## Stoke's Theorem Under Transformation of Coordinates

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

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
$\int \int_S (\mathbf{\nabla} \times \mathbf{F}) \cdot \mathbf{n} dS = \int \int_{S'} (\mathbf{\nabla} \times \mathbf{F}) \cdot ( \frac{\partial \mathbf{F}}{\partial u} \times \frac{\partial \mathbf{F}}{\partial v}) du dv$
.
where
$S'$
is the image of
$S$
. 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
$\oint_{C'} \mathbf{F} \cdot (\frac{\partial \mathbf{F}}{\partial u}du + \frac{\partial \mathbf{F}}{\partial v}dv=\int \int_{S'} (\mathbf{\nabla} \times \mathbf{F}) \cdot ( \frac{\partial \mathbf{F}}{\partial u} \times \frac{\partial \mathbf{F}}{\partial v}) du dv$