0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ Going from $(1,1,1)$ to $(1,1,0)$ would $γ(t)=-t\vec{k}$ and would the limits still remain the same? $\endgroup$ – TheMaster47x Jan 28 '17 at 16:10
  • $\begingroup$ No. You want $\gamma(0) = (1,1,1)$ and $\gamma(1) = (1,1,0)$ so the correct parametrization is $\gamma(t) = (1 - t)(1,1,1) + t(1,1,0) = (1,1,1-t) = (1,1,1) - t(0,0,1) = \vec{i} + \vec{j} + (1 - t)\vec{k}$. $\endgroup$ – levap Jan 28 '17 at 16:12
  • $\begingroup$ I'm going to try to get my head around that, the limits are all $0≤t≤1$? $\endgroup$ – TheMaster47x Jan 28 '17 at 16:26
  • $\begingroup$ 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 $\gamma(t) = (1 - t) \vec{p} + t \vec{q}$. $\endgroup$ – levap Jan 28 '17 at 16:31
  • $\begingroup$ Just to verify does $-\frac{1}{3}$ sound right for the second line and $-1$ for the third with a final answer of $-\frac{1}{2}$? $\endgroup$ – TheMaster47x Jan 28 '17 at 17:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.