Calculating the Mass of a Half Ball with Given Density δ(x, y, z)

In summary: The jacobian would be different in each coordinate system, but the integral would be the same. If you are changing coordinate systems, you must account for it somehow. If you don't, you will get an error.In summary, the student is trying to solve a problem using spherical coordinates, but does not know which integral to use. He is confused about which equation to use and does not answer which question.
  • #1
RaulTheUCSCSlug
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Homework Statement


If the density of the half-ball x ^2 + y ^2 + z ^2 ≤ 4 ; z ≥ 0 is given by δ(x, y, z) = ( x^ 2 + y^ 2 + z ^2)^(1/2) find its mass.

Homework Equations


∫∫F⋅ds
∫∫FoΦ(u,v)||Tu,Tv||dudv
∫∫FoΦ(u,v)⋅(Tu,Tv)dudv

The Attempt at a Solution


For the last problem I was asked to find the density of the shell and used the equation
∫∫FoΦ(u,v)||Tu,Tv||dudv, but now I don't know if I use the same thing or not. I know how to parametrize it and know that it will be a ball of radius 2 but don't know if I use the same equation or not.
 
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  • #2
No idea what your T stand for. Normally mass is ##\int \rho \, dV## and I'm missing that relevant equation...
 
  • #3
BvU said:
No idea what your T stand for. Normally mass is ##\int \rho \, dV## and I'm missing that relevant equation...
Tu stands for Φ(u,v) derivative with respect to u and Tv is Φ(u,v) with respect to v.
 
  • #4
What I did in the previous problem was parameterize the surface, but that was asking for mass of shell, now it is mass of ball. So am I supposed to use volume then divide by the density? ? ?
 
  • #5
RaulTheUCSCSlug said:
Tu stands for Φ(u,v) derivative with respect to u and Tv is Φ(u,v) with respect to v.
Displaces the unknown to u and v. Why not use the simple equation ?
 
  • #6
BvU said:
Displaces the unknown to u and v. Why not use the simple equation ?
Oops it's supposed to be x^2+y^2+z^2=/< 4, so it would be best to convert to spherical coordinates.
 
  • #7
Yep.
 
  • #8
doesnt answer my question though.
 
  • #9
That was a typ-o from copying and pasting, I am still stuck on what to do...
 
  • #10
RaulTheUCSCSlug said:
Oops it's supposed to be x^2+y^2+z^2=/< 4, so it would be best to convert to spherical coordinates.
This problem is a natural for spherical coordinates, as the density varies with the radius, and the object is half a sphere.
 
  • #11
RaulTheUCSCSlug said:
doesnt answer my question though.

What does not answer which question?
 
  • #12
Mark44 said:
This problem is a natural for spherical coordinates, as the density varies with the radius, and the object is half a sphere.
Yes, I know, there was a typo before, but the typo is no longer there, I already switched to spherical coordinates, hence the equation that I am using since I am perimeterizing.
 
  • #13
Ray Vickson said:
What does not answer which question?

BvU said:
Yep.

^^^^
 
  • #14
Okay, I know to use spherical coordinates. I just am not sure which integral to use. That is my confusion.
 
  • #15
How does this question differ from when they ask for the mass of a spherical shell. Because I can do the shell fine, but I just don't know how it jumps to volume... or when the object is full.
 
  • #16
RaulTheUCSCSlug said:
Okay, I know to use spherical coordinates. I just am not sure which integral to use. That is my confusion.
It wouldn't be any of the three you wrote in your relevant equations. As BvU wrote, the basic integral is ##\int \rho dV##, which in this problem turns out to be a triple integral in spherical coordinates. For the shape of the object and the density function, I think you might be able to get away with a double integral, although I haven't set up the problem.

In any case, the typical volume element would be a hemispherical shell, sort of like a layer in an onion. At each point in a given shell, all the points are the same distance from the origin, hence the density is the same for each point.
 
  • #17
so triple integral from the bounds being the same as if i was finding half a sphere (obviously) times the Jacobian integrated with the density function in terms of spherical coordinates. That would be this equations ∫∫FoΦ(u,v)||Tu,Tv||dudv as this part ∫∫FoΦ(u,v)(((((||Tu,Tv||)))))dudv is the Jacobian.

