal{C}_1$ retraces the circle above the $y$-axis, whereas $\mathcal{C}_2$ lies in the 3rd and 4th quadrant.

So the region common to both graphs lies inside $\mathcal{C}_1$ from $\theta=0$ to $\theta=\frac{\pi}{6}$, inside $\mathcal{C}_2$ from $\theta=\frac{\pi}{6}$ to $\theta=\frac{5\pi}{6}$, and inside $\mathcal{C}_1$ from $\theta=\frac{5\pi}{6}$ to $\theta=\pi$.

This analysis tells you what integrals you need to evaluate and add

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    @Nick: It's wrong because it's wrong. When $\theta$ is betweeN $\pi$ and $2\pi$, $\sin(\theta)$ is negative. That means that $r=2\sin(\theta)$ is negative. That means that the radius us pointing you in the direction **opposite** the half-line determined by $\theta$, which places the point in question on the top half of the plane. For example, when $\theta=3\pi/2$, you need to go $-2$ units in the direction of the negative $y$-axis. How do you go **negative** $2$ in the downward direction? By going **up**. So the point in question is actually on the **positive** $y$-axis, and is $(0,2)$.2012-06-11