Vector calculus fundamental theorem corollaries

[Adam Lewis]

Homework Statement

Prove

\int_{V}\nabla\ T d\tau\ = \oint_{S}Td\vec{a}

Homework Equations

Divergence theorem:
\int_{V}(\nabla\bullet\vec{A})d\tau\ = \oint_{S}\vec{A}\bullet\ d\vec{a}

The Attempt at a Solution

By using the divergence theorem with the product rule for divergences and setting \vec{A}\ = T\vec{c} where c is a constant vector, I've got it down to

\int_{V}\vec{c}\bullet\nabla\ Td \tau\ = \oint_{S}T\vec{c}\bullet\ d\vec{a}

Which is exactly what we're looking for except for that annoying c. You can work out the dot products in the integrals and cancel off the components of c, but this kills the vector nature of the expression. We need the gradient *vector* and the surface element *vector* to be in there. How do I get rid of c without turning everything scalar? Alternatively, how do I restore the expressions to vector-hood after getting rid of c?

 

[gabbagabbahey]

The Attempt at a Solution

By using the divergence theorem with the product rule for divergences and setting \vec{A}\ = T\vec{c} where c is a constant vector, I've got it down to

\int_{V}\vec{c}\bullet\nabla\ Td \tau\ = \oint_{S}T\vec{c}\bullet\ d\vec{a}

Which is exactly what we're looking for except for that annoying c. You can work out the dot products in the integrals and cancel off the components of c, but this kills the vector nature of the expression. We need the gradient *vector* and the surface element *vector* to be in there. How do I get rid of c without turning everything scalar? Alternatively, how do I restore the expressions to vector-hood after getting rid of c?

First, since T is a scalar function, T\vec{c} \cdot d\vec{a}=\vec{c} \cdot Td\vec{a}, from there, just use the fact that if \vec{c} is a constant vector it is uniform over space, and hence it is treated as a constant for the spatial integrations:

\int_{V}\vec{c}\cdot\vec{v}d \tau=\vec{c}\cdot\int_{V}\vec{v}d \tau

(for any vector \vec{v} ) and

\oint_{S}\vec{c}\cdot\ Td\vec{a}=\vec{c}\cdot\oint_{S} Td\vec{a}

If you haven't already proven these assertions in any of your calculus courses, it is a fairly straight forward process of writing the vectors in Cartesian coordinates (Since Cartesian unit vector are position independent, they can be pulled out of the integrals) and breaking up the integral into three pieces, and pulling out the constants c_x, c_y and c_z.

 

[Adam Lewis]

Hi again,
Thanks a bundle. I guess the dot when c is pulled out represents scalar multiplication? The integral without c is of course a scalar, which was confusing me.
Thanks!

 

[gabbagabbahey]

No, the 'dot' is still a vector dot product, the integral of a vector, is a vector.