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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.

1 Answers 1

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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

$$ \int_{\gamma} \vec{F} \,d\vec{r} = \int_0^1 \vec{F}(\gamma(t)) \cdot \dot{\vec{\gamma}}(t) \, dt = \int_0^1 ((t - t^2) \vec{i} + (t - t^2) \vec{j} + (t(1-t) + t^2) \vec{k}) \cdot (\vec{i} + \vec{j} + \vec{k}) \, dt \\ = \int_0^1 (t - t^2) + (t - t^2) + t(1-t) + t^2 \, dt = \int_0^1 (3t - 2t^2) \, dt = \left[ \frac{3}{2}t^2 - \frac{2}{3}t^3\right]^{t=1}_{t =0} = \frac{5}{6}.$$

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    Going from $(1,1,1)$ to $(1,1,0)$ would $γ(t)=-t\vec{k}$ and would the limits still remain the same? – 2017-01-28
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    No. You want $\gamma(0) = (1,1,1)$ and $\gamma(1) = (1,1,0)$ so the correct parametrization is $\gamma(t) = (1 - t)(1,1,1) + t(1,1,0) = (1,1,1-t) = (1,1,1) - t(0,0,1) = \vec{i} + \vec{j} + (1 - t)\vec{k}$. – 2017-01-28
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    I'm going to try to get my head around that, the limits are all $0≤t≤1$? – 2017-01-28
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    Yep. In general, a parametrization of a line which starts at $\vec{p}$ at time $t = 0$ and ends at $\vec{q}$ at time $t = 1$ is $\gamma(t) = (1 - t) \vec{p} + t \vec{q}$. – 2017-01-28
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    Just to verify does $-\frac{1}{3}$ sound right for the second line and $-1$ for the third with a final answer of $-\frac{1}{2}$? – 2017-01-28
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    @TheMaster47x: $-\frac{1}{3}$ sounds right for the second integral but if I haven't done a calculation error, the third integral should be zero. – 2017-01-28
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    My $γ(t)$ for the third one was $(1-t,1-t,0)$ resulting in an integral $\int_0^1 2t-2 \, dt$ which I got to equal $-1$, I apologise for not showing my full working, my Latex skills aren't the best. – 2017-01-28
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    @TheMaster47x: Yeah, you're right, I had a calculation error. – 2017-01-28