1. ## Line Integral.

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

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

Is this right?

2. Hi

You have forgotten to multiply by $\left| \underline{v} \right|= 3$

3. The correct is
$\int_0^1 (6t^2-3t+6)dt$

???

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

???
Yes.

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

6. 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.