Center of mass of one octant of a non-homogenous sphere

Question

Find the center of mass of that part of the sphere

$x^2+y^2+z^2 \le a^2$

having $x,y,z \ge 0$ (that is, the part in the first octant)

With density given by $\rho(x,y,z)=(x^2+y^2+z^2)^{3/2}$

It should be solved using a triple integral.

I get either $\frac{2a}3$ or $\frac{6a}7$, but I don't know whether any of these are correct. I used spherical coordinates. The problem is that the density function has a power of $\frac{3}{2}$.

Answer 1

You first need to find the total mass, which you could find by integrating over the whole sphere and dividing by $8$, but it is better to set up the problem as you will need for the coordinates. The first octant has intervals $\theta \in [0,\pi/2]$ and $\phi \in [0,\pi/2]$, so that the mass is

$$M=\int_0^a dr \, r^2 \rho(r) \, \int_0^{\pi/2} d\theta \, \sin{\theta} \, \int_0^{\pi/2} d\phi$$

The coordinates are then given by

$$\bar{x} = \frac1{M} \int_0^a dr \, r^3 \rho(r) \, \int_0^{\pi/2} d\theta \, \sin^2{\theta} \, \int_0^{\pi/2} d\phi\, \cos{\phi}$$

$$\bar{y} = \frac1{M} \int_0^a dr \, r^3 \rho(r) \, \int_0^{\pi/2} d\theta \, \sin^2{\theta} \, \int_0^{\pi/2} d\phi\, \sin{\phi}$$

$$\bar{z} = \frac1{M} \int_0^a dr \, r^3 \rho(r) \, \int_0^{\pi/2} d\theta \, \sin{\theta} \cos{\theta} \, \int_0^{\pi/2} d\phi$$

I get $$\left (\bar{x},\bar{y},\bar{z}\right ) = \left (\frac{3 a}{7},\frac{3 a}{7},\frac{3 a}{7} \right )$$

To compare, the analogous result for uniform density is

$$\left (\bar{x},\bar{y},\bar{z}\right ) = \left (\frac{3 a}{8},\frac{3 a}{8},\frac{3 a}{8} \right )$$

which is closer to the center.

Answer 2

the center of mass is $M = (x_m,y_m,z_m)$.

1. Compute the total mass: $$mass = \int_0^a r^3 (4\pi r^2 dr) = \frac{4\pi a^6}6 = \frac{2\pi a^6}3$$
2. Then, $$ x_m=y_m=z_m = \frac 1{mass} \int_V x dV\\ =\frac 1{mass} \int_0^a x dx \int_{y^2+z^2\le a^2-x^2}(x^2+y^2+z^2)^{3/2}dydz\\ \int_{y^2+z^2\le R^2}(x^2+y^2+z^2)^{3/2}dydz= \int_0^{R}(x^2+ u^2)^{3/2}2\pi udu=2\pi \frac {(x^2+R^2)^{5/2}}5\\ x_m=\frac 1{mass} \int_0^a x dx \times 2\pi \frac {a^5}5 = \frac 3{2\pi a^6}2\pi \frac {a^5}5\frac {a^2}2 =\frac{3a}{10}$$

I hope I did not make much mistakes.

Answer 3

First, we solve it for the unit sphere, since the solution is just scaled up by $a$.

Secondly, we observe that if we have a single octant, with center of mass at $(u, u, u)$, then if we combine the four positive-$z$ octants (say), then the center of mass will be at $(0, 0, u)$, by symmetry. So if we find the center of mass of the positive-$z$ hemisphere, we thereby obtain the center of mass of any octant.

We note that the center of mass of a unit hemispherical shell is at coordinate

$$ \frac{\int_{\theta=0}^{\pi/2} \cos\theta\sin\theta \, d\theta} {\int_{\theta=0}^{\pi/2} \cos\theta \,d\theta} = \frac{1}{2} $$

so that of a hemispherical shell of radius $r$ is at coordinate $\frac{r}{2}$.

In the sphere in question, we have shells of area proportional to $r^2$, multiplied by a density proportional to $r^3$, so the coordinate of the center of mass of the hemisphere is

$$ \frac{\int_{r=0}^1 \frac{r}{2} r^5 \, dr}{\int_{r=0}^1 r^5 \, dr} = \frac{\frac{1}{14}}{\frac{1}{6}} = \frac{3}{7} $$

For a uniform sphere, it would be

$$ \frac{\int_{r=0}^1 \frac{r}{2} r^2 \, dr}{\int_{r=0}^1 r^2 \, dr} = \frac{\frac{1}{8}}{\frac{1}{3}} = \frac{3}{8} $$

thus confirming the values obtained by Ron Gordon. It is mildly curious to me that that high a density gradient has such a small effect.