1. ## Line Integrals

If $\displaystyle F = (x+4)i +xyj + (xz-y)k$, evaluate $\displaystyle \int c F.dr$ along the following paths C:

the straight line joining (0,0,0) to (2,1,1)

I don't really understand what to do and what the limits are.

the straight line joining (0,0,0) to (2,1,1)

$\displaystyle \gamma (t)=(2t,t,t)\quad (t\in [0,1])$

3. You can write your integral in a bit more standard notation this way:

$\displaystyle \int_{C}\mathbf{F}\cdot d\mathbf{r}.$ The code to produce this, for your convenience, was

[TEX]\int_{C}\mathbf{F}\cdot d\mathbf{r}.[/TEX]

4. Originally Posted by FernandoRevilla
$\displaystyle \gamma (t)=(2t,t,t)\quad (t\in [0,1])$
Thanks i've done this and got the right answer.

What happens for the same question, if it is the straight lines from (0,0,0) to (0,0,1) then to (0,1,1) and then to (2,1,1)

5. Then,

$\displaystyle \displaystyle\int_C \vec{F}\cdot d\vec{r}=\displaystyle\int_{\gamma_1} \vec{F}\cdot d\vec{r}+\displaystyle\int_{\gamma_2} \vec{F}\cdot d\vec{r}$

where

$\displaystyle \gamma_1,\;\gamma_2$

are the corresponding lines.

6. Originally Posted by FernandoRevilla
Then,

$\displaystyle \displaystyle\int_C \vec{F}\cdot d\vec{r}=\displaystyle\int_{\gamma_1} \vec{F}\cdot d\vec{r}+\displaystyle\int_{\gamma_2} \vec{F}\cdot d\vec{r}$

where

$\displaystyle \gamma_1,\;\gamma_2$

are the corresponding lines.
Is gamma 1 (0,0,0) to (0,0,1)

And gamma 2 (0,0,1) and then to (2,1,1) ?

Is gamma 1 (0,0,0) to (0,0,1)

And gamma 2 (0,0,1) and then to (2,1,1) ?

Yes, those are the lines.

8. Originally Posted by FernandoRevilla
Yes, those are the lines.
Ok

Im not getting the same answer as my book

(1-t)(0,0,0) + t(0,0,1) = (0,0,t)

and (1-t)(0,1,1) + t(2,1,1) = (2t,1,1)

So for the first one
r' = (0,0,1)
F.r' = 0

So the integral of that is 0

Then for the second one

r' = (2,0,0)

F.r' = 4t + 8

so the integral between 1 and 0 = 10

But the answer in my book is 38/3

9. The second line is

$\displaystyle \gamma_2(t)=(2t,t,1)\quad (t\in [0,1])$

10. Originally Posted by FernandoRevilla
The second line is

$\displaystyle \gamma_2(t)=(2t,t,1)\quad (t\in [0,1])$
How?

(1-t)(0,1,1) + t(2,1,1) =(0+2t, 1-t+t, 1-t+t) = (2t,1,1)