1
$\begingroup$

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

$\textbf{f}(x,y,z)=\textbf{i}-\textbf{j}+\textbf{k}$

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

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?

Thanks.

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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

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.