So I'm not sure why you are all saying that the equations are wrong or unfamiliar? But if this is the case, which I do not think, where do we account for the change of radius from 0 to 2. We wouldn't account for it at all.
 
  • #18
There is no Jacobian. And why don't you start writing out ##dV## and ##\rho## ( your ##\delta## ) in spherical coordinates ? Who knows, the angular part can be integrated easily and you are left with one simple integral in ##r##...
 
  • #19
BvU said:
There is no Jacobian. And why don't you start writing out ##dV## and ##\rho## ( your ##\delta## ) in spherical coordinates ? Who knows, the angular part can be integrated easily and you are left with one simple integral in ##r##...

So just take the volume integral with the density formula and that will account for the change in density. But why is there no jacobian if we are changing coordinate systems we must account for this somehow...
 
  • #20
RaulTheUCSCSlug said:
So just take the volume integral with the density formula and that will account for the change in density. But why is there no jacobian if we are changing coordinate systems we must account for this somehow...
You aren't really changing coordinate systems, since you aren't starting from an iterated integral ##\int \int \int f(x, y, z) dx~dy~dz##, and then changing to a different integral with a different volume element.
 
  • #21
RaulTheUCSCSlug said:
What I did in the previous problem was parameterize the surface, but that was asking for mass of shell, now it is mass of ball. So am I supposed to use volume then divide by the density? ? ?
What's the formula for density?
 
  • #22
RaulTheUCSCSlug said:
How does this question differ from when they ask for the mass of a spherical shell. Because I can do the shell fine, but I just don't know how it jumps to volume... or when the object is full.
The density formula you are given is mass per unit volume. For a hemispherical shell whose volume is ##\Delta V##, the mass of that volume element is ##\Delta m = \rho \Delta V##.
 
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  • #23
RaulTheUCSCSlug said:
so triple integral from the bounds being the same as if i was finding half a sphere (obviously) times the Jacobian integrated with the density function in terms of spherical coordinates. That would be this equations ∫∫FoΦ(u,v)||Tu,Tv||dudv as this part ∫∫FoΦ(u,v)(((((||Tu,Tv||)))))dudv is the Jacobian.

So I'm not sure why you are all saying that the equations are wrong or unfamiliar? But if this is the case, which I do not think, where do we account for the change of radius from 0 to 2. We wouldn't account for it at all.
Because you keep referring to a two-dimensional integral whereas this is a three-dimensional problem.

The volume element in spherical coordinates is ##dV = \rho^2 \sin\theta\,dr\,d\phi\,d\theta##. If you're using spherical coordinates, you usually start here. You don't bother rederiving the factor of ##\rho^2 \sin\theta##, which comes from the Jacobian when you switch from cartesian to spherical coordinates.
 
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  • #24
vela said:
Because you keep referring to a two-dimensional integral whereas this is a three-dimensional problem.

The volume element in spherical coordinates is ##dV = \rho^2 \sin\theta\,dr\,d\phi\,d\theta##. If you're using spherical coordinates, you usually start here. You don't bother rederiving the factor of ##\rho^2 \sin\theta##, which comes from the Jacobian when you switch from cartesian to spherical coordinates.
Oh okay, this clarified it for me, redid the problem and got it correctly, thank you! :woot:
 

1. What is the formula for calculating the mass of a half ball?

The formula for calculating the mass of a half ball with given density is: m = (4/3)πr3δ, where m is the mass, r is the radius of the half ball, and δ is the density at a given point (x, y, z).

2. How do you determine the radius of the half ball?

The radius of the half ball can be determined by measuring the distance from the center of the half ball to the edge or by dividing the diameter by 2.

3. What is the unit of measurement for density?

The unit of measurement for density can vary depending on the system of measurement being used. In the SI system, the unit of measurement for density is kilograms per cubic meter (kg/m3), while in the imperial system, it is pounds per cubic foot (lb/ft3).

4. Can the density vary within the half ball?

Yes, the density can vary within the half ball. The formula for calculating the mass takes this into account by using the density at a given point (x, y, z).

5. How do you calculate the mass of a half ball with non-uniform density?

If the density is not uniform within the half ball, the mass can still be calculated by breaking the half ball into smaller sections and using the formula m = (4/3)πr3δ for each section. The total mass can then be determined by summing up the masses of each section.

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