Stoke's Theorem Under Transformation of Coordinates

Stoke's Theorem states that for a vector field  
  with continuous partial derivatives defined on a surface  
  with a boundary curve  
\[\oint_C \mathbf{F} \cdot d \mathbf{r} = \int \int_S (\mathbf{\nabla} \times \mathbf{F}) \cdot \mathbf{n} dS\]
Suppose we can write  
\[x=x(u,v), \; y=y(u,v), \; z=z(u,v)\]
  representing a change of coordinates. We can write  
A unit normal to  
\[\mathbf{n}= \frac { \frac{\partial \mathbf{F}}{\partial u} \times \frac{\partial \mathbf{F}}{\partial v} }{ \left| \frac{\partial \mathbf{F}}{\partial u} \times \frac{\partial \mathbf{F}}{\partial v} \right| } ,\; \left| \frac{\partial \mathbf{F}}{\partial u} \times \frac{\partial \mathbf{F}}{\partial v} \right| \neq 0. \]

and an element of surface area is  
\[\mathbf{n} dS =( \frac{\partial \mathbf{F}}{\partial u} \times \frac{\partial \mathbf{F}}{\partial v}) du dv \]

The right hand side of Stoke's equation becomes
\[ \int \int_S (\mathbf{\nabla} \times \mathbf{F}) \cdot \mathbf{n} dS = \int \int_{S'} (\mathbf{\nabla} \times \mathbf{F}) \cdot ( \frac{\partial \mathbf{F}}{\partial u} \times \frac{\partial \mathbf{F}}{\partial v}) du dv \]
  is the image of  
. ON the left hand side  
\[d \mathbf{r} = \frac{\partial \mathbf{F}}{\partial u}du + \frac{\partial \mathbf{F}}{\partial v}dv \]

The left hand side becomes  
\[\oint_{C'} \mathbf{F} \cdot (\frac{\partial \mathbf{F}}{\partial u}du + \frac{\partial \mathbf{F}}{\partial v}dv)\]

Stoke's Theorem is now  
\[\oint_{C'} \mathbf{F} \cdot (\frac{\partial \mathbf{F}}{\partial u}du + \frac{\partial \mathbf{F}}{\partial v}dv=\int \int_{S'} (\mathbf{\nabla} \times \mathbf{F}) \cdot ( \frac{\partial \mathbf{F}}{\partial u} \times \frac{\partial \mathbf{F}}{\partial v}) du dv \]

Add comment

Security code