# Line integral, incorrect setup

1. Aug 1, 2014

### jonroberts74

1. The problem statement, all variables and given/known data
$\int_\mathscr{C} \vec{F}(\vec{r})\cdot d\vec{r}; \vec{F}(x,y,z) = <sin z, cos \sqrt{y}, x^3>$ I am assuming $\vec{r}$ is the usual $\vec{c}$ used, so maybe this is where I am incorrect

3. The attempt at a solution

C goes from (1,0,0) to (0,0,3)

Parametrizing C

$\mathscr{C}: \vec{c}(t) = (1-t)<1,0,0> + t<0,0,3> = <1-t, 0 ,3t>; 0 \le t \le 1$

$\vec{c}\,\,'(t) = <-t, 0, 3>$

$\vec{F}(\vec{c}(t) = <\sin 3t, 1, (1-t)^3>$

$\displaystyle \int_{0}^{1} <\sin 3t, 1, (1-t)^3> \cdot <-t, 0, 3>dt$

$\displaystyle \int_{0}^{1} -t \sin 3t + 0 + 3(1-t)^3 dt$

I got this far and integrated it but got the wrong answer, I checked my integration already so I integrated this setup correctly but I screwed up on the setup somewhere.

2. Aug 1, 2014

Check $c'(t)$ again.

3. Aug 1, 2014

### jonroberts74

yeah I did the same mistake as the last one. thanks

4. Aug 2, 2014

### HallsofIvy

Staff Emeritus
Does the problem only say "C goes from (1,0,0) to (0,0,3)" or does it specifically say "the straight line from (1, 0, 0) to (0, 0, 3)?

5. Aug 2, 2014

### jonroberts74

says line segment from (1,0,0) to (0,0,3)

6. Aug 3, 2014

### SammyS

Staff Emeritus
What do you get for the answer?