How Can I Use Stokes' Theorem to Show Integral of fgrad(g)*dr=0?

You have to find a vector field F such that curl(F)= f grad(g).In summary, the conversation discusses using Stokes' Theorem to show that the integral of fgrad(g)*dr is equal to 0 when grad(f) is perpendicular to grad(g) x n everywhere on a simple parametrically defined surface with boundary C. The conversation also includes the use of vector identities to solve the problem.
  • #1
JaysFan31

Homework Statement


Let S be a simple parametrically defined surface with boundary C as in Stokes' Theorem. Let f and g be two continuously differentiable scalar fields defined on S. Let n be a choice of unit normal on S. If grad(f) is perpendicular to grad(g) x n everywhere on S, show that integral of fgrad(g)*dr=0.
Note: x is cross product and * is dot product


Homework Equations


Stokes' Theorem:
integral of F*dr=integral of curlF*ndS


The Attempt at a Solution


I'm pretty confused on this one.
First off, what does the perpendicular part mean? What does it tell me? Doesn't it just tell me that the cross product exists everywhere? How does this information help me solve the problem and show that the integration equals zero?
 
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  • #2
You do not know what "perpendicular" means? It means here exactly what it means in elementary geometry- that the two vectors are at right angles to one another. What you are really asking is "What does it tell me". No, it does not tell you that the cross product exists everywhere! As long as the two vectors exists, their cross product exists. What it tells you is much simpler than that: that the dot product of the two vectors is 0.
 
  • #3
To solve this (and the last post) all you need are some vector identities. Here they are at no extra charge. f is a scalar, a,b,c are vectors. (. is dot, * is scalar multiply).

curl(grad(f))=0
a.(bxc)=b.(cxa)=c.(axb)
curl(f*a)=grad(f)xa+f*curl(a)

I'm not charging for this service because I'm sure you already had them.
 
  • #4
Ok. So I get that
grad(f) . (grad(g) x n)=0 where x is cross product and . is dot product since the vectors are perpendicular.

Using the second identity shown,

n . (grad(f) x grad(g))=0.
This equals (grad(f) x grad(g)) . n=0.

How do I show however that (grad(f) x grad(g))=fgrad(g) to make it work?
 
  • #5
They AREN'T equal. Use one of your free identities to compute curl(f grad(g)). The point to Stokes is integrating curl(F) over the interior and integrating F over the boundary. curl(F) and F are TWO DIFFERENT FIELDS.
 

1. What is Stokes' Theorem?

Stokes' Theorem is a fundamental theorem in vector calculus that relates the surface integral of a vector field over a surface to the line integral of the same vector field around the boundary of the surface.

2. How is Stokes' Theorem used in mathematics?

Stokes' Theorem is used to evaluate integrals over surfaces in three-dimensional space. It is particularly useful in solving problems related to fluid flow, electromagnetism, and other physical phenomena.

3. What is the equation for Stokes' Theorem?

The equation for Stokes' Theorem is ∫∫S(∇×F)·dS = ∫CF·dr, where ∫∫S represents the surface integral, ∇×F is the curl of the vector field F, and ∫C represents the line integral along the boundary of the surface S.

4. How is Stokes' Theorem applied to show that ∫∫Sf∇g·dS = 0?

To show that ∫∫Sf∇g·dS = 0 using Stokes' Theorem, we first rewrite the equation as ∫Cf∇g·dr. Then, we use the fact that ∇×(f∇g) = 0 to simplify the equation. Finally, we apply Stokes' Theorem to show that the line integral around the boundary of the surface S is equal to 0, resulting in ∫Cf∇g·dr = 0 and proving the original equation.

5. What are the limitations of using Stokes' Theorem to evaluate integrals?

Stokes' Theorem can only be used to evaluate integrals over surfaces that are smooth and have a well-defined boundary. It also requires the vector field to be continuously differentiable over the surface. Additionally, it is only applicable in three-dimensional space and cannot be used for higher dimensions.

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