Use triple integral to find center of mass

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Homework Statement


Find the centre of mass of a uniform hemispherical shell of inner radius a and outer radius b.

Homework Equations


##r_{CoM} = \sum \frac{m\vec{r}}{m}##

The Attempt at a Solution


Using ##x(r,\theta,\phi)## for coordinates,
$$x_{CoM}=\frac{\int_{0}^{\frac{\pi}{2}}\int_{0}^{2\pi}\int_{a}^{b} \vec{x}\rho r^2\sin{\theta}drd\theta d\phi}{\int_{0}^{\frac{\pi}{2}}\int_{0}^{2\pi}\int_{a}^{b} \rho r^2\sin{\theta}drd\theta d\phi}$$
My vector calculus is rusty, how do I handle the ##\vec{x}## in this integral?
 

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  • #2
andrewkirk
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hvIt may help to first note that the LHS of the equation should be ##\vec x_{CoM}##. That is, it is a 3D vector that gives the location of the CoM.
Given that, treat the RHS as three separate integrals, one for each of the cartesian coordinates of the CoM: ##x,y## and ##z##.

##x_{CoM} = \int_{0}^{\frac{\pi}{2}}\int_{0}^{2\pi}\int_{a}^{b} x(r,\theta,\phi) \rho r^2\sin{\theta}\,dr\,d\theta\, d\phi\ /\ D##
##y_{CoM} = \int_{0}^{\frac{\pi}{2}}\int_{0}^{2\pi}\int_{a}^{b} y(r,\theta,\phi) \rho r^2\sin{\theta}\,dr\,d\theta\, d\phi\ /\ D##
##z_{CoM} = \int_{0}^{\frac{\pi}{2}}\int_{0}^{2\pi}\int_{a}^{b} z(r,\theta,\phi) \rho r^2\sin{\theta}\,dr\,d\theta\, d\phi\ /\ D##
where ##D## is the denominator in the formula in the OP.
Then use the formulas for converting from Cartesian to spherical coords to replace the ##x(r,\theta,\phi)## in the integrand by a function of ##r,\theta,\phi##. THen do the same for the integrals for ##y## and ##z##.
 
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  • #3
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hvIt may help to first note that the LHS of the equation should be ##\vec x_{CoM}##. That is, it is a 3D vector that gives the location of the CoM.
Given that, treat the RHS as three separate integrals, one for each of the cartesian coordinates of the CoM: ##x,y## and ##z##.

##x_{CoM} = \int_{0}^{\frac{\pi}{2}}\int_{0}^{2\pi}\int_{a}^{b} x(r,\theta,\phi) \rho r^2\sin{\theta}\,dr\,d\theta\, d\phi\ /\ D##
##y_{CoM} = \int_{0}^{\frac{\pi}{2}}\int_{0}^{2\pi}\int_{a}^{b} y(r,\theta,\phi) \rho r^2\sin{\theta}\,dr\,d\theta\, d\phi\ /\ D##
##z_{CoM} = \int_{0}^{\frac{\pi}{2}}\int_{0}^{2\pi}\int_{a}^{b} z(r,\theta,\phi) \rho r^2\sin{\theta}\,dr\,d\theta\, d\phi\ /\ D##
where ##D## is the denominator in the formula in the OP.
Then use the formulas for converting from Cartesian to spherical coords to replace the ##x(r,\theta,\phi)## in the integrand by a function of ##r,\theta,\phi##. THen do the same for the integrals for ##y## and ##z##.
I am confused about what Cartesian coordinates do here. I thought I am just looking for a point in the spherical coordinate?
 
