Line integral of straight lines path (quick question)

1. May 8, 2014

gl0ck

1. The problem statement, all variables and given/known data
Hello,

I have a quick question about the following problem
F = (2y+3)i+xzj+(yz-x)k
and straight lines from (0,0,0) to (0,0,1) to (0,1,1) to (2,1,1)

Considering C1 is the line from (0,0,0) to (0,0,1)
C2 is the line from (0,0,1) to (0,1,1)
and C3 is the line from (0,1,1) to (2,1,1)

Is it right to consider x , y , z as follows
C1 --> x = 0 , y = 0 , z = t
C2 --> x = 0 , y = t , z = 1
C3 --> x = 2t , y=1, z= 1
t from 0 to 1
r = 2ti + j + k
and dr/dt = 2i
c F.dr/dt = ∫ c ((2+3)i + 2tj +(1-2t)k).(2i) dt = [10t]01 = 10 ?
Which is the right answer just I am not sure about the t notation.

Thanks

2. May 8, 2014

xiavatar

You have to use the r associated with each curve in your calculations. So when your traversing $C_i$you use $r_i$. Equations below assuming the $C_1,C_2,C_3$ given by you.

$\vec{r_1} = (0,0,t)$
$\vec{r_2} = (0,t,1)$
$\vec{r_3} = (2t,1,1)$

Always remember that $\vec{r}$ is always equal to the line your integrating over.

3. May 8, 2014

gl0ck

Sorry, but just came accross something different.
I am supposed to find the line integral F along the arc of a circle x^2+y^2 = 1 in the 1st quadrant from (1,0) to (0,1) .
can I assume x = t-1 and y = t ?

Thanks

4. May 8, 2014

xiavatar

Remember that this is is an arc of the circle, not the line from (1,0) to (0,1). So our arc is simply
$\vec{r}=(cos(t),sin(t))$ $0\leq t\leq \frac{pi}{2}$.

Last edited: May 8, 2014
5. May 8, 2014

xiavatar

Do you understand why this is so?

6. May 8, 2014

gl0ck

Yes, I understood the interesting thing is that with x = t-1 and y = t I got the answer, but it seems its just a coincidence

7. May 8, 2014

xiavatar

That's peculiar. It might be because $\vec{F}$ is a conservative vector field. Too lazy to check.

8. May 8, 2014

LCKurtz

Yes. You can alway parameterize the straight line from $P_0$ to $P_1$ as $(1-t)P_0 + tP_1$ with $t: 0\to 1$, which looks like what you did.

9. May 8, 2014

LCKurtz

You mean $0\leq t\leq \frac \pi 2$.

10. May 8, 2014

xiavatar

Yes. That's what I meant. My post had a typo. Corrected.