$\int \nabla\times\vec{F}\cdot{\hat{n}}ds=\iint(-\frac{\partial z}{\partial x}(\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z})-\frac{\partial z}{\partial y}(\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x})+(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}))dxdy$
$\nabla\times\vec{F}=<\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z},\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x},\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}>$
${\hat{n}}ds=<-\frac{\partial z}{\partial x},-\frac{\partial z}{\partial y},1>dxdy$ I parameterize the surface $Z(x,y)=x+y$,$\vec{r}(x,y)=x\hat{i}+y\hat{y}+z(x,y)\hat{k}$,$d{s}=(\frac{\partial \vec{r}}{\partial x}\times\frac{\partial \vec{r}}{\partial y})dxdy$
${\hat{n}}ds=\frac{<-\frac{\partial z}{\partial x},-\frac{\partial z}{\partial y},1>}{\sqrt{(-\frac{\partial z}{\partial x})^2+(-\frac{\partial z}{\partial y})^2+1}}*\sqrt{(-\frac{\partial z}{\partial x})^2+(-\frac{\partial z}{\partial y})^2+1}dxdy=<-\frac{\partial z}{\partial x},-\frac{\partial z}{\partial y},1>dxdy$
I want to know how to convert to $\iint(\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z})dydz+(\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x})dzdx+(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})dxdy$
Can I make the following comparison？ $-\frac{\partial z}{\partial x}dxdy=dydz,-\frac{\partial z}{\partial y}dxdy=dzdx$
I don’t know if this is correct,I don’t see any intuition,I can’t think of how to connect.
Can someone correct my errors and explain if there is a connection between them？
Thank you very much for your help
I think you should look up exterior calculus.
$dxdy$ is actually $dx\wedge dy$. It represents infinitesimal (2-d) parallelogram patches which the integral evaluates the integrand on, much like $dx$ in single variable calculus are 1-dimensional "steps" in the $X$ direction.
When the parallelogram is formed by taking 1 step in $X$ and 1 step in $Y$, its signed area is $dx\wedge dy$. This area value is the same as the value of $\overrightarrow{x}\times\overrightarrow{y}$ being $|\overrightarrow{x}||\overrightarrow{y}|\sin\theta$ with the signed measurement of $\theta$ according to the right-hand-rule. So it is anticommutative just like the cross product.
The corresponding patch $dz\wedge dy$ is thus $\frac{\partial z}{\partial x}dx\wedge dy$.