A force field in 3-space is given by the formula $$\bar{F} (x, y, z) = (x − yz)\bar{i} + (y − xz)\bar{j} +(x(1−y) +z^2)\bar{k}$$ Calculate the work done by $\bar{F}$ in moving a particle once around the triangle with vertices $(0, 0, 0), (1, 1, 1), (1, 1, 0)$ in that order.
I'm really not sure how to go about doing this one. Any help would be much appreciated. Thank you.
You can split your triangle into three line segments, calculate the work of $F$ along each segment and add the results. I'll demonstrate the calculation for the first segment.
The line going from $(0,0,0)$ to $(1,1,1)$ can be parametrized as $$\gamma(t) = t(1,1,1) = t \vec{i} + t \vec{j} + t\vec{k}$$ for $0 \leq t \leq 1$. Hence
$$ \int_{\gamma} \vec{F} \,d\vec{r} = \int_0^1 \vec{F}(\gamma(t)) \cdot \dot{\vec{\gamma}}(t) \, dt = \int_0^1 ((t - t^2) \vec{i} + (t - t^2) \vec{j} + (t(1-t) + t^2) \vec{k}) \cdot (\vec{i} + \vec{j} + \vec{k}) \, dt \\ = \int_0^1 (t - t^2) + (t - t^2) + t(1-t) + t^2 \, dt = \int_0^1 (3t - 2t^2) \, dt = \left[ \frac{3}{2}t^2 - \frac{2}{3}t^3\right]^{t=1}_{t =0} = \frac{5}{6}.$$