Evaluate $\ \int_T \vec F \cdot d\vec r $ where $\vec F (x, y, z) = (2xy + 4xz)\vec i + (x^2 + 6yz)\vec j + (2x^2 + 3y^2) \vec k$ I'm a bit confused as to what I've to do for this exam question.  
$$\vec F (x, y, z) = (2xy + 4xz)\vec i + (x^2 + 6yz)\vec j + (2x^2 + 3y^2) \vec k , \qquad x, y, z ∈ \mathbb{R}. $$
Let $T$ denote the triangular path with vertices
$(1, 1, 1)$, $(2, 1, 1)$, and $(3, 2, 2)$,
traversed from $(1, 1, 1)$ to $(2, 1, 1)$ to $(3, 2, 2)$ to $(1, 1, 1)$. Evaluate
$$\ \int_T \vec F \cdot d\vec r. $$
Justify your answer.
I was thinking of parameterising with a set of lines but I'm not too sure how I would go about this? 
Thanks for any help you can give 
 A: Note that we can find a potential function $\phi$ for which we have $\vec F(x,y,z)=\nabla \phi(x,y,z)$.  To find $\phi$, we need only integrate the components of $\vec F$.
To proceed, we begin with the $\hat x$-component to find
$$\begin{align}
\phi(x,y,z)&=\int F_x(x,y,z)\,dx\\\\
&=\int (2xy+4xz)\,dx\\\\
&=x^2y+2x^2z+C_1(y,z) \tag 1
\end{align}$$
where $C_1(y,z)$ is an integration constant with respect to $x$, but can depend on both $y$ and $z$.
Next, we use $\phi(x,y,z)$ as given in $(1)$ to find $C_1(y,z)$.  Noting that $\frac{\partial \phi(x,y,z)}{\partial y}=F_y(x,y,z)$ yields
$$x^2+\frac{\partial C_1(y,z)}{\partial y}=x^2+6yz$$
from which we find $C_1(y,z)=3y^2z+C_2(z)$, where $C_2(z)$ is a second integration constant.  At this point, we can write the potential function as
$$\phi(x,y,z)=x^2y+2x^2z+3y^2z+C_2(z) \tag 2$$
Finally, taking the partial of $\phi(x,y,z)$, as given by $(2)$, with respect to $z$, and setting it equal to $F_z(x,y,z)$ yields
$$2x^2+3y^2+C_2'(z)=2x^2+3y^2$$
from which we find that $C_2(z)$ is a constant.  
Therefore, we can write the potential function as
$$\bbox[5px,border:2px solid #C0A000]{\phi(x,y,z)=x^2y+2x^2z+3y^2z+C}$$
for any constant $C$.

Since we have determined that $\vec F(x,y,z)=\nabla \phi(x,y,z)$, $\vec F$ is conservative, and its path integral depends only on the end points of the path.  We can write
$$\begin{align}
\int_{\vec r_1}^{\vec r_2}\vec F(x,y,z)\cdot \vec d\vec \ell&=\int_{\vec r_1}^{\vec r_2} \nabla \phi(x,y,z)\cdot d\vec \ell\\\\
&=\int_{\vec r_1}^{\vec r_2} d\phi\\\\
&=\phi(1,1,1)-\phi(1,1,1)\\\\
&=0\end{align}$$       
A: Say you want to integrate along the line from $(2,1,1)$ to $(1,1,1)$.
The vector that points from $(1,1,1)$ to $(2,1,1)$ is $(1,1,1) - (2,1,1) = (-1,0,0)$.
You start at $(2,1,1)$ and add some portion of $(-1,0,0)$ to that starting point to get a point along the path.  If it's all of $(-1,0,0)$, you get $(2,1,1) + (-1,0,0)$, and that's the point at which the journey ends.  If it's just half way, you start at $(2,1,1)$ and add half of $(-1,0,0)$ to get $(2,1,1)+ \frac 1 2 (-1,0,0)$.  And so on: what every fraction of the total trip is complete, you start at $(2,1,1)$ and add that fraction times $(-1,0,0)$.  Call that fraction  $t$.  So you start at $(2,1,1)$ and add $t$ times $(-1,0,0)$ to get a point on the path.  When $t=0$, you're still at the starting point.  When $t=1$, you're at the ending point.  So you have this:
\begin{align}
\text{The point } (2,1,1) + t(-1,0,0) & \text{ goes from } (2,1,1) \text{ to } (1,1,1) \\
& \text{ as $t$ goes from $0$ to $1$.}
\end{align}
It may also shed some light to write it like this:
\begin{align}
(2,1,1) + t(-1,0,0) & = (2,1,1) + t\Big( (1,1,1) - (2,1,1) \Big) \\[10pt]
& = (1-t)(2,1,1) + t( 1,1,1) \\[10pt]
& = (1-t)(\text{starting point}) + t(\text{ending point}).
\end{align}
A: Well it's pretty obvious when you are looking for it that $\vec F = \vec{\operatorname{grad}} (f)$ with$f(x,y,z)=x^2y+2x^2z+3y^2z$.
So $$\int \vec F \cdot \vec{dr} = f(1,1,1)-f(1,1,1) = 0$$
