The sides of a triangle $D$ are described by the intersections of plane $x + 3y + 4z = 6$ with the planes $x = 0, y = 0 $ and $z = 0$. Determine the integral over the triangle $D$ using Stokes' theorem.
$\int_{\partial D}^{} \! -2y^2 dx+zdy +x dz \, =$
My question:
I know that Stokes' theorem is defined as $\int_{C}^{} \! F\cdot ds \, =\int_{}^{} \! \int_{S}^{} \! (\nabla \times F) \, \, ds $. But I dont understand what excatly is $F$ in my example?
Let me rewrite Stokes' theorem as
$$ \int_{D} \boldsymbol{\nabla \times F \cdot dS} = \oint_{\partial D} \boldsymbol{F \cdot dr}. $$
In your case,
$$\boldsymbol{F \cdot dr} = -2y^2 dx + z dy + x dz = (-2y^2, z, x) \cdot (dx, dy, dz) $$
so $\boldsymbol{F}(x,y,z) = (-2y^2, z, x)$. Thus, in order to solve the question, you need to calculate $\boldsymbol{\nabla \times F}$ and then perform the surface integral of $\boldsymbol{\nabla \times F}$ over the (two-dimensional) triangle $D$.