Center of Mass of a Sphere with uniform density

I got the wrong answer for ##\bar z##. I should have got $$\bar z=\frac{3}{8}R.$$In summary, the conversation focused on finding the z-coordinate of the center of mass of the first octant of a sphere of radius R centered at the origin with uniform density. The solution involved using spherical polar coordinates and setting up a triple integral to calculate the mass and the z-coordinate of the center of mass. There were discussions about the correct limits and volume element for the first octant in spherical coordinates. The final answer for the z-coordinate of the center of mass was (3/8)*R.
  • #1
Mohamed Abdul

Homework Statement



Find the z -coordinate of the center of mass of the first octant of a sphere of radius R centered at the origin. Assume that the sphere has a uniform density.

Homework Equations


Mass = Integral of the density function
Center of mass for z = Integral of density * z divided by mass

The Attempt at a Solution


I have two questions with this problem, one being whether I should use a triple or double integral, and what the value inside my integral should be. Since the sphere has uniform density, I assume the integrand should be 1, but I'm not sure.

I at least know that my bounds should be 0<r<R and 0<theta<pi/4 since I'm operating in the first octant.
 
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  • #2
Well you can start by converting your expressions to spherical polar coordinates if you haven't already?
 
  • #3
JKC said:
Well you can start by converting your expressions to spherical polar coordinates if you haven't already?
So I would have the integral from 0 to pi/4 on the outside, then the integral of from r to R inside that, then since I'd have a sphere, would my innermost integral be from 0 to sqrt(r^2-x^2-y^2)?
 
  • #4
Well if you're integrating with respect to x, y and z it should be clear that this is a triple integral. All you need to do is convert x, y and z to their corresponding spherical forms in terms of r, θ and ρ respectively and then integrate with respect to these.

As for limits if you integrate in the order of dr dθ dρ you should have the theta terms on the second integral.
 
  • #5
JKC said:
Well if you're integrating with respect to x, y and z it should be clear that this is a triple integral. All you need to do is convert x, y and z to their corresponding spherical forms in terms of r, θ and ρ respectively and then integrate with respect to these.
So then my innermost integral would be from 0 to sqrt(r^2-xcostheta^2-ycostheta^2). But if xcostheta^2-ysintheta^2 = r^2, wouldn't I just get 0 for my upper bound?
 
  • #6
Your innermost should be with respect to "r". Your bounds for the radius are between 0 and R.

The middle integral is with respect to θ. What is your range for θ? Apply this to this integral.

The limits with respect to φ should be between 0 and 2π.
 
  • #7
JKC said:
Your innermost should be with respect to "r". Your bounds for the radius are between 0 and R.

The middle integral is with respect to θ. What is your range for θ? Apply this to this integral.

The limits with respect to φ should be between 0 and 2π.
I think the bounds of the middle would be between o and pi/4, but why is the outer integral from 0 to 2pi. We didn't really discuss cylindrical coordinates much in class so I assumed it was always from z to r to theta.
 
  • #8
Mohamed Abdul said:
I think the bounds of the middle would be between o and pi/4,

Seems about right.

Mohamed Abdul said:
We didn't really discuss cylindrical coordinates

These are not cylindrical coordinates. Spherical coordinates are in terms of (r, θ, φ) whereas cylindrical polar coordinates are in terms of (r, Φ, z). Since this is a sphere it is 2π. Also remember the Jacobians for spherical coordinates in your integral.
 
  • #9
JKC said:
Seems about right.
These are not cylindrical coordinates. Spherical coordinates are in terms of (r, θ, φ) whereas cylindrical polar coordinates are in terms of (r, Φ, z). Since this is a sphere it is 2π. Also remember the Jacobians for spherical coordinates in your integral.
So then I would have the outside integral from 0 to 2pi, the middle from 0 to pi/4, and the inner from 0 to r. And then the value inside myintegral would be rho*sin^2Φ since the problem said the sphere had uniform density. Is that the correct setup?
 
  • #10
In the first octant, none of the variables go from ##0## to ##2\pi##. And what about the ##z## variable in your calculations for ##\bar z##? You need it in your calculations. And spherical coordinates is the obvious choice. Also check what the ##dV## element is in spherical coordinates.
 
Last edited:
  • #11
LCKurtz said:
In the first octant, none of the variables go from ##0## to ##2\pi##. And what about the ##z## variable in your calculations for ##\bar z##? You need it in your calculations. And spherical coordinates is the obvious choice.
So for φ and theta, they would both go from 0 to pi/4? If it is constant density, we wouldn't need z in the mass equation, only for the equation that we divide mass by, right?
LCKurtz said:
In the first octant, none of the variables go from ##0## to ##2\pi##. And what about the ##z## variable in your calculations for ##\bar z##? You need it in your calculations. And spherical coordinates is the obvious choice.
A2MiiSM.jpg
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This is the integral I have so far, am I on the right track?
 

