# Line Integral.

• Apr 26th 2009, 03:59 AM
Showcase_22
Line Integral.
Quote:

Evaluate $\displaystyle \int_C(xy+y+z)~ds$ along the curve $\displaystyle \underline{r}(t)=2ti+tj+(2-2t)k$ where $\displaystyle 0 \leq t \leq 1$.
$\displaystyle \left| \underline{v} \right|= \left| \frac{d \underline{r}}{dt} \right|=\left| 2i+j-2k \right|=3$

$\displaystyle \int_0^1 2t^2+t+2-2t ~dt=\int_0^1 2t^2-t+2 ~dt= \left[ \frac{2t^3}{3}-\frac{t^2}{2}+2t \right]_0^1=\frac{2}{3}-\frac{1}{2}+2=\frac{13}{6}$

Is this right?(Wondering)
• Apr 26th 2009, 05:18 AM
running-gag
Hi

You have forgotten to multiply by $\displaystyle \left| \underline{v} \right|= 3$ (Wink)
• Apr 26th 2009, 06:15 AM
Apprentice123
The correct is
$\displaystyle \int_0^1 (6t^2-3t+6)dt$

???
• Apr 26th 2009, 06:26 AM
mr fantastic
Quote:

Originally Posted by Apprentice123
The correct is
$\displaystyle \int_0^1 (6t^2-3t+6)dt$

???

Yes.
• Apr 26th 2009, 06:34 AM
Apprentice123
Thank you.
Can I post my problem and the solution to see if it is correct in a new topic ?
• Apr 26th 2009, 06:42 AM
mr fantastic
Quote:

Originally Posted by Apprentice123
Thank you.
Can I post my problem and the solution to see if it is correct in a new topic ?

Yes. Post it in a new thread.