What do the volume calculations look like, and how would one go about calculating the center of mass/geometric centroid of a spherical plot? More after the break…

So, using our good buddy, Wolfram Mathworld, we know that the spherical coordinate volume element, basically the volume we will be using when integrating the solid, is equal to:

$latex dV = \rho^2 \sin(\phi) \, d\rho \, d\phi \, d\theta$

Note that we are going to use the ISO 31-11 convention for our coordinate system, meaning that $latex \theta$ is the “longitude” and ranges from $latex 0$ to $latex 2\pi$, $latex \phi$ is the “latitude” and ranges from $latex 0$ to $latex \pi$, and $latex \rho$ is the radius at any given point.

Integrating this around a sphere gives us the volume of the sphere:

$latex V = \int_0^{2\pi} \! \int_o^{\pi} \! \int_0^{r} \! \rho^2 \sin(\phi) \, d\rho \, d\phi \, d\theta$

$latex V = \int_0^{2\pi} \! \int_o^{\pi} \! \frac{1}{3} r^3 \sin(\phi) \, d\phi \, d\theta$

$latex V = \int_0^{2\pi} \! \frac{2}{3} r^3 \, d\theta$

$latex V = \frac{4}{3} \pi r^3$

To convert from spherical to Cartesian coordinates, we have to simply use the following:

$latex x = \rho\sin(\phi)\cos(\theta)$

$latex y = \rho\sin(\phi)\sin(\theta)$

$latex z = \rho\cos(\phi)$

So, to solve for the x, y, and z centers of mass, we will need to convert from spherical cooridnates to Cartesian coordinates. We will do this while calculating the function centroid for each axis. The basic formula is:

$latex COM = \frac{\int \! x f(x) \, dx}{\int \! f(x) \, dx}$

Our $latex f(x) \, dx$ in this case is the $latex dV$ from above and so the basic forms for the COM for each of the three axes is:

$latex COM_{x} = \frac{\int_0^{2\pi} \! \int_o^{\pi} \! \int_0^{r} \! x \, dV}{\int_0^{2\pi} \! \int_o^{\pi} \! \int_0^{r} dV}$

$latex COM_{y} = \frac{\int_0^{2\pi} \! \int_o^{\pi} \! \int_0^{r} \! y \, dV}{\int_0^{2\pi} \! \int_o^{\pi} \! \int_0^{r} dV}$

$latex COM_{z} = \frac{\int_0^{2\pi} \! \int_o^{\pi} \! \int_0^{r} \! z \, dV}{\int_0^{2\pi} \! \int_o^{\pi} \! \int_0^{r} dV}$

So, substituting our Cartesian coordinate conversions and our $latex dV$ from above:

$latex COM_{x} = \frac{\int_0^{2\pi} \! \int_o^{\pi} \! \int_0^{r} \! \rho\sin(\phi)\cos(\theta) \rho^2 \sin(\phi) \, d\rho \, d\phi \, d\theta}{\int_0^{2\pi} \! \int_o^{\pi} \! \int_0^{r} d\rho \, d\phi \, d\theta}$

$latex COM_{y} = \frac{\int_0^{2\pi} \! \int_o^{\pi} \! \int_0^{r} \! \rho\sin(\phi)\sin(\theta) \rho^2 \sin(\phi) \, d\rho \, d\phi \, d\theta}{\int_0^{2\pi} \! \int_o^{\pi} \! \int_0^{r} d\rho \, d\phi \, d\theta}$

$latex COM_{z} = \frac{\int_0^{2\pi} \! \int_o^{\pi} \! \int_0^{r} \! \rho\cos(\phi) \rho^2 \sin(\phi) \, d\rho \, d\phi \, d\theta}{\int_0^{2\pi} \! \int_o^{\pi} \! \int_0^{r} d\rho \, d\phi \, d\theta}$

Simplified:

$latex COM_{x} = \frac{\int_0^{2\pi} \! \int_o^{\pi} \! \int_0^{r} \! \rho^3 \sin(\phi)^2 \cos(\theta) \, d\rho \, d\phi \, d\theta}{V}$

$latex COM_{y} = \frac{\int_0^{2\pi} \! \int_o^{\pi} \! \int_0^{r} \! \rho^3 \sin(\phi)^2 \sin(\theta) \, d\rho \, d\phi \, d\theta}{V}$

$latex COM_{z} = \frac{\int_0^{2\pi} \! \int_o^{\pi} \! \int_0^{r} \! \rho^3 \sin(\phi) \cos(\phi) \, d\rho \, d\phi \, d\theta}{V}$

Solving these for a constant (i.e., $latex r$) leads to:

$latex COM_{x} = 0$

$latex COM_{y} = 0$

$latex COM_{z} = 0$

No big surprise, because, as we would assume, the center of mass for a sphere should be at the origin, $latex (0,0,0)$.

