Divergence of a Curl - Then Integrate By Parts

[Chingon]

Homework Statement

∫Bdot[∇×A]dV=∫Adot[∇×B]dV

Prove this by integration by parts. A(r) and B(r) vanish at infinity.

Homework Equations

I'm getting stuck while trying to integrate by parts - I end up with partial derivatives and dV, which is dxdydz?

The Attempt at a Solution

I can break things down to Cartesian components, but integrating by parts is where I get stuck.

Essentially, I'm simplifying by stating the identity that
BdotCurlA - AdotCurlB = Div(A×B) = 0 in this case (Subtract right hand side from left and combine under one ∫dV)

The components look like this:
∂x(AyBz−AzBy)=(∂xAy)Bz+Ay(∂xBz)−(∂xAz)By−Az(∂xBy)
plus the y,z terms as well.

How would one integrate ∫[(∂xAy)Bz]dV by parts? I have Ay and Bz which are potentially functions of x,y,z and a partial wrt x, and dxdydz...

Thanks!

 

[dextercioby]

Do you know cartesian tensors ? If so, the problem is trivial.

\iiint A_k \epsilon_{klm} \partial_{l}B_{m} dV = ...

 

[lurflurf]

Integrations by parts are equivalent to product rules.
use
\nabla \cdot (\mathbf{A} \times \mathbf{B}) = \mathbf{B} \cdot (\nabla \times \mathbf{A}) - \mathbf{A} \cdot (\nabla \times \mathbf{B})
and the divergence theorem to reduces the problem to showing a surface integral at infinity is zero.

 

[Chingon]

Do you know cartesian tensors ? If so, the problem is trivial.

\iiint A_k \epsilon_{klm} \partial_{l}B_{m} dV = ...

No, we haven't studied those so I don't believe that's a path I can take.

Integrations by parts are equivalent to product rules.
use
\nabla \cdot (\mathbf{A} \times \mathbf{B}) = \mathbf{B} \cdot (\nabla \times \mathbf{A}) - \mathbf{A} \cdot (\nabla \times \mathbf{B})
and the divergence theorem to reduces the problem to showing a surface integral at infinity is zero.

I have used that identity to reduce the problem to ∫Div(A×B)=0

This reduces to:
∫[∂_xA_y*B_z-∂_xA_z*B_y+∂_yA_z*B_y-∂_yA_x*B_Z+∂_ZA_x*B_y-∂_zA_y*B_x]dV

 

[Chingon]

So I guess I'm wondering how to handle the partial derivatives inside the integrand. Take the first term for example:
∫∂/∂_x(AyBz)dxdydz

Do the ∂_x and dx cancel out?

 

[dextercioby]

No, no, no. Do you know about the Gauß-Ostrogradskii theorem ? It turns a volume integral into a surface integral.

 

[Chingon]

Yes, but I was focused on the brute force of integration by parts method as opposed to using this theorem. I think I'm just going to go the easy route and use this method - hopefully I won't lose too many points!

 

[dextercioby]

You are doing vector integrals, you should use the vectors not their components, and you can use the Gauß-Ostrogradskii theorem in case of vectors.

<In vector calculus, the divergence theorem, also known as Gauss's theorem or Ostrogradsky's theorem,[1] is a result that relates the flow (that is, flux) of a vector field through a surface to the behavior of the vector field inside the surface.>

No reason to lose points.