How do I solve this problem?

math_N00b

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May 21, 2020
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I have no idea where to begin or how do I go on about solving this, need desperate help. Thanks. You guys been absolute legends for helping me promptly for my previous 2 questions. God bless.

The question:
Screenshot_20200523-151024.jpg
 
First, have you graphed it? If you don't mind "cheating" a little, you can use https://www.desmos.com/calculator.
Just enter r=3cos(θ)\displaystyle r= 3 cos(\theta) for the first function and r=3sin(2θ)\displaystyle r= 3 sin(2\theta) for the second.

One thing you will need to know is where the two graphs intersect. That is, solve 3cos(θ)=3sin(2θ)\displaystyle 3 cos(\theta)= 3 sin(2\theta). That is, of course, equivalent to cos(θ)=sin(2θ)\displaystyle cos(\theta)= sin(2\theta). There is a "double angle identity that says that sin(2θ)=2sin(θ)cos(θ)\displaystyle sin(2\theta)= 2sin(\theta)cos(\theta) (that's probably in your text book) so the equation is cos(θ)=2sin(θ)cos(θ)\displaystyle cos(\theta)= 2 sin(\theta)cos(\theta). If cos(θ)0\displaystyle cos(\theta)\ne 0 (i.e. for θπ/2\displaystyle \theta\ne \pi/2) we can divide both sides by cos(θ)\displaystyle cos(\theta) to get 2sin(θ)=1\displaystyle 2sin(\theta)= 1, sin(θ)=12\displaystyle sin(\theta)= \frac{1}{2}. That's true for θ=π6\displaystyle \theta= \frac{\pi}{6} radians. You can see from the graph that this is symmetric about the x-axis so you want to integrate 3cos(θ)3sin(2θ)\displaystyle 3 cos(\theta)- 3 sin(2\theta) from π6\displaystyle -\frac{\pi}{6} to π6\displaystyle \frac{\pi}{6}.
 
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