I am trying to solve this problem but I am having difficulties to finish it. I would appreciate of someone can advice me on how to continue
Problem:
Let S be the graph of $$f(x,y) = 4x-8y+30$$ over the rectangle $R =\{(x,y,z)\in\mathbb{R^3}\mid-2\lt x \lt3, 0\lt y\lt2\}$.
Consider the vector field
$$F(x,y)= -x^2i+xzj+yxk$$
use Stoke's Theorem to Evaluate
$$\int_{C} F.\mathrm dr$$
where C is the boundary of S, oriented in the counterclockwise direction when viewed from above.
Solution From Stoke's Theorem $$\int_{C} F.\mathrm dr = \iint_{S} \hat n. CurlF \mathrm dS $$
$$Curl F = \nabla XF = -yj+zk$$ $$4x-8y+30=0$$ $$8y=4x+30$$ $$y=4x+30$$ $$y=\frac{1}{2}x+\frac{15}{4}$$
Please how do I proceed from Here
We need a parametric representation of $S$. Under the given circumstances we just take $${\bf f}:\quad R\to{\mathbb R}^3,\quad(x,y)\mapsto(x,y,4x-8y+30)\ .$$ From this we obtain the vectorial surface element $$d\vec S=({\bf f}_x\times{\bf f}_y){\rm d}(x,y)=(1,0,4)\times(0,1,-8)\>{\rm d}(x,y)=(-4,8,1)\>{\rm d}(x,y)\ ,$$ which is already directed upwards. Stokes' theorem now gives $$Q:=\int_{\partial S}{\bf F}\cdot d{\bf r}=\int_S{\rm curl}({\bf F})\cdot d\vec S=\int_R{\rm curl}\bigl({\bf F}({\bf f}(x,y)\bigr)\cdot(-4,8,1)\>{\rm d}(x,y)\ .$$ It remains to compute$${\rm curl}\bigl({\bf F}({\bf f}(x,y)\bigr)=(0,-y,z)_{z=4x-8y+30}=(0,-y,4x-8y+30)\ ,$$ so that we now have $$Q=\int_R(-8y+4x-8y+30)\>{\rm d}(x,y)\ ,$$ which I may leave to you.