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)  If you like seeing points visually, attract a graph to aid you watch the picture more clearly.

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 Varun IyerVarun Iyer
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