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Math Help - Polar Coordinate..

  1. #1
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    Polar Coordinate..

    Find the area inside the larger loop and outside the smaller loop of the limacon
    r= 1/2 + Cos[x]
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  2. #2
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    The region between the loops would be:

    2\left[\int_{0}^{\frac{2\pi}{3}}\frac{1}{2}(\frac{1}{2}+c  os{\theta})^{2}d{\theta}-\int_{\frac{2\pi}{3}}^{\pi}\frac{1}{2}(\frac{1}{2}  +cos{\theta})^{2}d{\theta}\right]
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by galactus View Post
    The region between the loops would be:

    2\left[\int_{0}^{\frac{2\pi}{3}}\frac{1}{2}(\frac{1}{2}+c  os{\theta})^{2}d{\theta}-\int_{\frac{2\pi}{3}}^{\pi}\frac{1}{2}(\frac{1}{2}  +cos{\theta})^{2}d{\theta}\right]
    could we not use \int_{ - 2 \pi / 3}^{ 2 \pi / 3} \frac {1}{2} \left( \frac {1}{2} + \cos \theta \right)^2 ~d \theta ?

    or, the equivalent answer would be, i think, \int_{0}^{2 \pi} \frac {1}{2} \left( \frac {1}{2} + \cos \theta \right)^2~d \theta - \int_{2 \pi / 3}^{4 \pi / 3} \frac {1}{2} \left( \frac {1}{2} + \cos \theta \right)^2~d \theta

    or am i too rusty on polar areas?
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  4. #4
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    the graph
    Attached Thumbnails Attached Thumbnails Polar Coordinate..-july4.gif  
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  5. #5
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    so what is the answer right?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by camherokid View Post
    so what is the answer right?
    I never calculated the answer, I assume you are capable of doing the integration on your own, so I don't know what answer you got or whether or not it is correct.

    I do believe galactus' solution would give a different answer from mine though, but i think i'm right (i may be wrong )

    i'd go with the first formula i used, not the difference of two integrals, but the single one
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  7. #7
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    Talking

    Quote Originally Posted by Jhevon View Post
    I never calculated the answer, I assume you are capable of doing the integration on your own, so I don't know what answer you got or whether or not it is correct.

    I do believe galactus' solution would give a different answer from mine though, but i think i'm right (i may be wrong )

    i'd go with the first formula i used, not the difference of two integrals, but the single one
    galactus has used the symmetry of the region, but you haven't. Both solutions are right.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by curvature View Post
    You have used the symmetry of the region, but galactus hasn't. Both solutions are right.
    i don't think so. i think the " -\int_{\frac{2\pi}{3}}^{\pi}\frac{1}{2}(\frac{1}{2}  +cos{\theta})^{2}d{\theta}" in galactus' solution makes it different from mine. without it, they would be the same
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  9. #9
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    Quote Originally Posted by Jhevon View Post
    i don't think so. i think the " -\int_{\frac{2\pi}{3}}^{\pi}\frac{1}{2}(\frac{1}{2}  +cos{\theta})^{2}d{\theta}" in galactus' solution makes it different from mine. without it, they would be the same
    I have corrcted my last post.
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  10. #10
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    Quote Originally Posted by galactus View Post
    The region between the loops would be:

    2\left[\int_{0}^{\frac{2\pi}{3}}\frac{1}{2}(\frac{1}{2}+c  os{\theta})^{2}d{\theta}-\int_{\frac{2\pi}{3}}^{\pi}\frac{1}{2}(\frac{1}{2}  +cos{\theta})^{2}d{\theta}\right]
    The above calculation (using symmetry of the region) is verified by the following Maple-generated graph.

