Compute the line integral of the triangular region with vertices $\left ( 0,0 \right ),(2,0),\left ( 0,2 \right )$
with the function $\vec{v}=xy\hat{x}+\left ( 2yz \right )\hat{y}+\left ( 3xz \right )\hat{z}$
Forget the vertical and horizontal path. Those are trivial.
Diagonal path integral is annoying.
Along the diagonal line integral, $d\vec{l}=\left \langle 0,dy,dz \right \rangle$ evidently, x=0 and z=2-y
Then the line integral over this path is
$\int_{diagonal}\left \langle -2y,3z,-x \right \rangle\cdot\left \langle 0,dy,dz \right \rangle=\int_{0}^{y=2-z}3z~dy-\int_{0}^{z=2-y}x~dz $
This produces an answer with z variables which cannot be.
Where am I going wrong?
By the Stokes theorem: $$ \oint_C \vec{F}\cdot d\vec{r} = \iint_S \nabla\times \vec{F}\cdot d\vec{S}, $$ where $C$ represents your triangle, $S$ the surface bounded by this triangle, and $\vec{F}$ is the field $(xy,2yz,3xz)$.
Assuming the triangle is orientated counterclockwise, a normal unit vector to $S$ is given by $\vec{n}=(0,0,1)$, and it follows that $$ \iint_S \nabla\times \vec{F}\cdot d\vec{S} = \iint_S -x \;dS = \int_{0}^2\int_{0}^{2-x}-x \;dydx=-\frac{4}{3}. $$
Otherwise, directly with the line integral. You need to parametrize the diagonal path $C_1$, which you can do as follows: $$ (x,y,z)=(1-t)(2,0,0)+t(0,2,0)=(2-2t,2t,0), \quad 0\le t \le 1 $$ Then, $$ \int_{C_1}\vec{F}\cdot d\vec{r} = \int_{0}^1 \vec{F}(t)\cdot (-2,2,0) \;dt =\\ \int_{0}^1 ((2-2t)2t,0,0)\cdot (-2,2,0) \;dt = -4 \int_{0}^1 (2-2t)t \;dt = -4/3. $$