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math_proof
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Center of mass for the region bounded by the paraboloid z=4x^2y^2 and z=0 with p(x,y,z)=5z
 2 years ago
 2 years ago
math_proof Group Title
Center of mass for the region bounded by the paraboloid z=4x^2y^2 and z=0 with p(x,y,z)=5z
 2 years ago
 2 years ago

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tkhunny Group TitleBest ResponseYou've already chosen the best response.0
Well, youhave three integral sin your future. Can you find the first? Usually "M"?
 2 years ago

math_proof Group TitleBest ResponseYou've already chosen the best response.1
first we have to find the mass. but i have trouble setting up the integrals
 2 years ago

math_proof Group TitleBest ResponseYou've already chosen the best response.1
would it be \[\int\limits_{0}^{2\pi}\int\limits_{0}^{2}\int\limits_{0}^{4r}(6z) rdzdrd \Theta\]
 2 years ago

malevolence19 Group TitleBest ResponseYou've already chosen the best response.0
Okay, first notice that z goes between: \[0 \le z \le 4(x^2+y^2)\] To me this looks like a good time to switch to cylindrical. So we have: \[0 \le z \le 4r^2\] \[0 \le \phi \le 2 \pi \] Now we only need r. So look at z=4r^2 if z=0 r=2 and at the max height r will be zero. So we have what you have :D
 2 years ago

malevolence19 Group TitleBest ResponseYou've already chosen the best response.0
except the top bound should be 4r^2 not 4r
 2 years ago

math_proof Group TitleBest ResponseYou've already chosen the best response.1
malevolence19 is there easier way to find \[\frac{ }{ x }\]
 2 years ago

math_proof Group TitleBest ResponseYou've already chosen the best response.1
instead of going through all of the integrals etc, since the difference is that you use only x
 2 years ago

malevolence19 Group TitleBest ResponseYou've already chosen the best response.0
Well careful inspection tells you that x and y are both zero. By symmetry and the fact that: \[x=r \cos(\phi);y=r \sin(\phi) \implies \iiint r \sin(\phi) \rho dV; \iiint r \cos(\phi) \rho dV\] But if you realize that integrating sine or cosine over a full period (i.e., 2 pi) are just ZERO. That is: \[\int\limits_0^{2 \pi}\sin(x)dx=\int\limits_0^{2 \pi} \cos(x)dx=0\] So they are zero by symmetry and by the math.
 2 years ago

malevolence19 Group TitleBest ResponseYou've already chosen the best response.0
Since z is just z is cylindrical you see that there is no sine or cosine so the integral is NON zero, as we'd expect.
 2 years ago
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