    A:=plot(cos(theta)+1/2,theta=0..2*Pi/3,coords=polar):
    B:=plot(cos(theta)+1/2,theta=2*Pi/3..Pi,coords=polar,color=blue):
    display(A,B);
    Attached Thumbnails Attached Thumbnails Polar Coordinate..-july6.gif  
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  11. #11
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    Quote Originally Posted by Jhevon View Post
    could we not use \int_{ - 2 \pi / 3}^{ 2 \pi / 3} \frac {1}{2} \left( \frac {1}{2} + \cos \theta \right)^2 ~d \theta ?

    or, the equivalent answer would be, i think, \int_{0}^{2 \pi} \frac {1}{2} \left( \frac {1}{2} + \cos \theta \right)^2~d \theta - \int_{2 \pi / 3}^{4 \pi / 3} \frac {1}{2} \left( \frac {1}{2} + \cos \theta \right)^2~d \theta

    or am i too rusty on polar areas?
    the whole graph

    A:=plot(cos(theta)+1/2,theta=-2*Pi/3..2*Pi/3,coords=polar):
    B:=plot(cos(theta)+1/2,theta=2*Pi/3..4*Pi/3,coords=polar,color=blue):
    display(A,B);
    Attached Thumbnails Attached Thumbnails Polar Coordinate..-july7.gif  
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  12. #12
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    I agree with Galactus . . .

    Quote Originally Posted by galactus
    The region between the loops would be:

    2\,\left[\frac{1}{2}\int_{0}^{\frac{2\pi}{3}}\left(\frac{1}  {2}+\cos{\theta}\right)^{2}d{\theta} \,-\,\frac{1}{2}\int_{\frac{2\pi}{3}}^{\pi}\left(\fra  c{1}{2}+\cos{\theta}\right)^{2}d{\theta}\right]

    I would use: . \int_0^{\frac{2\pi}{3}}\left(\frac{1}{2}+\cos\thet  a\right)^2 d\theta \:-\:\int_{\pi}^{\frac{4\pi}{3}}\left(\frac{1}{2} + \cos\theta\right)^2d\theta


    The integral is: . \int\left(\frac{1}{4} + \cos\theta + \cos^2\theta\right)d\theta \;=\;\int\left(\frac{1}{4} + \cos\theta + \frac{1 + \cos2\theta}{2}\right)d\theta

    . . = \;\int\left(\frac{3}{4} + \cos\theta + \frac{1}{2}\cos2\theta\right)d\theta \;= \;\frac{3}{4}\theta + \sin\theta + \frac{1}{4}\sin2\theta


    The "outer" area is: . \frac{3}{4}\theta + \sin\theta + \frac{1}{4}\sin2\theta\,\bigg|^{\frac{2\pi}{3}}_0

    . . = \;\left[\frac{3}{4}\!\cdot\!\frac{2\pi}{3} + \sin\frac{2\pi}{3} + \frac{1}{4}\sin\frac{4\pi}{3}\right] - \left[\frac{3}{4}\!\cdot\!0 + \sin 0 + \frac{1}{4}\sin0\cos0\right]

    . . = \;\left[\frac{\pi}{2} + \frac{\sqrt{3}}{2} + \frac{1}{4}\left(-\frac{\sqrt{3}}{2}\right)\right] - \left[0+ 0 + 0\right] \;\;=\;\;\frac{\pi}{2} + \frac{3\sqrt{3}}{8}


    The "inner" area is: . \frac{3}{4}\theta + \sin\theta + \frac{1}{4}\sin2\theta\,\bigg|^{\frac{4\pi}{3}}_{\  pi}

    . . = \;\left[\frac{3}{4}\!\cdot\!\frac{4\pi}{3} + \sin\frac{4\pi}{3} + \frac{1}{4}\sin\frac{8\pi}{3}\right] - \left[\frac{3}{4}\!\cdot\!\pi + \sin\pi + \frac{1}{4}\sin2\pi\right]

    . . = \;\left[\pi - \frac{\sqrt{3}}{2} + \frac{1}{4}\cdot\frac{\sqrt{3}}{2}\right] - \left[\frac{3}{4}\pi + 0 + 0\right] \;\;=\;\;\frac{\pi}{4} - \frac{3\sqrt{3}}{8}


    The desired area is: . \left(\frac{\pi}{2} + \frac{3\sqrt{3}}{8}\right) - \left(\frac{\pi}{4} - \frac{3\sqrt{3}}{8}\right) \;\;= \;\;{\bf{\color{blue}\boxed{\frac{\pi + 3\sqrt{3}}{4}}}}


    Check my work . . . please!

    I've check and double-checked.
    . . If there are errors, I can't find them.

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  13. #13
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    That's what I got, Soroban.
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