Find the center of mass of the solid $$ R = {(x,y,z) \in \mathbb{R}^3 : x^2 + y^2 \leq 1^2, 0 \leq z \leq \frac \pi2 }$$ if the density is given by $$\rho(x,y,z) = 5\bigg(\sqrt{x^2 + y^2} + sin(z)\bigg). $$

First i tried solving for the mass with the triple integral $$\int_0^{2\pi} \int_0^{\frac\pi2} \int_0^15r\big(r+sin(z)\big) = \frac{10\pi^2}{6} $$ Which gave me the $z$ coordinate $\frac{3+\pi}{\pi}$ Not sure if this is correct though. Also, i don't know how to find the other coordinates now seeing as the density is not uniform.

  • $\begingroup$ set up is ok but check the derivation, by wolfy the mass should be different wolframalpha.com/input/… $\endgroup$ – gimusi Mar 9 '18 at 15:01

Note that the method is correct since $\rho(x,y,z)$ is symmetric with respect to z axis the center of mass has coordinates $(0,0,z_G)$ with (I didn't check the detail of calculation):

$$z_G=\frac{\int_V z\cdot \rho(x,y,z)dV}{M}$$

  • $\begingroup$ How do you see that $\rho(x,y,z)$ is symmetric about the z-axis? Doesen't the $\sqrt{x^2 + y^2}$ term mean that it also depends on $x$ and $y$ ? $\endgroup$ – Pame Mar 9 '18 at 15:08
  • $\begingroup$ @Pame since $\rho(x,y,z)=\rho(-x,y,z)=\rho(x,-y,z)=\rho(-x,-y,z)$ $\endgroup$ – gimusi Mar 9 '18 at 15:10
  • $\begingroup$ @Pame or also note that for z=constant $\rho=\rho(r)$ $\endgroup$ – gimusi Mar 9 '18 at 15:17
  • $\begingroup$ Ok i see that it must be symmetric about the $z$-axis, but i don't quite get why it means the center of mass must lie on the $z$-axis. $\endgroup$ – Pame Mar 9 '18 at 15:31
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    $\begingroup$ @Pame for the mass integral the integrand is positive thus $\int_V \rho(x,y,z)dV>0$ whereas for $x_G=\frac{\int_V x\cdot \rho(x,y,z)dV}{M}$ we have a term x wich makes one half integral positive and one half equal but negative thus the sum is zero. $\endgroup$ – gimusi Mar 9 '18 at 16:20

In general, the coordinates of the center of mass of the region $\Omega$ are given by

$$ \overline{{\bf x}} = \frac{\int_\Omega{\rm d}{\bf x}~{\bf x}\rho({\bf x})}{\int_\Omega{\rm d}{\bf x}~\rho({\bf x})} \tag{1} $$

In your case $\Omega = \{(x,y,z) | x^2+y^2 < 1, 0 < z < \pi/2 \}$

We can then calculate

$\int_\Omega{\rm d}{\bf x}~\rho({\bf x})$

\begin{eqnarray} \int_\Omega{\rm d}{\bf x}~\rho({\bf x}) &=& \int_0^{2\pi}{\rm d}\phi\int_{0}^{\pi/2}{\rm dz}\int_0^1{\rm dR}~R\rho(R,\phi,z) \\ &=& \int_0^{2\pi}{\rm d}\phi\int_{0}^{\pi/2}{\rm dz}\int_0^1{\rm dR}~R [5(R + \sin z)] \\ &=& 10\pi \int_{0}^{\pi/2}{\rm dz}\int_0^1{\rm dR}~R (R + \sin z) \\ &=& (\dots) \end{eqnarray}

$\int_\Omega{\rm d}{\bf x}~x\rho({\bf x})$

\begin{eqnarray} \int_\Omega{\rm d}{\bf x}~x\rho({\bf x}) &=& \int_0^{2\pi}{\rm d}\phi\int_{0}^{\pi/2}{\rm dz}\int_0^1{\rm dR}~Rx\rho(R,\phi,z) \\ &=& \int_0^{2\pi}{\rm d}\phi\int_{0}^{\pi/2}{\rm dz}\int_0^1{\rm dR}~R [R\cos\phi][5(R + \sin z)] &=& (\dots) \end{eqnarray}

$\int_\Omega{\rm d}{\bf x}~y\rho({\bf x})$

\begin{eqnarray} \int_\Omega{\rm d}{\bf x}~y\rho({\bf x}) &=& \int_0^{2\pi}{\rm d}\phi\int_{0}^{\pi/2}{\rm dz}\int_0^1{\rm dR}~Ry\rho(R,\phi,z) \\ &=& \int_0^{2\pi}{\rm d}\phi\int_{0}^{\pi/2}{\rm dz}\int_0^1{\rm dR}~R [R\sin\phi][5(R + \sin z)] &=& (\dots) \end{eqnarray}

$\int_\Omega{\rm d}{\bf x}~z\rho({\bf x})$

\begin{eqnarray} \int_\Omega{\rm d}{\bf x}~z\rho({\bf x}) &=& \int_0^{2\pi}{\rm d}\phi\int_{0}^{\pi/2}{\rm dz}\int_0^1{\rm dR}~Rz\rho(R,\phi,z) \\ &=& \int_0^{2\pi}{\rm d}\phi\int_{0}^{\pi/2}{\rm dz}\int_0^1{\rm dR}~R [z][5(R + \sin z)] &=& (\dots) \end{eqnarray}

Everything together

When you calculate these integrals, just use Eq. (1)


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