Geometrical Center: Proving if Area-Centers Lie on X-Y Origin

[navalstudent]

Hey, this is actually a question arising from physics, but it is actually only mathematical.

Let's say you have a 3 dimensional object, with the origin in the volume-center.

-the object is symmetric about the x-z axis

-If we look at slices in the x-y-plane(z=constant). Will then the area-center of each slice be in the x-y origin?

If this is true then my engineering text-books makes sense. If not, I need to ask the professors about something. Can you guys help me?

It would be cool if one could prove this. I

PS: If the above is not true, can it be true that the area-center allways will lie along the y-axis of each slice?

 

[HallsofIvy]

What do you mean by "x-z axis"? I know the x-axis and the z-axis but I do not recognize an "x-z axis". Or do you mean xz plane?

The concept here is "centroid". The center, or "centroid", of a three dimensional figure has coordinates $\left(\overline{x}, \overline{y}, \overline{z}\right)$ where
$$\overline{x}= \frac{\int\int\int x dxdydz}{\int\int\int dxdydz}$$
$$\overline{y}= \frac{\int\int\int y dxdydz}{\int\int\int dxdydz}$$
$$\overline{z}= \frac{\int\int\int z dxdydz}{\int\int\int dxdydz}$$

The denominator in each case is just the volume of the figure, which we can call "V".

If the figure is symmetrical about the z- axis, then we can think of it as a "volume of rotation" of some curve r= f(z) about the z. In that case, we can put the problem in cylindrical coordinates and the integrals become
$$\overline{x}= \frac{\int_{z=z_0}^z_1\int_{r= 0}^{r(z)} r cos(\theta) r drd\theta}{V}$$
$$\overline{y}= \frac{\int_{z=z_0}^z_1\int_{r= 0}^{r(z)} r sin(\theta) r drd\theta}{V}$$
$$\overline{x}= \frac{\int_{z=z_0}^z_1\int_{r= 0}^{r(z)} z r drd\theta}{V}$$

But now it is easy to see that the integral with respect to $\theta$
is
$$\int_{\theta= 0}^{2\pi} cos(\theta) d\theta= sin(\theta)\right|_0^{2\pi}= 0$$
and
$$\int_{\theta= 0}^{2\pi} sin(\theta) d\theta= cos(\theta)\right|_0^{2\pi}= 0$$

so that the centroid is, indeed, on the axis of symmetry.

For each cross section at a given z value, we would find the centroid in exactly the same way except that we would not do the "z" integral and the denominator of each fraction would be the area of the cross section. For a given z, the area is bounded by $r= f(z)$ rotated around the origin so we would again have
$$\overline{x}= \frac{\int_{r= 0}^{f(z)}\int_{\theta= 0}^{2\pi} r cos(\theta)d\theta dr}{A(z)}$$
and
$$\overline{y}= \frac{\int_{r= 0}^{f(z)}\int_{\theta= 0}^{2\pi} r sin(\theta) d\theta dr}{A(z)}$$
where "A(z)" is the area of that particular cross section.

Again, the "$\theta$" integrals are 0 irrespective of the "r" integrals so the centoid of each cross section is again at (0, 0).

 

[navalstudent]

Sorry I meant about the x-z plane not object symmetric about the z-axis. Is it still valid then?

Thanks for your reply!