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:
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 $5\pi/4$
edited Dec 23 "15 at 12:16
reply Jun 28 "14 at 20:52
Varun IyerVarun Iyer
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