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  • #12
Neither your limits for the angles nor your volume element is correct for the first octant in spherical coordinates.
 
  • #13
LCKurtz said:
Neither your limits for the angles nor your volume element is correct for the first octant in spherical coordinates.
Ok so I redid my work and saw that the two outer integrals range from 0 to pi/2 in the first octant. However, how would I find the inner volume element if density is constant? I don't have the constant density element.
 
  • #14
You can use a constant density of ##1## or ##\sigma##. It doesn't matter because it will cancel out when you divide to calculate ##\bar z##. But you haven't got ##dV## correct. Look it up. Then show your work for calculating ##\bar z## and what answer you get.
 
  • #15
LCKurtz said:
You can use a constant density of ##1## or ##\sigma##. It doesn't matter because it will cancel out when you divide to calculate ##\bar z##. But you haven't got ##dV## correct. Look it up. Then show your work for calculating ##\bar z## and what answer you get.
5hcq3cg.jpg

This is what I got
92Ckl
 

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  • #16
That's correct for the mass. Now how about ##\bar z##?
 
  • #17
LCKurtz said:
That's correct for the mass. Now how about ##\bar z##?
OHDg3Do.jpg

Are there any mistakes with this calculation? If I divide this by the mass, I get (3/8)*r for my z center. I'm just concerned that it's not a non-variable number.
 

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  • #18
##\frac 3 8 r## is correct for ##\bar z##. I don't know what you mean by a "non-variable" number. It makes sense that the coordinates of the center of mass should depend on the radius doesn't it? The bigger the sphere, the larger the coordinates would be, no? It's late here so I'm signing off for today.
 
  • #19
LCKurtz said:
##\frac 3 8 r## is correct for ##\bar z##. I don't know what you mean by a "non-variable" number. It makes sense that the coordinates of the center of mass should depend on the radius doesn't it? The bigger the sphere, the larger the coordinates would be, no? It's late here so I'm signing off for today.
No problem, thank you. You helped me a lot haha
 
  • #20
Mohamed Abdul said:

Homework Statement



Find the z -coordinate of the center of mass of the first octant of a sphere of radius R centered at the origin. Assume that the sphere has a uniform density.

Homework Equations


Mass = Integral of the density function
Center of mass for z = Integral of density * z divided by mass

The Attempt at a Solution


I have two questions with this problem, one being whether I should use a triple or double integral, and what the value inside my integral should be. Since the sphere has uniform density, I assume the integrand should be 1, but I'm not sure.

I at least know that my bounds should be 0<r<R and 0<theta<pi/4 since I'm operating in the first octant.

Using some symmetry properties, you can reduce the problem to a 1-dimensional integration.

It is "obvious" (and provable) that the centroid ##(x_0,y_0,z_0)## of your octant is equidistant from the ##x\,z## and ##y\,z## planes, so lies on the vertical plane through the ##z##-axis which is at an angle of ##45^{\circ}## from the ##x## and ##y##-axes. So, the only remaining task is to find the z-coordinate ##z_0##. You can decompose the octant into quarter-circular plates parallel to the ##x\,y## plane, and figure out the mass of each such plate as a function of its altitude ##z##. Then just integrate over ##z##.
 

Related to Center of Mass of a Sphere with uniform density

1. What is the formula for calculating the center of mass of a sphere with uniform density?

The formula for calculating the center of mass of a sphere with uniform density is (4/3)πr^3, where r is the radius of the sphere.

2. How does the distribution of mass affect the location of the center of mass in a sphere?

In a sphere with uniform density, the center of mass will always be located at the exact center of the sphere, regardless of the distribution of mass within the sphere.

3. Can the center of mass of a sphere with non-uniform density be calculated using the same formula?

No, the formula for calculating the center of mass of a sphere with uniform density only applies to spheres with a constant density throughout. For spheres with non-uniform density, the center of mass must be calculated using integration.

4. How does the center of mass of a sphere with uniform density relate to its surface area and volume?

The center of mass of a sphere with uniform density is located at the geometric center of the sphere, which is also the center of its surface area and volume. This means that the center of mass will always be equidistant from all points on the surface of the sphere and will also divide the volume of the sphere into equal halves.

5. What are the practical applications of knowing the center of mass of a sphere with uniform density?

Knowing the center of mass of a sphere with uniform density is useful in various fields such as physics, engineering, and astronomy. It can help in calculating the stability and balance of objects, designing structures and machines, and understanding the motion and behavior of celestial bodies.

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