- #1
jhartc90
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1. The problem
I am trying to prove the following relation in cartesian coordinates. We were given a hint to use integration by parts, as well as the fact that we know $d \vec r = dx\,dy\,dz$ (volume integral).
$$\int f(\vec r)\ \nabla \cdot \vec A(\vec r) \, d \vec r = -\int \vec A(\vec r)\ \cdot \nabla f(\vec r)\,d\vec r$$
If I use integration by parts, we have:
$$uv - \int(v\ du)$$
Does anyone see the way in which to prove this relation?
I can start by saying:
$$u=f(r)$$
$$du = \nabla fdr$$
$$dv = \nabla \cdot A(r)dr$$
$$v = A(r)$$
Does this look correct?
I am not sure if I am doing integration by parts correctly, because I am not sure how to handle gradient/divergence in this manner. Is there anyone out there who can provide some insight?
m statement, all variables and given/known data
[/B]
I am trying to prove the following relation in cartesian coordinates. We were given a hint to use integration by parts, as well as the fact that we know $d \vec r = dx\,dy\,dz$ (volume integral).
$$\int f(\vec r)\ \nabla \cdot \vec A(\vec r) \, d \vec r = -\int \vec A(\vec r)\ \cdot \nabla f(\vec r)\,d\vec r$$
Homework Equations
The Attempt at a Solution
If I use integration by parts, we have:
$$uv - \int(v\ du)$$
Does anyone see the way in which to prove this relation?
I can start by saying:
$$u=f(r)$$
$$du = \nabla fdr$$
$$dv = \nabla \cdot A(r)dr$$
$$v = A(r)$$
Does this look correct?
I am not sure if I am doing integration by parts correctly, because I am not sure how to handle gradient/divergence in this manner. Is there anyone out there who can provide some insight?
m statement, all variables and given/known data
[/B]