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Math Help - Center of Mass through density

  1. #1
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    Center of Mass through density

    Question:

    A wire takes the shape of the semicircle y= x^2 + y^2, y \geq  0 . Find the center of mass of the wire if the linear density at any point is proportional to its distance from the line y=1.


    So far I have this:

    p(x,y) = K(\sqrt(1-y)^2); polarized = K(1-rsin\theta)

    The Double Integral I have is:

    \int\int(K(1-rsin\theta))rdrd\theta with 0\leq r\leq 1;   0\leq \theta \leq \pi

    Which works out to K\pi/2

    I then attempt to solve the \overline{y} expression:
    <br />
2/K\pi \int\int (rsin\theta)(K)(1-sin\theta)rdrd\theta

    I end up with the answer of:

    (4-\pi)/(3\pi) whereas the correct answer is: (4-\pi)/(2(2\pi-2))

    Any ideas where I went wrong?

    Thanks,
    Mike
    Last edited by Mei Liu; December 28th 2008 at 07:35 PM.
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  2. #2
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    Quote Originally Posted by Mei Liu View Post
    Question:

    A wire takes the shape of the semicircle y= x^2 + y^2, y \geq  0 . Find the center of mass of the wire if the linear density at any point is proportional to its distance from the line y=1.


    So far I have this:

    p(x,y) = K(sqrt(1-y^2)); polarized = K(1-rsin\theta)

    The Double Integral I have is:

    \int\int(K(1-rsin\theta))rdrd\theta with 0\leq r\leq 1;   0\leq \theta \leq \pi

    Which works out to K\pi/2

    I then attempt to solve the \overline{y} expression:
    <br />
2/K\pi \int\int (rsin\theta)(K)(1-sin\theta)rdrd\theta

    I end up with the answer of:

    (4-\pi)/(3\pi) whereas the correct answer is: (4-\pi)/(2(2\pi-2))

    Any ideas where I went wrong?

    Thanks,
    Mike
    What happened to the square root when you polarised?
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  3. #3
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    Quote Originally Posted by Mush View Post
    What happened to the square root when you polarised?
    Shoot, the entire expression was supposed to be squared. Sorry, here's what it should look like:

    p(x,y) = K(\sqrt(1-y)^2); polarized = K(1-rsin\theta)


    OK, so I figured out what I was doing wrong. I was apparently supposed to use Line Integrals to solve this problem instead of a double integral. Double integrals solve for a solid lamina with region D. The wire does not represent a lamina. The equation I would use is:

    \int f(x(t),y(t)) \sqrt(((dx/dt)^2)+((dy/dt)^2)) after converting to parametric equations.

    In case anyone encounters a similar question.

    Thanks,
    Mike
    Last edited by Mei Liu; December 28th 2008 at 08:28 PM. Reason: Found Mistake
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