Probably simple proof (Stoke's Theorem)

[theBTMANIAC]

Homework Statement

Prove that ∫u∇v dl = - ∫v∇u dl

Both integrals are over closed surfaces.

Homework Equations

The question is being asked in a chapter over Stoke's Theorem. However, I'm confused because I think found the solution without invoking the theorem... Which leads to...

The Attempt at a Solution

I used integration by parts to derive

∫u∇v dl = uv - ∫v∇u dl

However since it's over a closed surface, I believe (uv) goes to zero. Is this the correct proof? Thanks.

 

[mark726]

Thank you for your question. Your solution using integration by parts is a valid approach to proving this statement. However, there is another way to prove it using Stoke's Theorem, which may be more in line with the chapter you are currently studying.

First, let's define the closed surface as S and the vector field as F = u∇v. Using Stoke's Theorem, we can rewrite the left side of the equation as:

∫u∇v dl = ∫∫∇×u∇v dS

Next, using the vector identity ∇×(f∇g) = ∇f×∇g, we can simplify the above equation to:

∫u∇v dl = ∫∫∇u×∇v dS

Similarly, we can rewrite the right side of the equation as:

-∫v∇u dl = -∫∫∇×v∇u dS = -∫∫∇v×∇u dS

Since the cross product is anti-commutative, we can see that ∇u×∇v = -∇v×∇u. Therefore, the right side of the equation becomes:

-∫v∇u dl = ∫∫∇v×∇u dS

Combining this with the left side of the equation, we get:

∫u∇v dl = ∫∫∇u×∇v dS = -∫v∇u dl

This proves that the two integrals are equal. I hope this helps and clarifies the use of Stoke's Theorem in this proof. Keep up the good work in your studies!



Scientist