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

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math_proof
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1352954819183:dw

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.0First, you'll need the total area. What do you get?

math_proof
 2 years ago
Best ResponseYou've already chosen the best response.0you have to do double integral to find mass right?

math_proof
 2 years ago
Best ResponseYou've already chosen the best response.0so it will be (e1)p?

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.0Right, I assumed the very common uniform density. Sorry abuot that. "e1"? No. How did you get that?

math_proof
 2 years ago
Best ResponseYou've already chosen the best response.0im confused about setting out the integrals limits

math_proof
 2 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{1}^{e}\int\limits_{0}^{\ln(x)}p dydx\]

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.0\[\int\ln(x)\;dx = x\ln(x)  x + C\] (e*ln(e)  e)  (1*ln(1)  1) = e*ln(e)  e  0 + 1 = e(1)  e + 1 = ee+1 = 1

math_proof
 2 years ago
Best ResponseYou've already chosen the best response.0what is that? is that a mass?

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.0\[\int_{1}^{e}\int_{0}^{ln(x)}\rho\;dy\;dx = \rho\]

math_proof
 2 years ago
Best ResponseYou've already chosen the best response.0for the center mass the integral will be the same?

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.0It'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,

math_proof
 2 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{1}^{e}\int\limits_{0}^{\ln(x)}xpdydx\]

math_proof
 2 years ago
Best ResponseYou've already chosen the best response.0just wanna know if this is the integration

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.0Yes, that is one of them.
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