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math_proof

  • 3 years ago

Find center of mass

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  1. math_proof
    • 3 years ago
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    Of the upper half (\[y \ge 0\] of the disk bounded by the circle x^2+y^2=4 with p(x,y)=1+y/2

  2. math_proof
    • 3 years ago
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    i got the mass of 8/3 +2pi is that right?

  3. amistre64
    • 3 years ago
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    \[\int_{0}^{2} 1+\frac12y~dy\] 2 + 2/4 is what i see along the y axis

  4. amistre64
    • 3 years ago
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    the center of mass is a point at which the total mass can be depicted

  5. amistre64
    • 3 years ago
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    or in this case, maybe a line for the y cg and another for the xcg; where they cross would be the center of mass

  6. math_proof
    • 3 years ago
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    what i was taught is first to calculate the mass and then calculate integrals for moments with respect to the y-axis and x-axis and then divide My/M (my moment by mass ) to get x

  7. amistre64
    • 3 years ago
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    hmmm, that does ring a bell

  8. amistre64
    • 3 years ago
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    http://tutorial.math.lamar.edu/Classes/CalcII/CenterOfMass.aspx yeah, i might need to read this for a refresher :)

  9. amistre64
    • 3 years ago
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    this is your density function? p(x,y)=1+y/2

  10. math_proof
    • 3 years ago
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    i get the concept but i have trouble setting up the integral

  11. math_proof
    • 3 years ago
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    i found the mass \int\limits_{-2}^{2}\int\limits_{0}^{\sqrt{4-x^2}}1+\frac{ y }{ 2 }dydx = \frac{ 8 }{ 3 }+2\pi

  12. math_proof
    • 3 years ago
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    \[\int\limits_{-2}^{2}\int\limits_{0}^{\sqrt{4-x^2}}1+\frac{ y }{2 } dydx =\frac{ 8 }{ 3 }+2\pi\]

  13. amistre64
    • 3 years ago
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    wrap that with a: ``` \[ \] ``` yeah, there ya go

  14. math_proof
    • 3 years ago
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    would the integral for my be the same just with extra y? like Mx\[\int\limits_{-2}^{2}\int\limits_{0}^{\sqrt{4-x^2}}y(1+\frac{ y }{ 2 })dydx\]

  15. math_proof
    • 3 years ago
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    would that be Mx?

  16. amistre64
    • 3 years ago
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    http://people.usd.edu/~jflores/MultiCalc02/WebBook/Chapter_16/Graphics16/MI_5/Html16_5/16.5%20Applications%20of%20Double%20Integrals.htm that does look appriopriate to me

  17. math_proof
    • 3 years ago
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    then i would get 0, ooo it should be zero since circle is centered at 0

  18. amistre64
    • 3 years ago
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    \[\int\limits_{-2}^{2}\int\limits_{0}^{\sqrt{4-x^2}}y(1+\frac{ y }{ 2 })dydx\] \[\int\limits_{-2}^{2}\int\limits_{0}^{\sqrt{4-x^2}}y+\frac12 y^2~dy~dx\] \[\int\limits_{-2}^{2}\frac12(\sqrt{4-x^2})^2+\frac16 (\sqrt{4-x^2})^3~dx\] \[k\int\limits_{-2}^{2}~dx\] \[k(2-(-2))=4k\] right?

  19. amistre64
    • 3 years ago
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    lol .... i totally looked over the x parts in the second go around :/

  20. math_proof
    • 3 years ago
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    whats k?

  21. amistre64
    • 3 years ago
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    i thought id clean it up by using a constant k, because i overran the xs and simply forgot to integrate them \[\frac12\int\limits_{-2}^{2}(\sqrt{4-x^2})^2~dx+\frac16\int\limits_{-2}^{2} (\sqrt{4-x^2})^3~dx\] \[\frac12\int\limits_{-2}^{2}4-x^2~dx+\frac16\int\limits_{-2}^{2} (4-x^2)\sqrt{4-x^2}~dx\] \[8-\frac{8}3+6\pi\]

  22. amistre64
    • 3 years ago
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    http://www.wolframalpha.com/input/?i=integrate+y%281%2By%2F2%29+from+y%3D0..sqrt%284-x%5E2%29 http://www.wolframalpha.com/input/?i=integrate+%28integrate+y%281%2By%2F2%29+from+y%3D0..sqrt%284-x%5E2%29%29+from+x%3D-2+to+2 i seem to have mismathed it someplace .. ah well. i have to get going so ill try to take another look at this tomorrow; good luck

  23. math_proof
    • 3 years ago
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    thanks i think i got it, just the integration is painful

  24. Zarkon
    • 3 years ago
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    I get \[\int\limits_{-2}^{2}\int\limits_{0}^{\sqrt{4-x^2}}y\left(1+\frac{ y }{ 2 }\right)dydx=\pi+\frac{16}{3}\]

  25. Zarkon
    • 3 years ago
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    it is easy to integrate if you switch to polar coordinates

  26. math_proof
    • 3 years ago
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    oo I didn't think of that, and yea it seems right

  27. math_proof
    • 3 years ago
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    thanks a lot

  28. Zarkon
    • 3 years ago
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    \[\int\limits_{-2}^{2}\int\limits_{0}^{\sqrt{4-x^2}}y\left(1+\frac{ y }{ 2 }\right)dydx=\pi+\frac{16}{3}\] \[\int\limits_{0}^{\pi}\int\limits_{0}^{2}r\sin(\theta)\left(1+\frac{ r\sin(\theta) }{ 2 }\right)rdrd\theta=\pi+\frac{16}{3}\]

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