discover the area of the region that lies inside both curves $r=3+2\cos\theta ; r=3+2\sin\theta$

The clues of intersection need to be $\frac \pi4 and also \frac 5\pi4 $

I don"t think this graphs space symmetrical and also I am lost setting up the problem. Any help would be significantly appreciated.

You are watching: Find the area of the region that lies inside both curves. r = 3 cos(θ), r = sin(θ)


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First point to do is to set the two equations equal to one another and also solve because that theta.

So, as we can see $sin\theta = cos\theta$, for this reason the intersection points would be $\frac \pi4 and \frac 5\pi4 $.

Now, let"s draw the graph:

http://www.wolframalpha.com/input/?i=r+%3D+3+%2B+2*sin%28x%29%2C+r+%3D+3+%2B+2*cos%28x%29+polar

As we deserve to see, the area would certainly be the area for $3 + 2cos\theta$ from $\pi/4$ come $5\pi/4.$

Therefore, the integral would be $(1/2)*(3 + 2cos\theta)^2$ indigenous $\pi/4$ to $5\pi/4$


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edited Dec 23 "15 at 12:16
reply Jun 28 "14 at 20:52
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Varun IyerVarun Iyer
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