- #1

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Well this problem started off simply enough. I was given this function:

[tex]r=2cos(3\theta)[/tex]

And I had to find the area bound by it. I sketched it out from zero to 2pi and got this:

[PLAIN]http://www4d.wolframalpha.com/Calculate/MSP/MSP58019c987a3dd6i59cg00000i1ghg0cgb4b0a00?MSPStoreType=image/gif&s=6&w=364&h=302 [Broken]

So I did that correctly. I then proceeded to integrate the function like so:

[tex]\frac{1}{2}\int_{0}^{2\pi}{(2cos(3\theta))^{2} d\theta}[/tex]

Got this:[tex]2[\theta+sin(6\theta)][/tex] from 0 to 2pi. My final answer was 2pi...but the correct answer is pi. Where did I go wrong. From looking at my notes I followed the right steps. I think I chose the limits incorrectly, however they make sense to me because the function runs from 0 to 2pi.

[tex]r=2cos(3\theta)[/tex]

And I had to find the area bound by it. I sketched it out from zero to 2pi and got this:

[PLAIN]http://www4d.wolframalpha.com/Calculate/MSP/MSP58019c987a3dd6i59cg00000i1ghg0cgb4b0a00?MSPStoreType=image/gif&s=6&w=364&h=302 [Broken]

So I did that correctly. I then proceeded to integrate the function like so:

[tex]\frac{1}{2}\int_{0}^{2\pi}{(2cos(3\theta))^{2} d\theta}[/tex]

Got this:[tex]2[\theta+sin(6\theta)][/tex] from 0 to 2pi. My final answer was 2pi...but the correct answer is pi. Where did I go wrong. From looking at my notes I followed the right steps. I think I chose the limits incorrectly, however they make sense to me because the function runs from 0 to 2pi.

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