Solving $\oint_c \textbf{F}\centerdot d\textbf{r}$ using Stokes' Theorem

Question

It would be great if someone can help me with this problem:

$S$ is the graph of \begin{equation*} f(x,y)=4x-8y+30 \end{equation*} Over the rectangle $R$ \begin{equation*}R=\left \{ (x,y)|-2<x<3,0<y<2 \right \}\end{equation*} And the vector field \begin{equation*}\mathbf{F}(x,y,z) =-x^{2}\mathbf{i}+xz\mathbf{j}+yx\mathbf{k}\end{equation*}

Evaluate with Stokes \begin{equation*} \oint_c \textbf{F}\centerdot d\textbf{r}\end{equation*}

Where $C$ is the boundary of $S,$ oriented in the counter clockwise direction when viewed from above.

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I assume that I have to calculate the $\mathrm{curl}\,{(F)}$ but how to proceed from there?

Answer 1

Remember $d\textbf{S} = \textbf{n}(u,v) du \ dv$ and so if you parametrize $S$ then you have by Stokes';

$$\oint_c \textbf{F} \cdot d\textbf{s} = \iint_D \textrm{curl}(\textbf{F}) \cdot d \textbf{S} = \iint_D \textrm{curl}(\textbf{F}) \cdot \textbf{n}(u,v) \ du \ dv$$

You have to parametrize the graph $z = 4x - 8y + 30$, so it'll be simple to just use the map $G(x,y) = (x,y, 4x-8y+30)$ and $D = \{(x,y): x \in (-2,3), y \in (0,2)\}$. Here recall that $\textbf{n}(x,y) = G_x \times G_y$.