Given $F=(3x^2+6y)\mathbf{\hat{x}}-14yz\mathbf{\hat{y}}+20xz^2\mathbf{\hat{z}}$, I am trying to evaluate the line integral $\int_C{A}\cdot d\mathbf{r}$ from $(0,0,0)$ to $(1,1,1)$. C is given as three different cases, where

  1. $x=t, y=t^2, z=t^3$
  2. straight line from $(0,0,0)$ to $(1,1,1)$
  3. straight lines from $(0,0,0)$ to $(1,0,0)$, then $(1,1,0)$ to $(1,1,1)$

Can anyone help me start this? I'm having a difficult time remembering exactly how this is done.

  • 1
    $\begingroup$ Maybe you should look it up in your notes and come back? Then you can post what you don't understand. $\endgroup$ – MrYouMath Sep 17 '15 at 16:01
  • $\begingroup$ I concur with MrYouMath. If it's a question of "what-means-what": In this case, $A=F$, and $\bf{r}$ represents the parameterization of a given path. The first one, for instance, would have $d{\bf{r}}=\frac{d}{dt}[t{\bf{\hat{x}}}+t^2{\bf{\hat{y}}}+t^3{\bf{\hat{z}}}]$ $\endgroup$ – user170231 Sep 17 '15 at 16:04
  • $\begingroup$ Thanks user170231, that helped. I'm wondering about part 3 though, would I do essentially the same as what I did in part 2 except three times? $\endgroup$ – A4Treok Sep 17 '15 at 17:50

For the first integral, the path is outlined clearly: ${\bf r}=\langle t,t^2,t^3\rangle$ (excuse the sudden change in notation, it's more convenient to type). $$\int_CA\cdot d{\bf{r}}=\int_a^b \langle 9t^2,-14t^5,20t^7\rangle\cdot d\langle t,t^2,t^3\rangle=\int_a^b \langle 9t^2,-14t^5,20t^7\rangle\cdot \langle 1,2t,3t^2\rangle\,dt=\cdots$$ where $[a,b]$ is the domain you seem to have left out.

Based on your comment, you seem to have a handle on the second question. With your third question, it's unclear whether the path is continuous (i.e. from $(0,0,0)$ to $(1,0,0)$ to $(1,1,0)$ to $(1,1,1)$) or if there are two disconnected paths. Assuming the former, the work is similar to what you did for (2): You have to parameterize the four distinct paths, then integrate along them accordingly.


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