Line integral, incorrect setup

[jonroberts74]

Homework Statement

$\int_\mathscr{C} \vec{F}(\vec{r})\cdot d\vec{r}; \vec{F}(x,y,z) = <sin z, cos \sqrt{y}, x^3>$ I am assuming $\vec{r}$ is the usual $\vec{c}$ used, so maybe this is where I am incorrect

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.

 

[Quesadilla]

Check $c'(t)$ again.

 

[jonroberts74]

Check $c'(t)$ again.

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

 

[HallsofIvy]

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)?

 

[jonroberts74]

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)?

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

 

[SammyS]

Homework Statement

$\int_\mathscr{C} \vec{F}(\vec{r})\cdot d\vec{r}; \vec{F}(x,y,z) = <sin z, cos \sqrt{y}, x^3>$ I am assuming $\vec{r}$ is the usual $\vec{c}$ used, so maybe this is where I am incorrect

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.

What do you get for the answer?