Using Stokes' Theorem to evaluate $\int_{C}{(xyz)dx+(xy)dy+(x)dz}$ Let $C$ be the closed, piecewise smooth curve formed by traveling in straight lines between the points $(0,0,0),(2,1,5),(1,1,3)$ and back to the origin, in that order. Use Stokes' theorem to evaluate the integral:
$$\displaystyle\int_{C}{(xyz)dx+(xy)dy+(x)dz}$$
First, Stokes' theorem is:
$$\displaystyle\int_{∂S}{F \cdot dS} = \displaystyle\iint_{S}{(\nabla \times F)\cdot dS} = \displaystyle\iint_{D}{((\nabla \times F)\cdot n)\,dx\,dy}$$
I started by creating two vectors by subtracting the vertex points of the triangle. The I took the cross product of those two vectors to get $n = (-2,-1,1)$. Which is the normal vector to the surface. Taking $n \cdot (x,y,z) = 0$ will give me the tangent plane. Which will be:
$$-2x - y + z = 0$$ so, $$z = 2x + y$$
I also can deduce from the question that (correct me if I'm wrong)
$$F = (xyz,xy,x)$$ So, $$\nabla \times F = (0,xy-1,y-xz)$$
so now the integral is:
$$\displaystyle\iint_{D}{((1-xy+y-xz)\,dx\,dy}$$
However, I'm not sure what to do for the boundaries of the integral. In the xy-plane, I believe D would be the triangle with the vertices (0,0), (2,1), (1,1). But that doesn't really do me much good. Would I need to do some sort of transformation to make D into something easier to work with and then just include the Jacobian determinant of that transformation? Or is there a better way to go about this? Thanks!
 A: You know that two vectors that you could use to build the triangular region are: $\langle2,1,5\rangle$ and $\langle1,1,3\rangle$. So a point $P$ on the triangular region we are interested in will have coordinates $\langle2,1,5\rangle u+\langle1,1,3\rangle v$. Where $0\leq u\leq1$. How do we obtain the limits of $v$? The line joining the heads of the vectors will be defined by the vector $\langle-1,0,-2\rangle$ (verify this). So if we want to have a vector with its head on the line joining the heads of the two vectors we have selected to parameterize the region we will have $\langle-1,0,-2\rangle u+\langle2,1,5\rangle$ (considering $u$ limits). Now you can verify that $v$ limits will be $0\leq v \leq 1-u$.
From here you can use a parameterization $\mathbf r(u,v)=\langle2u+v,u+v,5u+3v\rangle$ with the given limits on $u$ and $v$ and use Stokes' theorem as
$$\int_{C}\mathbf F\cdot d\mathbf r = \iint_S \text{curl}\mathbf F\cdot d\mathbf S=\iint_D\text{curl}\mathbf F(\mathbf r(u,v))\cdot (\mathbf r_u\times \mathbf r_v)\,dA.$$ 
In the last equation do verify the orientation of the surface and the orientation of the curve to check if you are using the correct signs.
What the parameterization does is to take a point $(u,v)\in D$ and send it to a point on the triangular region $S$. With respect to this language of transformating a region into another I see that you are commiting a mistake, you don't make $D$ into something easier to work with, you make something easier to work with into $D$, the transformation transforms the easier region into the difficult region that you begin to work with, so it's the other way round.
