Use triple integral to find center of mass

[BearY]

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?

 

[andrewkirk]

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 $z$.

 

[BearY]

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?

 

[andrewkirk]

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:

$r_{CoM} = \int_{0}^{\frac{\pi}{2}}\int_{0}^{2\pi}\int_{a}^{b} r \,\rho r^2\sin{\theta}\,dr\,d\theta\, d\phi\ /\ D$
$\theta_{CoM} = \int_{0}^{\frac{\pi}{2}}\int_{0}^{2\pi}\int_{a}^{b} \theta\, \rho r^2\sin{\theta}\,dr\,d\theta\, d\phi\ /\ D$
$\phi_{CoM} = \int_{0}^{\frac{\pi}{2}}\int_{0}^{2\pi}\int_{a}^{b} \phi\, \rho r^2\sin{\theta}\,dr\,d\theta\, d\phi\ /\ D$

 

[BearY]

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 dimension 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.

 

[vela]

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 $r_{CoM} = \frac 38 R$.

 

[LCKurtz]

@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.

 

[Ray Vickson]

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).