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Math Help - Determine the area enclosed by the curve y = cos ^2 fi

  1. #1
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    Determine the area enclosed by the curve y = cos ^2 fi

    Hi guys you can see my problem attaced to the pictuse + my attempt to solve it.
    and my question is how I can go further from there.Determine the area enclosed by the curve y = cos ^2 fi-eul-001.jpg
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  2. #2
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by Riazy View Post
    Hi guys you can see my problem attaced to the pictuse + my attempt to solve it.
    and my question is how I can go further from there.Click image for larger version. 

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    If you know what the curve looks like it becomes a bit easiar to evaluate the area. Look at cos^2 x - Wolfram|Alpha and let us know how you want to approach this now.
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  3. #3
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    If you can see my attempt, i would like if someone could complete and show me the euler step there, I want to learn how to solve this analytically. Thanks
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  4. #4
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    Hello, Riazy!

    \text{A polar curve is given: }\:r \,=\,\cos^2\!\theta,\;0 \le \theta \le 2\pi

    \text{Calculate the area enclosed by the curve.}

    \displaystyle \text{Formula: }\:A \;=\;\tfrac{1}{2}\!\!\int^{\beta}_{\alpha}\!\! r^2\,d\theta


    \displaystyle \text{We have: }\;A \;=\;\tfrac{1}{2}\!\!\int^{2\pi}_0\!\! \cos^4\!\theta\,d\theta


    Why involve complex variables? . . . Why not standard Trig identities?


    \displaystyle \tfrac{1}{2}\cos^4\!\theta \;=\;\tfrac{1}{2}(\cos^2\!\theta)^2 \;=\;\tfrac{1}{2}\left(\frac{1+\cos2\theta}{2}\rig  ht)^2 \;=\;\tfrac{1}{8}(1 + 2\cos2\theta + \cos^2\!2\theta)

    . . =\;\frac{1}{8}\left(1 + 2\cos2\theta + \dfrac{1 + \cos4\theta}{2}\right) \;=\;\frac{1}{8}\left(\frac{3}{2} + 2\cos2\theta + \frac{1}{2}\cos4\theta\right)

    . . =\;\frac{1}{16}(3 + 4\cos2\theta + \cos4\theta)


    \displaystyle \text{We have: }\:A \;=\;\tfrac{1}{16}\!\int^{2\pi}_0 \!\!\left(3 + 4\cos2\theta + \cos4\theta \right)\,d\theta

    . . =\;\frac{1}{16}\!\bigg[3\theta + 2\sin2\theta + \frac{1}{4}\sin4\theta\bigg]^{2\pi}_0

    . . =\;\frac{1}{16}\bigg[\left(6\pi + 2\sin4\pi + \frac{1}{4}\sin8\pi\right) - \left(0 + 2\sin0 + \frac{1}{4}\sin0\right)\bigg]

    . . =\;\;\dfrac{3\pi}{8}

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  5. #5
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    Really helpfull, Soroban, Thanks a lot.
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  6. #6
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    Okay there is one thing I didnt understand, sorry but i dont consider this to be a bump as I have got my answer, but I would like to ask

    how you got 1/8 into 1/16 I can see that you multiplied everything within the brackets with 2. but what about 1/16?

    Thanks for answers!
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  7. #7
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    What Soroban did was to change 2 to 4/2, then factored out a 1/2 .
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