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My lecture notes look to prove Stokes' theorem for the special case where a surface can be represented as the graph of some function, so $z=f(x,y)$.

The surface $S$ is parametrized as $r(x,y)=(x,y,f(x,y))$, where $(x,y)$ is in the region $U$ in the $xy$ plane. Now assume $U$ has a boundary curve $C_u$ and $S$ has a boundary curve $C_s$.

My first question is where it is said that the line integral of the vector field $v=(v_1,v_2,v_3)$ over the curve $C_s$ is equal to the line integral of the same vector field over the curve $C_u$. Why is this the case?

Lecture notes for my first question

My second question is I believe about a total derivative but I'm not sure. My lecturer has written that since $z=f(x,y)$, $dz=(f_x)dx+(f_y)dy$ where $f_x$ and $f_y$ denote the partial derivatives of $f$ with respect to $x$ and $y$ respectively and $dx, dy, dz$ denote normal differentials, not partial ones. Can someone also explain this equality to me, please?

Lecture notes for my second question

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    $\begingroup$ @anomaly: thank you for correcting the spelling of Stokes $\endgroup$ Mar 20, 2020 at 16:53

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The boundary $\partial U$ of the set $U$ in the $(x,y)$-plane has some parametrization $$\partial U:\quad t\mapsto\bigl(x(t),y(t)\bigr)\qquad(0\leq t\leq 1)\ ,$$ and the boundary $\partial S$ of your surface $S\subset{\mathbb R}^3$ then has the parametrization $$\partial S:\quad t\mapsto\bigl(x(t),y(t),f\bigl(x(t),y(t)\bigr)\bigr)\qquad(0\leq t\leq 1)\ .$$ Expanding the integrals $$\int_{\partial S}{\bf v}\cdot d{\bf r},\qquad \int_{\partial U}\bigl(v_1 dx+v_2 dy+v_3 dz\bigr)$$ with $$dx=x'\>dt,\quad dy=y'(t)dt,\quad dz=f_x\bigl(x(t),y(t)\bigr)\>x'(t)dt+f_y\bigl(x(t),y(t)\bigr)\> y'(t) dt$$ then shows that they have the same value.

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