Diffential elements and analysis dimensional

[Jhenrique]

The definition for volume element is simples, is $dV=dxdydz$, ok. But, if you integrate this you'll have problems, because $\int dV = \int dxdydx$ no make sense in the right side of equation and, on the other hand, $\iiint dV =\iiint dxdydx$ no make sense in the left side of equation... so, this problem is eliminated if you define the volume element like $d^3V = dxdydz$, now the tiple integral make sense: $$\iiint d^3V = \iiint dxdydz$$ However, to think if a quantity physical, in infinitesimal size, have simple, double, triple, ..., differential is non-intuitive, is a concept very analytical. But, ignore this information is metematically wrong.

So, which is correct form for deal with this?

 

[Mark44]

The definition for volume element is simples, is $dV=dxdydz$, ok. But, if you integrate this you'll have problems,

Not really, if you work with definite integrals. As a triple integral, the one on the left would be $\int_R dV$, where R is the three-dimensional region over which integration takes place. A triple integral can be rewritten as three iterated integrals, such as you show on the right, with suitable limits of integration.

I'm not sure that your integral with dV makes any sense as an indefinite integral. What would you have as the antiderivative?

because $\int dV = \int dxdydx$ no make sense in the right side of equation and, on the other hand, $\iiint dV =\iiint dxdydx$ no make sense in the left side of equation... so, this problem is eliminated if you define the volume element like $d^3V = dxdydz$, now the tiple integral make sense: $$\iiint d^3V = \iiint dxdydz$$ However, to think if a quantity physical, in infinitesimal size, have simple, double, triple, ..., differential is non-intuitive, is a concept very analytical. But, ignore this information is metematically wrong.

So, which is correct form for deal with this?

 

[Jhenrique]

I omitted the limits...

BTW, which is the correct definition for volume element, d³V or dV ?

 

[Mark44]

I omitted the limits...

BTW, which is the correct definition for volume element, d³V or dV ?

It's dV, a differential volume element.

 

[Jhenrique]

Not really, if you work with definite integrals. As a triple integral, the one on the left would be $\int_R dV$, where R is the three-dimensional region over which integration takes place. A triple integral can be rewritten as three iterated integrals, such as you show on the right, with suitable limits of integration.

I'm not sure that your integral with dV makes any sense as an indefinite integral. What would you have as the antiderivative?

Are you saing that $$\int\limits_{V} = \iiint\limits_{x\;y\;z}$$ ?

 

[Mark44]

Are you saing that $$\int\limits_{V} = \iiint\limits_{x\;y\;z}$$ ?

Yes, but the three iterated integrals on the right side don't have to be in that particular order. They can appear in any of six ways.

What I said was that a triple integral ($\int_R f(x, y, z) dV$) can be written as three iterated integrals - one possible order is $\int \int \int f(x, y, z) dx~dy~dz$.

 

[Jhenrique]

What I said was that a triple integral ($\int_R f(x, y, z) dV$) can be written as three iterated integrals - one possible order is $\int \int \int f(x, y, z) dx~dy~dz$.

If this notation proceeds: $$\iiint\limits_{x\;y\;z}f(x,y,z)dxdydz$$ (attention for the argument of the function f)

So this too proceeds: $$\int\limits_{V}f(V)dV$$ What is a function of a tridimensional argument?

 

[Mark44]

If this notation proceeds: $$\iiint\limits_{x\;y\;z}f(x,y,z)dxdydz$$ (attention for the argument of the function f)

Based on the order of the differentials, it would be written as
$$\iiint\limits_{z\;y\;x}f(x,y,z)dxdydz $$
The innermost integration is with respect to x, so the inner limits of integration are x values. Next, the integration is with respect to y. Finally, the outer integral is with respect to z, so the limits of integration are z values.

So this too proceeds: $$\int\limits_{V}f(V)dV$$

No, it doesn't. We started with a function of three variables. You can't just switch it be be a function of one variable.

For example, if f(x, y, z) = 2xy + e^3z, how does f(V) make any sense at all?

What is a function of a tridimensional argument?

See just above for an example of a function with three arguments.