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anonymous

  • 5 years ago

Find the center of mass of a thin plate covering the triangular region cut from the first quadrant by the line y = -x + 3, with density δ(x) = x.

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  1. anonymous
    • 5 years ago
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    Take rectangular strips of length dx at a distance x from origin perpendicular to y axis dm becomes density*V. Find V in terms of x and dx(You know the equation of the line). Then integrate from 0 to 3. y coordinate is same as that of a normal triangular plate.

  2. anonymous
    • 5 years ago
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    |dw:1326959348751:dw|find side length and multiply by dx to find V

  3. anonymous
    • 5 years ago
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    ok.thanks.

  4. anonymous
    • 5 years ago
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    i wanna ask one more question to u

  5. anonymous
    • 5 years ago
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    \[\int\limits_{0}^{3}\]\[^{?}\sqrt{9-x^2}\]\[x^{2}\]dx

  6. anonymous
    • 5 years ago
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    err i did not get that properly do you want to integrate sqrt(9-x^2) , what is that x^2 below it?

  7. anonymous
    • 5 years ago
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    x^2.square root of 9-x^2 dx integrate this

  8. anonymous
    • 5 years ago
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    from 0 to 3

  9. anonymous
    • 5 years ago
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    put x=3cos(theta) i think just give me a minute.

  10. anonymous
    • 5 years ago
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    ok

  11. anonymous
    • 5 years ago
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    put x= 3cos t dt=-3sin t so the integral becomes -3(sin t) (3cos t)^2 sqrt(9 sin^2 t) = -81 sin^2 t cos ^2 t . sin ^2 t cos ^2 t=(sin ^2 (2t))/4.then sin ^2 (2t)=(1-cos 4t)/2 now you can integrate this easily.

  12. anonymous
    • 5 years ago
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    in the first line it was dx=-3sint dt

  13. anonymous
    • 5 years ago
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    Did you get it?

  14. anonymous
    • 5 years ago
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    thnx

  15. anonymous
    • 5 years ago
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    No problem.

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