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math_proof

  • 3 years ago

find centroid of the region bounded by y=ln(x) the axis and x=e

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  1. math_proof
    • 3 years ago
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    |dw:1352954819183:dw|

  2. tkhunny
    • 3 years ago
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    First, you'll need the total area. What do you get?

  3. math_proof
    • 3 years ago
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    you have to do double integral to find mass right?

  4. math_proof
    • 3 years ago
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    so it will be (e-1)p?

  5. tkhunny
    • 3 years ago
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    Right, I assumed the very common uniform density. Sorry abuot that. "e-1"? No. How did you get that?

  6. math_proof
    • 3 years ago
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    im confused about setting out the integrals limits

  7. math_proof
    • 3 years ago
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    is it e(loge-1)

  8. math_proof
    • 3 years ago
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    \[\int\limits_{1}^{e}\int\limits_{0}^{\ln(x)}p dydx\]

  9. tkhunny
    • 3 years ago
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    \[\int\ln(x)\;dx = x\ln(x) - x + C\] (e*ln(e) - e) - (1*ln(1) - 1) = e*ln(e) - e - 0 + 1 = e(1) - e + 1 = e-e+1 = 1

  10. math_proof
    • 3 years ago
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    what is that? is that a mass?

  11. tkhunny
    • 3 years ago
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    \[\int_{1}^{e}\int_{0}^{ln(x)}\rho\;dy\;dx = \rho\]

  12. math_proof
    • 3 years ago
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    for the center mass the integral will be the same?

  13. tkhunny
    • 3 years ago
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    It's not a matter of guesing or whether we can predict it or how we feel about it. It's just how it works out on this one,

  14. math_proof
    • 3 years ago
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    \[\int\limits_{1}^{e}\int\limits_{0}^{\ln(x)}xpdydx\]

  15. math_proof
    • 3 years ago
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    just wanna know if this is the integration

  16. tkhunny
    • 3 years ago
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    Yes, that is one of them.

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