I am currently in Calculus 3, or Multivariable Calculus and need to prove this special case of Stokes' theorem. Please forgive me as I do need this simplified to the bones to understand the explanations.
This version is below.
$$ \int_{\partial S}\mathbf{F}(x,y,z)\cdot d \mathbf{r} = \iint_S(\nabla\times\mathbf{F})\cdot \mathbf{n} dS $$
The proof starts with the conditions of
$ S= \{ (x,y,z)\vert z=f(x,y),(x,y)\in R \} $
where R is the region in the
$ xy $
-plane with piecewise-smooth boundary
$ \partial R $
, where
$ f(x,y) $
has continuous first partial derivatives and for which
$ \partial R $
is the projection of the boundary
$ \partial S $
of the surface S onto the
$ xy $
-plane.
The first step called for the curl of F where
$ F(x,y,z) = \langle M(x,y,z),N(x,y,z),P(x,y,z) \rangle $
which I found.
$$ curl F = \nabla\times\mathbf{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ M(x,y,z) & N(x,y,z) & P(x,y,z) \\ \end{vmatrix} = (\frac{\partial P}{\partial y} -\frac{\partial N}{\partial z})\hat{i} + (\frac{\partial M}{\partial z} -\frac{\partial P}{\partial x})\hat{j} + (\frac{\partial N}{\partial x} -\frac{\partial M}{\partial z})\hat{k} $$
Of course, we're less than halfway done with the steps.
The second step had the condition where
$ G(x,y,z) = z – f(x,y) $
and called for the exterior unit normal vector
$ \frac{\nabla G}{\vert \vert \nabla G \vert \vert} $
to any point on the surface S. Now this might be a great jump like a joke flying above my head but for some reason I keep on thinking this leads to what is seen below.
$$ n = \frac{\nabla G}{\vert \vert \nabla G \vert \vert} = \frac{\langle 0,0,0 \rangle}{\sqrt{0^2+0^2+0^2}} = undefined $$
This is because one of the initial conditions is
$ z=f(x,y) $
so I believe they cancel and I know this should not be the case because this would nullify the entire proof (unless I'm mistaken). I think this is a major oversight and yet I can't figure out why. If anybody could help fix this misconception, I would appreciate it. And I also have no idea as to why a separate function
$ G(x,y,z) $
is necessary in order to prove this theorem.
If anybody has extra time to aid me in solving the rest, I will list the next steps.
The third step asks to express
$ \int_{\partial S}\mathbf{F}(x,y,z)\cdot d \mathbf{r} = \iint_S(\nabla\times\mathbf{F})\cdot \mathbf{n} dS $
in terms of M, N, and P with a hint that
$ dS = \vert \vert \frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v} \vert \vert dA $
where
$ \vert \vert \frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v} \vert \vert = \sqrt{ (\frac{\partial z}{\partial x})^2 + \frac{\partial z}{\partial y})^2 + 1} $
Having not done this yet, I believe the left side of the equation could be rewritten using the condition in the first step of the proof where
$ F(x,y,z) = \langle M(x,y,z),N(x,y,z),P(x,y,z) \rangle $
so that
$ \int_{\partial S}\mathbf{F}(x,y,z)\cdot d \mathbf{r} = \int_{\partial S} M(x,y,z)\hat{i} + N(x,y,z)\hat{j} + P(x,y,z)\hat{k}\cdot d \mathbf{r} $
which I do not believe can be simplified (correct me if I'm wrong). As for the right side of the equation, I simply do not remember how to manipulate it to be in terms of M, N, and P but I do believe the second step and finding the exterior unit normal vector
$ n $
is quite important.
The fourth step expects us to show that
$ \int_{\partial S} M(x,y,z)dx = – \iint_R(\frac{\partial M}{\partial y} + \frac{\partial M}{\partial z}f_y) _{z=f(x,y)}dA $
,
$ \int_{\partial S} N(x,y,z)dy = \iint_R(\frac{\partial N}{\partial x} + \frac{\partial N}{\partial z}f_x) _{z=f(x,y)}dA $
, and
$ \int_{\partial S} P(x,y,z)dz = \iint_R(\frac{\partial P}{\partial x}f_y + \frac{\partial P}{\partial y}f_x) _{z=f(x,y)}dA $
. This comes with a hint to let the boundary of R be described parametrically by
$ \partial R = \{ (x,y)\vert x=x(t),y=y(t),a \le t \le b \} $
which implies that the boundary of S is described parametrically by
$ \partial R = \{ (x,y,z)\vert x=x(t),y=y(t),z=(x(t),y(t)),a \le t \le b \} $
. Use Green's Theorem and the Chain Theorem to prove the given equations.
The fifth step (also the last) asks us to explain how the results prove Stokes' Theorem.
As I said, I am not that fluent in the language of math and hope that you are able to break it down for me if possible. Thank you and I hope you are doing well!
Best Answer
$G(x,y,z)=z-f(x,y)=0$ on the surface (indeed, this is the definition of the surface - the set of points (x,y,z) on which $G$ vanishes), but is positive above it and negative below it, meaning that $\nabla G$ points perpendicularly away from the surface in the direction of increasing $z$. Explicitly,
$$\nabla G = \left\langle -\frac{\partial f}{\partial x},-\frac{\partial f}{\partial y},1\right\rangle$$ $$\Vert \nabla G\Vert = \sqrt{1 + \left(\frac{\partial f}{\partial x}\right)^2 + \left(\frac{\partial f}{\partial y}\right)^2}$$
and the unit vector pointing away from the surface is given by
$$\hat n = \frac{\nabla G}{\Vert \nabla G \Vert}$$