Calculate $\int_C \textbf{f} \ d \textbf{r}$ for given vector field and curve $C$.


$C: x=3t, y=2t, z=t, 0 \leq t \leq 1$

Using Stokes' Theorem, how would you find bounds of the surface to solve this? And since it would be a double integral, what would $d\textbf{S}$ be?


  • $\begingroup$ This is not a closed curved, hence it cannot bound a surface. You can't apply Stokes's theorem. $\endgroup$ – Mark Fantini May 24 '14 at 0:57
  • $\begingroup$ Notice that all three coordinates $x$, $y$, and $z$ are functions of one parameter $t$, so $C$ is one-dimensional (a curve). $\endgroup$ – Sammy Black May 24 '14 at 0:59
  • $\begingroup$ How would you solve it then? Thanks again. $\endgroup$ – user7000 May 24 '14 at 1:00

Notice that: $$\mathrm{d}x = 3 \mathrm{d}t , \quad \mathrm{d}y = 2 \mathrm{d}t, \quad \mathrm{d}z = \mathrm{d}t$$

We can write your integral as: $$\int_C 1 \mathrm{d}x + (-1) \mathrm{d}y + 1 \mathrm{d}z = \int_{0}^{1} 3 \mathrm{d}t - 2 \mathrm{d}t + \mathrm{d}t$$

Go for it.

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  • $\begingroup$ Thank you! What theorem are you using to write the integral like that? $\endgroup$ – user7000 May 24 '14 at 1:15
  • $\begingroup$ Another way to write your curve is $r(t) = (3t, 2t, t)$, and, by definition, $$\int_C \mathbb{f} \mathrm{d}\mathbb{r} = \int_{0}^{1} \mathbb{f} (r(t)) \cdot r'(t) \mathrm{d}t$$ $\endgroup$ – Ivo Terek May 24 '14 at 1:24
  • $\begingroup$ And a nice mnemonic is: if $r = (x,y,z)$, then $\mathrm{d}r = \left(\frac{\mathrm{d}x}{\mathrm{d}t}, \frac{\mathrm{d}y}{\mathrm{d}t}, \frac{\mathrm{d}z}{\mathrm{d}t} \right) \mathrm{d}t$ $\endgroup$ – Ivo Terek May 24 '14 at 1:31

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