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.
integration multivariable-calculus
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add a comment |
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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.
integration multivariable-calculus
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How did you come to this conclusion?
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– John Keeper
Dec 15 '18 at 23:01
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The ellipsoids are obtained from the unit sphere by scaling along the coordinate axes.
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– amd
Dec 16 '18 at 0:08
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I wrote the constants wrongly. The right transformation simplifies the first inequality to $r^2leq1.$ It is $x=rcos tcos phi, 2y=rsin t cos phi, 3z=rsin phi.$
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– user376343
Dec 16 '18 at 0:20
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math.stackexchange.com/questions/3043727/…
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– Hans Lundmark
Dec 17 '18 at 10:05
add a comment |
$begingroup$
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.
integration multivariable-calculus
$endgroup$
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
integration multivariable-calculus
asked Dec 15 '18 at 21:56
John KeeperJohn Keeper
532315
532315
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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
add a comment |
$begingroup$
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
add a comment |
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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