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Math Help - find area of polar graph

  1. #1
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    find area of polar graph

    i'll use x for theta because i don't know how to make the theta symbol

    find the area inside the oval limacon r=4+2cos(x)

    so i've squared r to get 16 +16cos(x) +4cos(x)^2

    factor out a 4 to get 4(4+4cos(x)+cos(x)^2)

    then change to 4(4+4cos(x)+(1+cos2x)/2
    factor out a 1/2 and integrate, which comes to
    2(3x + 2sin(x)+(sin2x)/2

    so i think the limits of integration are from 0 to pi. but i'm not getting the correct answer using those. can anyone help me out?
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  2. #2
    MHF Contributor Calculus26's Avatar
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    4[4+4cos(x)+(1+cos2x)/2]


    = 18 + 16cos(x) + 2cos(2x)

    1/2 integral {18 + 16cos(x) + 2cos(2x)dx} = 9x + 8sin(x) +1/2 sin(2x)


    also the graph is generated by letting theta vary from 0 to 2pi

    so you can integrate from 0 to pi and double it
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  3. #3
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    Quote Originally Posted by isuckatcalc View Post
    i'll use x for theta because i don't know how to make the theta symbol

    find the area inside the oval limacon r=4+2cos(x)

    so i've squared r to get 16 +16cos(x) +4cos(x)^2

    factor out a 4 to get 4(4+4cos(x)+cos(x)^2)

    then change to 4(4+4cos(x)+(1+cos2x)/2
    factor out a 1/2 and integrate, which comes to
    2(3x + 2sin(x)+(sin2x)/2

    so i think the limits of integration are from 0 to pi. but i'm not getting the correct answer using those. can anyone help me out?
    A = \int_0^{2\pi} \frac{(4+2\cos{\theta})^2}{2} \, d\theta<br />

    A = \int_0^{2\pi} \frac{16+16\cos{\theta}+4\cos^2{\theta}}{2} \, d\theta

    A = \int_0^{2\pi} 8+8\cos{\theta}+2\cos^2{\theta} \, d\theta

    A = \int_0^{2\pi} 9+8\cos{\theta}+\cos(2\theta) \, d\theta

    A = \left[9\theta + 8\sin{\theta} + \frac{\sin(2\theta)}{2}\right]_0^{2\pi} = 18\pi
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  4. #4
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    how do you get that 18? i don't see where that's coming from. thanks for the help. replying to calculus 26.
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  5. #5
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    or to skeeter, how do you get the 9 there? i'm guessing it's some kind of identity that i'm not familiar with. or i'm just dumb today.
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  6. #6
    MHF Contributor Calculus26's Avatar
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    distribute the 4 through 4[4+4cos(x)+(1+cos2x)/2]


    you get 16 + 4(1/2) = 18
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  7. #7
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    ok thank you very much to both of you, much appreciated
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