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!


1 Answer 1


This is because one of the initial conditions is $z=f(x,y)$ so I believe they cancel [...]

$G(x,y,z)=z-f(x,y)=0$ 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

$$\hat n = \frac{\nabla G}{\Vert \nabla G \Vert}$$


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