# 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.