Now, instead of a perfect circle, let’s try a different shape, say defined by:

$latex r(\phi,\theta) = 1+\sin(\frac{\theta}{2})\sin(\phi)$

A shape that looks somewhat like a peach and leads to COM values of:

$latex COM_{x} = \frac{-\frac{784}{225}-\frac{41\pi^2}{128}}{5\pi} \approx -0.423$

$latex COM_{y} = 0$

$latex COM_{z} = 0$

Or a spiraling shell:

$latex r(\phi,\theta) = 1+\theta$

$latex COM_{x} = \frac{6\pi^2(-3+6\pi+4\pi^2)}{-1+(1+2\pi)^4} \approx 1.165$

$latex COM_{y} = \frac{6\pi^2(1+\pi)(-5+2\pi+2\pi^2)}{-1+(1+2\pi)^4} \approx -1.833$

$latex COM_{z} = 0$

Or perhaps something really out there:

$latex r(\phi,\theta) = 1+\sin(3\theta)\sin(4\phi)+\sin(\theta)\sin(\frac{\phi}{3})$

With COM values of:

$latex COM_{x} = 0$

$latex COM_{y} = \frac{32383290752127}{62456794017280} \approx 0.518$

$latex COM_{z} = -\frac{359881023}{2416474112} \approx -0.149$

Unfortunately, it wasn’t as easy to derive as I assumed, but the steps toward that end are pretty straightforward.  After the break you can see the general equation I ultimately derived for calculating the COM in 3D-space.

A conical frustum will be defined with attributes $latex R_{1}$ (the top radius), $latex R_{2}$ (the bottom radius), and $latex h$ (the height of the truncated cone).

The radius of the frustum at any given height $latex z$ is:

$latex r_{fr}(z)=\frac{h-z}{h} R_{1} + \frac{z}{h} R_{2}$

The area of this slice at any given height $latex z$ is:

$latex A(z) = \int_0^{2\pi} \! \int_0^{r_{fr}(z)} \! r \, dr \, d\theta = \pi r_{fr}(z)^2$

Then the volume can be calculated by integrating across all heights:

$latex V = \int_0^{h} \! A(z) , dz$

Combine these three equations together and we have:

$latex V = \int_0^{h} \! \pi (\frac{h-z}{h} R_{1} + \frac{z}{h} R_{2})^2 \, dz = \frac{1}{3} h \pi (R_{1}^2 + R_{1} R_{2} + R_{2}^2)$

Ok, so now we have the radius, area, and volume of the conical frustum. Using this information, we can now calculate the center of mass.

$latex COM_{x} = \frac{1}{V} \int_0^h \! \int_0^{2\pi} \! \int_0^{r_{fr}(z)} \! r^2 \cos(\theta) A(z) \, dr \, d\theta \, dz = 0$

$latex COM_{y} = \frac{1}{V} \int_0^h \! \int_0^{2\pi} \! \int_0^{r_{fr}(z)} \! r^2 \sin(\theta) A(z) \, dr \, d\theta \, dz = 0$

$latex COM_{z} = \frac{1}{V} \int_0^h \! z A(z) \, dz$

Doing a little substitution with $latex COM_{z}$:

$latex COM_{z} = \frac{1}{V} \int_0^{h} \! z \pi (\frac{h-z}{h} R_{1} + \frac{z}{h} R_{2})^2 \, dz$

$latex = \frac{1}{V} \frac{h^2 \pi (R_{1}^2 + 2 R_{1} R_{2} + 3 R_{2}^2)}{12}$

$latex = \frac{h (R_{1}^2 + 2 R_{1} R_{2} + 3 R_{2}^2)}{4(R_{1}^2 + R_{1} R_{2} + R_{2}^2)}$

Note that because the conical frustum is radially symmetric around the z-axis, the COM for both x and y will be 0. This obviously will not be the case for shapes with more complex surfaces of revolution, but is a nice little proof.

Complex, you want complex??

