Line integral of straight lines path (quick question)

[gl0ck]

Homework Statement

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

 

[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.

 

[gl0ck]

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.

Sorry, but just came across 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

 

[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
$\vec{r}=(cos(t),sin(t))$ $0\leq t\leq \frac{pi}{2}$.

 

[xiavatar]

Do you understand why this is so?

 

[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

 

[xiavatar]

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

 

[LCKurtz]

Homework Statement

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

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.

 

[LCKurtz]

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

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

 

[xiavatar]

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