Proving the relation using multivariable calculus

[vela]

If only uv disappears, we have proved the relation. How will doing it for each component change anything? Have you figured that part out? I am stumped

For right now, let's just assume the uv term vanishes. I don't see how you can conclude you've proved the relation, especially in light of your questions.

How will the dy and dz terms affect this? Wouldnt they just turn into yz?

No because the integral of $f(\vec{r})\partial_x A_x$ is generally still a function of $y$ and $z$.

It seems to me that you're reluctant to do any calculations because you can't see yet how it's going to work out. Sometimes you just have to try stuff and then it becomes clear. So deal with the other terms and see what you get.

 

[jhartc90]

For right now, let's just assume the uv term vanishes. I don't see how you can conclude you've proved the relation, especially in light of your questions.No because the integral of $f(\vec{r})\partial_x A_x$ is generally still a function of $y$ and $z$.

It seems to me that you're reluctant to do any calculations because you can't see yet how it's going to work out. Sometimes you just have to try stuff and then it becomes clear. So deal with the other terms and see what you get.

You are right, I am reluctant because I do not see how its going to work out. The issue I see is that We have a triple integral here. The first one we are integrating wrt x. From there, we get the uv - int(vdu) term.

Then we have two outside integrals. I am lost on how this will affect the inside integral

 

[vela]

You're not doing any integration. You still have integrals with respect to each coordinate. Note that that's what you have on the righthand side as well.

 

[jhartc90]

In the integral you wrote, we would get:

$$\int\int((f(x,y,z)A_x-\int(A_x*\frac{df}{dx}*dx)dydz)$$

Is this the final answer for the x-component, and we just repeat with y and z? And then add them all together?

 

[Orodruin]

Yes. Note that we are assuming boundary conditions such that the first term (the one with only two integrations) vanishes.

 

[jhartc90]

What sort of BCs? Just as r->infinity, f(r)A(r) goes to 0?

 

[Orodruin]

Yes. That would be sufficient.