So, we will add a simple part onto the shape (like the one in the image above), defined using both a radial function, $latex f_{rad}(\theta)$, and a vertical function, $latex f_{vert}(z)$. The functions themselves can be any analytical functions, as long as they are continuous. And frankly, it doesn’t matter for this proof, so we’ll leave them undefined for the time being.

Remember how simple the radius at any given height was? (Just to remind you: $latex r_{fr}(z)=\frac{h-z}{h} R_{1} + \frac{z}{h} R_{2}$). Now we need to add our radial and vertical functions, making the function dependent upon not only height, but also angle:

$latex r_{fr}(z,\theta)=\frac{h-z}{h} R_{1} + \frac{z}{h} R_{2} + f_{vert}(z) f_{rad}(theta)$

Given that the radius is now a bit more complex, the area of this slice at any given height $latex z$ is now:

$latex A(z) = \int_0^{2\pi} \! \int_0^{r_{fr}(z,\theta)} \! r \, dr \, d\theta$

Then the volume is now a bit more complicated:

$latex V = \int_0^{h} \! \int_0^{2\pi} \! \int_0^{r_{fr}(z,\theta)} \! r \, dr \, d\theta \, dz$

And now we can calculate the COMs:

$latex COM_{x} = \frac{1}{V} \int_0^h \! \int_0^{2\pi} \! \int_0^{r_{fr}(z,\theta)} \! r^2 \cos(\theta) (\int_0^{2\pi} \! \int_0^{r_{fr}(z,\theta)} \! r \, dr \, d\theta) \, dr \, d\theta \, dz$

$latex COM_{y} = \frac{1}{V} \int_0^h \! \int_0^{2\pi} \! \int_0^{r_{fr}(z,\theta)} \! r^2 \sin(\theta) (\int_0^{2\pi} \! \int_0^{r_{fr}(z,\theta)} \! r \, dr \, d\theta) \, dr \, d\theta \, dz$

$latex COM_{z} = \frac{1}{V} \int_0^h \! z (\int_0^{2\pi} \! \int_0^{r_{fr}(z,\theta)} \! r \, dr \, d\theta) \, dz$

The good news, all you really need to define is $latex r_{fr}(z,theta)$ and h. The bad news, pentuple integrals. Substituting all of the necessary equations, we get these monstrosities:

$latex COM_{x} = \frac{\int_0^h \! \int_0^{2\pi} \! \int_0^{\frac{h-z}{h} R_{1} + \frac{z}{h} R_{2} + f_{vert}(z) f_{rad}(\theta)} \! r^2 \cos(\theta) (\int_0^{2\pi} \! \int_0^{\frac{h-z}{h} R_{1} + \frac{z}{h} R_{2} + f_{vert}(z) f_{rad}(\theta)} \! r \, dr \, d\theta) \, dr \, d\theta \, dz}{\int_0^{h} \! \int_0^{2\pi} \! \int_0^{\frac{h-z}{h} R_{1} + \frac{z}{h} R_{2} + f_{vert}(z) f_{rad}(\theta)} \! r \, dr \, d\theta \, dz}$

$latex COM_{y} = \frac{\int_0^h \! \int_0^{2\pi} \! \int_0^{\frac{h-z}{h} R_{1} + \frac{z}{h} R_{2} + f_{vert}(z) f_{rad}(\theta)} \! r^2 \sin(\theta) (\int_0^{2\pi} \! \int_0^{\frac{h-z}{h} R_{1} + \frac{z}{h} R_{2} + f_{vert}(z) f_{rad}(\theta)} \! r \, dr \, d\theta) \, dr \, d\theta \, dz}{\int_0^{h} \! \int_0^{2\pi} \! \int_0^{\frac{h-z}{h} R_{1} + \frac{z}{h} R_{2} + f_{vert}(z) f_{rad}(\theta)} \! r \, dr \, d\theta \, dz}$

$latex COM_{z} = \frac{\int_0^h \! z (\int_0^{2\pi} \! \int_0^{\frac{h-z}{h} R_{1} + \frac{z}{h} R_{2} + f_{vert}(z) f_{rad}(\theta)} \! r \, dr \, d\theta) \, dz}{\int_0^{h} \! \int_0^{2\pi} \! \int_0^{\frac{h-z}{h} R_{1} + \frac{z}{h} R_{2} + f_{vert}(z) f_{rad}(\theta)} \! r \, dr \, d\theta \, dz}$

Not very pretty, ehh? Well, in the meantime, I’ll leave the final calculations for you to work on. Suffice to say, it is a pain in the butt.