Is this an acceptable proof for Stokes' theorem? I'm currently creating my own Multivariable Calculus course and this is what I have for the proof.$\newcommand{\curl}{\operatorname{curl}}$
Also, I copied this from my TeX file so apologies if some of the notation doesn't work, and equation numbers don't work as well. Firstly, assume that $\vec{F}$ only has a vertical component, hence $$\vec{F}(x,y,z)=P\vec{k}$$ where $P$ is a function. We want to prove that $$\oint_C P(x,y,z)\: dz=\iint_S \curl (P\vec{k})\cdot \vec{\hat{n}}\: dS \tag{1}\label{std}$$ Assume that the surface $S$ is given by the equation $z=f(x,y)$ over a closed region $\mathcal{R}$ in the $xy$ axis. Let $C$ be the boundary curve for $R$ and let us define a new curve $C'$ such that $C'$ is the $\textbf{boundary}$ of $\mathcal{R}$. Moreover, assume that the unit normal vector $\vec{\hat{n}}$ is pointing in the positive $z$-direction.
To verify Eq. \eqref{std}, we'll have to change the left hand side into a line integral along $C'$. Likewise, we'll also have to change the right hand side to a double integral around $\mathcal{R}$. After doing both of these steps, we'll finally use Green's theorem to show that both sides are equal!
By parametrizing both of the curves, we can compute the required line integrals, so let $$C' \: : \: \quad x=x(t), \qquad y=y(t)$$ and $$C' \: : \: \quad x=x(t), \qquad y=y(t), \qquad z=f(x(t),y(t))$$ both for $t_1\leq t \leq t_2$. It's particularly important to note that \textbf{$C$ lies on $S$}. The next line can get a bit fiddly, but all it essentially is, is a total differential, so $$ \oint_C P\: dz=\oint_{C'} P(x,y,f(x,y))\underbrace{\left(\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy\right)}_{=\:dz} \tag{2}\label{li2}$$ By using the parametrizations, we can verify that both line integrals are indeed equal. So, \begin{align*} \oint_C P\: dz&=\int^{t_2}_{t_1} P(x(t),y(t),z(t))\frac{dz}{dt}\: dt \\ &=\int^{t_2}_{t_1} P(x(t),y(t),z(t))\left(\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy\right)\: dt \qquad \text{(by the chain rule)}\\ &=\oint_{C'} P(x,y,f(x,y))\left(\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy\right) \end{align*} This verifies Eq. \eqref{li2}. To make the notation easier, let's define $\displaystyle A_{1}(x,y,f)=\dfrac{\partial P}{\partial x}$ - note that $A_{1}(x,y,f)\neq A_{x}(x,y,z)$, for $A_{1}(x,y,f)$, we calculate the first partial derivative after substituting $z=f(x,y)$. Also, $$\frac{\partial P}{\partial x}=A_1(x,y,f)+A_3(x,y,f)$$ Now, computing the surface integrals is literally just bookwork! Recall that \begin{equation} d\vec{S}=\left\langle -\frac{\partial f}{\partial x},-\frac{\partial f}{\partial y},1\right\rangle dA \end{equation} Computing the curl, we get \begin{align*} \curl (P\vec{k})=\begin{vmatrix} \vec{i}&\vec{j}&\vec{k} \\ \partial_x & \partial_y & \partial_z \\ 0&0&P \end{vmatrix}=A_2(x,y,z)\vec{i}-A_1(x,y,z)\vec{j} \end{align*} By using this, \begin{align} \iint_S \curl (P\vec{k})\cdot d\vec{S}&=\iint_S \left(-A_2(x,y,z)\frac{\partial f}{\partial x}+A_1(x,y,z)\frac{\partial f}{\partial y}\right)\: dxdy \nonumber \\ &=\iint_R \left(-A_2(x,y,z)\frac{\partial f}{\partial x}+A_1(x,y,z)\frac{\partial f}{\partial y}\right)\: dxdy \label{killmepls} \end{align} Recall that Green's theorem states \begin{equation} \boxed{ \oint_{C'} M\: dx+N\: dy=\iint_R \left( \frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}\right)\: dxdy } \end{equation} Define $M$ and $N$ such that \begin{align*} M=P(x,y,f(x,y))\: \frac{\partial f}{\partial y} \qquad \text{and} \qquad N=P(x,y,f(x,y))\: \frac{\partial f}{\partial x} \end{align*} Hence, computing the first order partial derivatives, we get \begin{align*} M_x&=\frac{\partial M}{\partial x}=\left(A_1+A_3 \frac{\partial f}{\partial x}\right)\frac{\partial f}{\partial y}+P(x,y,f)\frac{\partial^2 f}{\partial y\partial x} \\ N_y&=\frac{\partial N}{\partial y}=\left(A_1+A_3 \frac{\partial f}{\partial y}\right)\frac{\partial f}{\partial x}+P(x,y,f)\frac{\partial^2 f}{\partial x \partial y} \end{align*} Since $\displaystyle \frac{\partial^2 f}{\partial x \partial y}=\frac{\partial^2 f}{\partial y \partial x}$, several terms will cancel out, giving us \begin{align*} N_x-M_y=P_1(x,y,f)\frac{\partial f}{\partial y}-P_2(x,y,f)\frac{\partial f}{\partial x} \end{align*} This verifies Stokes' theorem for $\vec{F}(x,y,z)=P\vec{k}$.
To verify Stokes' theorem for a general vector field $\vec{F}(x,y,z)=A\vec{i}+B\vec{j}+C\vec{k}$, repeat the procedure above for $\vec{F}(x,y,z)= A\vec{i}$, then for $\vec{F}(x,y,z)= B\vec{j}$ - finally, add them all up!