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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. –  Arturo Magidin Jun 10 '12 at 1:49
    
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) –  Nick Jun 10 '12 at 1:55
    
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 $\theta=2\pi$. –  Arturo Magidin Jun 10 '12 at 1:56
    
But I have overlapping occuring in all 4 quadrants... wouldnt that mean I have to account for the full 2pi? –  Nick Jun 10 '12 at 2:03
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The answer you got cannot possibly be correct as it is a negative number... –  DonAntonio Jun 10 '12 at 2:05
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1 Answer

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? –  Nick Jun 10 '12 at 19:49
    
@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. –  Arturo Magidin Jun 10 '12 at 19:51
    
Yes that is how I graphed it, howcome it is wrong? When I plot out the values I seem to be correct? –  Nick Jun 10 '12 at 21:38
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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 $(0,2)$. –  Arturo Magidin Jun 11 '12 at 1:08
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