Calculating the center of mass of a body in $mathbb{R^3}$












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I want to calculate the center of mass of the body defined by $$begin{cases} x^2+4y^2+9z^2leq 1 \x^2+4y^2+9z^2leq 6z end{cases}$$ where the density of mass is proportional to the distance to the plane $xy$. First of all I will have to calculate the mass, meaning I have to solve $$M=iiint dm=iiint_V lambda z dV$$ My problem is that I don't know what to do when the body is given in that way, I've done similar problems where it was clear I had to do a change of coordinates (spherical, cylindrical...) however in this case I don't know how to set the integral.










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  • $begingroup$
    How did you come to this conclusion?
    $endgroup$
    – John Keeper
    Dec 15 '18 at 23:01










  • $begingroup$
    The ellipsoids are obtained from the unit sphere by scaling along the coordinate axes.
    $endgroup$
    – amd
    Dec 16 '18 at 0:08










  • $begingroup$
    I wrote the constants wrongly. The right transformation simplifies the first inequality to $r^2leq1.$ It is $x=rcos tcos phi, 2y=rsin t cos phi, 3z=rsin phi.$
    $endgroup$
    – user376343
    Dec 16 '18 at 0:20












  • $begingroup$
    math.stackexchange.com/questions/3043727/…
    $endgroup$
    – Hans Lundmark
    Dec 17 '18 at 10:05
















0












$begingroup$


I want to calculate the center of mass of the body defined by $$begin{cases} x^2+4y^2+9z^2leq 1 \x^2+4y^2+9z^2leq 6z end{cases}$$ where the density of mass is proportional to the distance to the plane $xy$. First of all I will have to calculate the mass, meaning I have to solve $$M=iiint dm=iiint_V lambda z dV$$ My problem is that I don't know what to do when the body is given in that way, I've done similar problems where it was clear I had to do a change of coordinates (spherical, cylindrical...) however in this case I don't know how to set the integral.










share|cite|improve this question









$endgroup$












  • $begingroup$
    How did you come to this conclusion?
    $endgroup$
    – John Keeper
    Dec 15 '18 at 23:01










  • $begingroup$
    The ellipsoids are obtained from the unit sphere by scaling along the coordinate axes.
    $endgroup$
    – amd
    Dec 16 '18 at 0:08










  • $begingroup$
    I wrote the constants wrongly. The right transformation simplifies the first inequality to $r^2leq1.$ It is $x=rcos tcos phi, 2y=rsin t cos phi, 3z=rsin phi.$
    $endgroup$
    – user376343
    Dec 16 '18 at 0:20












  • $begingroup$
    math.stackexchange.com/questions/3043727/…
    $endgroup$
    – Hans Lundmark
    Dec 17 '18 at 10:05














0












0








0





$begingroup$


I want to calculate the center of mass of the body defined by $$begin{cases} x^2+4y^2+9z^2leq 1 \x^2+4y^2+9z^2leq 6z end{cases}$$ where the density of mass is proportional to the distance to the plane $xy$. First of all I will have to calculate the mass, meaning I have to solve $$M=iiint dm=iiint_V lambda z dV$$ My problem is that I don't know what to do when the body is given in that way, I've done similar problems where it was clear I had to do a change of coordinates (spherical, cylindrical...) however in this case I don't know how to set the integral.










share|cite|improve this question









$endgroup$




I want to calculate the center of mass of the body defined by $$begin{cases} x^2+4y^2+9z^2leq 1 \x^2+4y^2+9z^2leq 6z end{cases}$$ where the density of mass is proportional to the distance to the plane $xy$. First of all I will have to calculate the mass, meaning I have to solve $$M=iiint dm=iiint_V lambda z dV$$ My problem is that I don't know what to do when the body is given in that way, I've done similar problems where it was clear I had to do a change of coordinates (spherical, cylindrical...) however in this case I don't know how to set the integral.







integration multivariable-calculus






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asked Dec 15 '18 at 21:56









John KeeperJohn Keeper

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532315












  • $begingroup$
    How did you come to this conclusion?
    $endgroup$
    – John Keeper
    Dec 15 '18 at 23:01










  • $begingroup$
    The ellipsoids are obtained from the unit sphere by scaling along the coordinate axes.
    $endgroup$
    – amd
    Dec 16 '18 at 0:08










  • $begingroup$
    I wrote the constants wrongly. The right transformation simplifies the first inequality to $r^2leq1.$ It is $x=rcos tcos phi, 2y=rsin t cos phi, 3z=rsin phi.$
    $endgroup$
    – user376343
    Dec 16 '18 at 0:20












  • $begingroup$
    math.stackexchange.com/questions/3043727/…
    $endgroup$
    – Hans Lundmark
    Dec 17 '18 at 10:05


















  • $begingroup$
    How did you come to this conclusion?
    $endgroup$
    – John Keeper
    Dec 15 '18 at 23:01










  • $begingroup$
    The ellipsoids are obtained from the unit sphere by scaling along the coordinate axes.
    $endgroup$
    – amd
    Dec 16 '18 at 0:08










  • $begingroup$
    I wrote the constants wrongly. The right transformation simplifies the first inequality to $r^2leq1.$ It is $x=rcos tcos phi, 2y=rsin t cos phi, 3z=rsin phi.$
    $endgroup$
    – user376343
    Dec 16 '18 at 0:20












  • $begingroup$
    math.stackexchange.com/questions/3043727/…
    $endgroup$
    – Hans Lundmark
    Dec 17 '18 at 10:05
















$begingroup$
How did you come to this conclusion?
$endgroup$
– John Keeper
Dec 15 '18 at 23:01




$begingroup$
How did you come to this conclusion?
$endgroup$
– John Keeper
Dec 15 '18 at 23:01












$begingroup$
The ellipsoids are obtained from the unit sphere by scaling along the coordinate axes.
$endgroup$
– amd
Dec 16 '18 at 0:08




$begingroup$
The ellipsoids are obtained from the unit sphere by scaling along the coordinate axes.
$endgroup$
– amd
Dec 16 '18 at 0:08












$begingroup$
I wrote the constants wrongly. The right transformation simplifies the first inequality to $r^2leq1.$ It is $x=rcos tcos phi, 2y=rsin t cos phi, 3z=rsin phi.$
$endgroup$
– user376343
Dec 16 '18 at 0:20






$begingroup$
I wrote the constants wrongly. The right transformation simplifies the first inequality to $r^2leq1.$ It is $x=rcos tcos phi, 2y=rsin t cos phi, 3z=rsin phi.$
$endgroup$
– user376343
Dec 16 '18 at 0:20














$begingroup$
math.stackexchange.com/questions/3043727/…
$endgroup$
– Hans Lundmark
Dec 17 '18 at 10:05




$begingroup$
math.stackexchange.com/questions/3043727/…
$endgroup$
– Hans Lundmark
Dec 17 '18 at 10:05










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