Why is $dA$in Green’s Theorem replaced by $dS$in Stokes’ Theorem?

$dA$in Green’s Theorem replaced by $dS$in Stokes’ Theorem.

Green's Theorem states that;
"Let $R$be a region in the plane with smooth boundary curve $C$oriented counterclockwise by $\mathbf{r}\left(t\right)=⟨\left(x\right(t),y(t\left)\right)⟩$for$a\le t\le b$.
If a vector field is defined by $R$, then, ${\int }_C\mathbf{F}·d\mathbf{r}={\iint }_R\left(\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}\right)dA."$
Stokes' Theorem states that;
"Let be $S$an oriented, smooth or piecewise-smooth surface bounded by a curve $C$. Suppose that $\mathit{n}$is an oriented unit normal vector of $S$and $C$has a parametrization that traverses $C$in the counterclockwise direction with respect to$\mathit{n}$.
If a vector field$\mathbf{F}(x,y,z)=F_1(x,y,z)\mathbf{i}+F_2(x,y,z)\mathbf{j}+F_3(x,y,z)\mathbf{k}$is defined on$S$, then,${\int }_C\mathbf{F}(x,y,z)·d\mathbf{r}={\iint }_{S}curl\mathbf{F}(x,y,z)·\mathbf{n}dS"$.

Both the notations $dA$and$dS$ refer to area.
In the case of Green's Theorem,$dA$ refers to the ordinary area.
In the case of Stokes' Theorem, $dS$refers to the surface area. In this case, $dS$keeps track of any distortion to the area of the surface $S$caused by transforming a patch of the $uv-$plane to parametrize$S$.
Hence,$dA$in Green's Theorem can be replaced by$dS$in Stokes' Theorem.