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math_proof

  • 3 years ago

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

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  1. tkhunny
    • 3 years ago
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    Well, youhave three integral sin your future. Can you find the first? Usually "M"?

  2. math_proof
    • 3 years ago
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    first we have to find the mass. but i have trouble setting up the integrals

  3. math_proof
    • 3 years ago
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    would it be \[\int\limits_{0}^{2\pi}\int\limits_{0}^{2}\int\limits_{0}^{4-r}(6-z) rdzdrd \Theta\]

  4. malevolence19
    • 3 years ago
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    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: \[0 \le z \le 4-r^2\] \[0 \le \phi \le 2 \pi \] Now we only need r. So look at z=4-r^2 if z=0 r=2 and at the max height r will be zero. So we have what you have :D

  5. malevolence19
    • 3 years ago
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    except the top bound should be 4-r^2 not 4-r

  6. math_proof
    • 3 years ago
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    thanks

  7. math_proof
    • 3 years ago
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    malevolence19 is there easier way to find \[\frac{ }{ x }\]

  8. math_proof
    • 3 years ago
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    instead of going through all of the integrals etc, since the difference is that you use only x

  9. malevolence19
    • 3 years ago
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    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.

  10. malevolence19
    • 3 years ago
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    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.

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