Volume average of mass function

[Apashanka]

Homework Statement

Suppose mass function in 3-D is given as m(x,y,z) and mass density as $$\rho(x,y,z)$$

Relevant Equations

$<m>=\frac{\int_vm(x,y,z)dv}{\int v} $

I want to express <m(x,y,z)> over a sphere of radius R in terms of $$<\rho(x,y,z)>$$
e.g $$<m>=\frac{\int_{sphere R}m(x,y,z)dv}{\int_{sphere}dv}$$
$$<m>=\frac{\int_{sphereR}(\int \rho(x,y,z)dv)dv}{\int_{sphere R}dv}$$

 

[archaic]

$\rho(x,\,y,\,z)=\dfrac{dm}{dv}\Leftrightarrow m=?$
By the way, I do not think that there is a "mass function". When you will carry out that integration, what you will get is a constant, not a function of position (correct me if I'm wrong).

 

[Apashanka]

$\rho(x,\,y,\,z)=\dfrac{dm}{dv}\Leftrightarrow m=?$
By the way, I do not think that there is a "mass function". When you will carry out that integration, what you will get is a constant, not a function of position (correct me if I'm wrong).

$$dm=\rho(x,y,z) dv$$
$$m(x,y,z)=\int \rho(x,y,z)dv$$

 

[archaic]

$$dm=\rho(x,y,z) dv$$
$$m(x,y,z)=\int \rho(x,y,z)dv$$

https://en.wikipedia.org/wiki/Density#Heterogeneous_materials

 

[Cutter Ketch]

$$dm=\rho(x,y,z) dv$$
$$m(x,y,z)=\int \rho(x,y,z)dv$$

If you integrate the right side over x, y, and z, why is the left side still a function of x, y, and z? Not that it can’t be, but you will have to explain what that means to you.

For example, say I had a function f(x) and I asked you for the area under the curve. You might write:

A = $\int_a^b {f(x)} \ dx$

You see that the area is not a function of x.

The integration limits a and b define a domain along the x axis. You can make A a function of x by defining a and b as a functions of x. For example if f(x) is a probability distribution, you might want to know the cumulative probability and integrate from -$\infty$ to x.
In the case of mass and density, density is already the variation of mass over x y and z. When you integrate density you get the total mass m, not m(x,y,z). I don’t know of a generic “mass function” that would be a function of x, y, and z (other than the density itself). You could certainly define such a thing (like in spherical coordinates people sometimes ask how much mass is inside radius r of a sphere). But in general I really don’t know what you could mean by a “mass function” m(x, y, z)

 

[Apashanka]

If you integrate the right side over x, y, and z, why is the left side still a function of x, y, and z? Not that it can’t be, but you will have to explain what that means to you.

The left side is an indefinite integral (without limits) ,of course there will be some integration constt.
For exm $$\frac{df}{dx}=x$$
$$f=\int x dx+c$$
$$f(x)=\frac{x^2}{2}+c$$

 

[Apashanka]

I really don’t know what you could mean by a “mass function” m(x, y, z)

As if m(x,y,z) defines mass at a coordinate x,y,z.

 

[Apashanka]

The left side

**right side

 

[Chestermiller]

As if m(x,y,z) defines mass at a coordinate x,y,z.

What the heck is that supposed to mean?

 

[Apashanka]

What the heck is that supposed to mean?

Say point mass defined at every coordinate given by m(x,y,z)

 

[Chestermiller]

Say point mass defined at every coordinate given by m(x,y,z)

That would mean that the density at each point is infinite.

 

[Apashanka]

That would mean that the density at each point is infinite.

Would you please explain ...
mass at a distance x,y,z is m(x,y,z) within volume dxdydz ,then density will be $$\rho(x,y,z)=m(x,y,z)/dxdydz$$

 

[Apashanka]

Would you please explain ...
mass at a distance x,y,z is m(x,y,z) within volume dxdydz ,then density will be $$\rho(x,y,z)=m(x,y,z)/dxdydz$$

if dx,dy,dz negligibly small tends to 0

 

[Chestermiller]

if dx,dy,dz negligibly small tends to 0

So, if m is finite and dxdydz goes to zero...

 

[haruspex]

The left side is an indefinite integral (without limits) ,of course there will be some integration constt.
For exm $$\frac{df}{dx}=x$$
$$f=\int x dx+c$$
$$f(x)=\frac{x^2}{2}+c$$

The shorthand involved in those algebraic statements hides what is really going on. Not all of the x occurrences there mean the same thing.
Filling in some details and removing some puns:
$\frac{df}{dx}=x$ means $\frac{df(t)}{dt}\Bigr|_{t=x}=x$
We can swap x and t to arrive at:
$$\frac{df(x)}{dx}\Bigr|_{x=t}=t$$
Which leads to
$$f(x)=\int^{t=x} t dt$$
And integrate to get
$$f(x)=\frac{x^2}{2}+c$$
See section 3 of https://www.physicsforums.com/insights/conceptual-difficulties-roles-variables/.
If you apply the same formal process to your volume integral you will find that x, y, and z do not make it through the integration.