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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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    It is not correct, unless it so happens that the second graph is completely contained inside the first graph (which it is not). Try graphing the curves, and find the points of intersection.2012-06-10
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    I did graph them... they intersect at (1,1) and (-1,1). For the sin graph, I flipped at about the x axis as well, so there are two circles centered at (0,1) and (0,-1)2012-06-10
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    Which correspond to certain values of $\theta$. So you want to do the integral that "sweeps" **only** the area of intersection. That is most certainly not the integral from $\theta=0$ to $\theta=2\pi$.2012-06-10
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    But I have overlapping occuring in all 4 quadrants... wouldnt that mean I have to account for the full 2pi?2012-06-10
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    The answer you got cannot possibly be correct as it is a negative number...2012-06-10
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    @Nick: No, because that includes regions where we do **not** have area enclosed in both graphs. You need to think about it a lot more carefully than you are.2012-06-10
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    @Nick: $r=2\sin\theta$ is a circle of radius $1$ centered at $(0,1)$. It does not even *go* through quadrants III and IV, so how could you possibly have overlap "in all 4 quadrants"?2012-06-10
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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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    I don't quite follow... looking at the graphs I drew, when 2sin(theta) is from pi to 2pi, it is completed contained within the cardioid... so why dont you include that for your final solution?2012-06-10
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    @Nick: The graph of $2\sin(\theta)$ is a **single** circle, above the $x$-axis. When $\theta$ is between $\pi$ and $2\pi$, the values of $r=2\sin(\theta)$ are negative, so the graph occurs *above* the $x$-axis. You trace the circle twice. Did you graph it as *two* circles, tangent to one another, one above the $x$-axis and one below? Then you graphed it wrong.2012-06-10
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    Yes that is how I graphed it, howcome it is wrong? When I plot out the values I seem to be correct?2012-06-10
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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