force field work done 
A force field in 3-space is given by the formula
  $F(x,y,z)=(x+yz,y+xz,x(y+1)+z^2)$. Calculate the work done by F in
  moving a particle once around the triangle with vertices $(0,0,0)$,
  $(1,1,1)$, $(1,1,0)$ in that order.

So $W=\int \limits_{C} Fdr$ but how do I make my $r(t)$ function? Do I do it part by part as in let A, B, C be the vertices of the triangle respectively with the order they are given. First find the vector equation of AB intersecting with A and B which would be say $r_1(t)= (0,0,0)+(1,1,1)t=(t,t,t)$ right? But this is just a guess. Even if this is right, what would the limits even be.
I missed one lecture on this module so I am unsure. Please help.
 A: In a vector setting
$$
W=\int F\cdot\mathrm{d}r
$$
We can parametrize the first side of the triangle by $r=(t,t,t)$ for $t$ from $0$ to $1$:
$$
\begin{align}
W_1
&=\int_0^1(t+t^2,t+t^2,t+2t^2)\cdot(1,1,1)\,\mathrm{d}t\\
&=\int_0^1(3t+4t^2)\,\mathrm{d}t\\
&=\frac32+\frac43\\
&=\frac{17}6
\end{align}
$$
We can parametrize the second side of the triangle by $r=(1,1,1-t)$ for $t$ from $0$ to $1$:
$$
\begin{align}
W_2
&=\int_0^1(2-t,2-t,2+(1-t)^2)\cdot(0,0,-1)\,\mathrm{d}t\\
&=\int_0^1(-2-(1-t)^2)\,\mathrm{d}t\\
&=-2-\frac13\\
&=-\frac73
\end{align}
$$
The third side is similar.
A: One could as well evaluate it via the Kelvin-Stokes theorem:
$$
W = \int\limits_{\partial T} F \cdot du 
= \int\limits_{T} \mbox{rot } F \cdot dA
$$
where $\mbox{rot } F = (\epsilon_{ijk}\partial_j F_k) = (0, -1, 0)^T$ and $dA = \frac{1}{\sqrt{2}}(-1,1,0)^T\,\lVert dA \rVert$ which gives the constant $-\frac{1}{\sqrt{2}}$ as integrand.
The orientation of $T$ is given by the right hand rule.
That leaves the proper 2D integration over the triangle as calculating its area.
It should have $\lVert A \rVert = \frac{\sqrt{2} \times 1}{2} = \frac{1}{\sqrt{2}}$
That should give $W = -\frac{1}{2}$ if I did not make a mistake.
Checking
Integrating along $u = (1, 1, 1-t)^T$ gives $du = (0, 0, -1)^T dt$ and
\begin{align}
W_2 
&= \int\limits_0^1 (*,*,2+(1-t)^2)^T\cdot(0, 0, -1)^T dt \\
&= \int\limits_0^1 (-t^2 + 2t - 3)dt \\
&= \left[-\frac{1}{3} t^3 + t^2 - 3t\right]_0^1 \\
&= -\frac{7}{3}
\end{align}
Integrating along $u = (1-t, 1-t, 0)^T$ gives $du = (-1, -1, 0)^T dt$ and
\begin{align}
W_3 
&= \int\limits_0^1 (1-t,1-t,*)^T\cdot(-1,-1,0)^T dt \\
&= \int\limits_0^1 (2t-2)dt \\
&=  [t^2 - 2t]_0^1 \\
&= -1
\end{align}
This would give
$$
W = W_1 + W_2 + W_3 = \frac{17}{6} - \frac{7}{3} - 1 =  -\frac{1}{2}
$$
A: If you have $\mathbf r_1(t)=\left\langle t,t,t\right\rangle$ to go from $A$ to $B$, at $t=0$ you are at $\langle 0,0,0\rangle=A$, if $t=1$ you have reached $\langle 1,1,1\rangle =B$. Normally when you construct your parameterizations this way, for example $\mathbf r_2(t)=B+(B-C)t=\langle 1,1,1\rangle + \langle 0,0,-1\rangle t = \langle 1,1,-t\rangle$, you will have the limits $0\le t\le 1$.