  • #4
andrewkirk
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I am confused about what Cartesian coordinates do here. I thought I am just looking for a point in the spherical coordinate?
Quite right. Sorry, I was distracted by the use of the character 'x' in the vector ##\vec x##.
So we can do it more simply as:

##r_{CoM} = \int_{0}^{\frac{\pi}{2}}\int_{0}^{2\pi}\int_{a}^{b} r \,\rho r^2\sin{\theta}\,dr\,d\theta\, d\phi\ /\ D##
##\theta_{CoM} = \int_{0}^{\frac{\pi}{2}}\int_{0}^{2\pi}\int_{a}^{b} \theta\, \rho r^2\sin{\theta}\,dr\,d\theta\, d\phi\ /\ D##
##\phi_{CoM} = \int_{0}^{\frac{\pi}{2}}\int_{0}^{2\pi}\int_{a}^{b} \phi\, \rho r^2\sin{\theta}\,dr\,d\theta\, d\phi\ /\ D##
 
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  • #5
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Quite right. Sorry, I was distracted by the use of the character 'x' in the vector ##\vec x##.
So we can do it more simply as:

##r_{CoM} = \int_{0}^{\frac{\pi}{2}}\int_{0}^{2\pi}\int_{a}^{b} r \,\rho r^2\sin{\theta}\,dr\,d\theta\, d\phi\ /\ D##
##\theta_{CoM} = \int_{0}^{\frac{\pi}{2}}\int_{0}^{2\pi}\int_{a}^{b} \theta\, \rho r^2\sin{\theta}\,dr\,d\theta\, d\phi\ /\ D##
##\phi_{CoM} = \int_{0}^{\frac{\pi}{2}}\int_{0}^{2\pi}\int_{a}^{b} \phi\, \rho r^2\sin{\theta}\,dr\,d\theta\, d\phi\ /\ D##
When I was studying vector calculus we only did parameterizing dimention x, y, and z with t and then integrate w.r.t dt only.
So without parameterization, we still integrate each dimension separately? That makes sense. I suppose I will have to redo vector calculus to know for sure.
 
  • #6
vela
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Quite right. Sorry, I was distracted by the use of the character 'x' in the vector ##\vec x##.
So we can do it more simply as:

##r_{CoM} = \int_{0}^{\frac{\pi}{2}}\int_{0}^{2\pi}\int_{a}^{b} r \,\rho r^2\sin{\theta}\,dr\,d\theta\, d\phi\ /\ D##
##\theta_{CoM} = \int_{0}^{\frac{\pi}{2}}\int_{0}^{2\pi}\int_{a}^{b} \theta\, \rho r^2\sin{\theta}\,dr\,d\theta\, d\phi\ /\ D##
##\phi_{CoM} = \int_{0}^{\frac{\pi}{2}}\int_{0}^{2\pi}\int_{a}^{b} \phi\, \rho r^2\sin{\theta}\,dr\,d\theta\, d\phi\ /\ D##
I don't think this works because generally ##\langle \cos\theta \rangle \ne \cos\langle \theta \rangle##. Also, for a hemisphere of radius ##R##, the center of mass is on the z-axis at ##z=\frac {3}{16}{R}##, but the integral above with ##a=0## and ##b=R## gives ##r_{CoM} = \frac 38 R##.
 
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  • #7
LCKurtz
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@BearY: If that was a typical homework problem I would expect the student to consider the upper hemisphere at the origin, notice immediately that ##\bar x = \bar y = 0## from symmetry, and calculate$$
\bar z = \frac {\iiint_V z~dV}{\iiint_V 1~dV} =
\frac {\int_0^{2\pi}\int_0^{\frac \pi 2} \int_a^b \rho\cos\phi \rho^2\sin\phi~d\rho d\phi d\theta}
{\int_0^{2\pi}\int_0^{\frac \pi 2} \int_a^b 1 \rho^2\sin\phi~d\rho d\phi d\theta}$$
Notice the integral is conveniently worked in spherical coordinates, but you are calculating the center of mass ##(\bar x, \bar y, \bar z)##. And it generally is not true, for example, that ##\bar z = \bar \rho \cos\bar\phi## just because ##z =\rho \cos\phi##. You must set up the rectangular integral then change coordinates.
 
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  • #8
Ray Vickson
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I am confused about what Cartesian coordinates do here. I thought I am just looking for a point in the spherical coordinate?
Your spherical coordinates ARE relative to a particular Cartesian coordinate system! In your convention---basically, the "physics" rather than "mathematics" convention---##\theta## is the azimuthal angle (between the ##z##-axis and the point ##(x,y,z)##) while ##\phi## is the "longitude" ( the counter-clockwise angle between the ##+x## axis and ##(x,y)##, as viewed from ##z > 0## looking down).
 
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