# Stokes' Theorem - How to find the surface?

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.

• This is not a closed curved, hence it cannot bound a surface. You can't apply Stokes's theorem. – Mark Fantini May 24 '14 at 0:57
• Notice that all three coordinates $x$, $y$, and $z$ are functions of one parameter $t$, so $C$ is one-dimensional (a curve). – Sammy Black May 24 '14 at 0:59
• How would you solve it then? Thanks again. – 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: $$\int_C 1 \mathrm{d}x + (-1) \mathrm{d}y + 1 \mathrm{d}z = \int_{0}^{1} 3 \mathrm{d}t - 2 \mathrm{d}t + \mathrm{d}t$$
• Another way to write your curve is $r(t) = (3t, 2t, t)$, and, by definition, $$\int_C \mathbb{f} \mathrm{d}\mathbb{r} = \int_{0}^{1} \mathbb{f} (r(t)) \cdot r'(t) \mathrm{d}t$$ – Ivo Terek May 24 '14 at 1:24
• And a nice mnemonic is: if $r = (x,y,z)$, then $\mathrm{d}r = \left(\frac{\mathrm{d}x}{\mathrm{d}t}, \frac{\mathrm{d}y}{\mathrm{d}t}, \frac{\mathrm{d}z}{\mathrm{d}t} \right) \mathrm{d}t$ – Ivo Terek May 24 '14 at 1:31