## anonymous 4 years ago Evaluate the line integral $\int_c F\cdot dr$ where c is the vector function $\hat{r}(t)=t^3\hat i-t^2 \hat j+ t \hat k$ and $F(x,y,z)=sinx\hat i + cosy \hat j+zx \hat k$ Here is how far I have gotten so far: $\hat{r}'(t)=3t^2\hat i-2t \hat j+ \hat k$

1. lgbasallote

you might want to use \cdot rather than \bullet hehe

2. anonymous

I'll change the Function F in terms of t's in just a second

3. anonymous

thanks @lgbasallote. Better? :P

4. lgbasallote

yes. yes it is.

5. anonymous

oh my, I did I similar problem 28 days ago http://openstudy.com/study#/updates/50060b97e4b06241806745b8 and I'm also looking at http://tutorial.math.lamar.edu/Classes/CalcIII/LineIntegralsVectorFields.aspx but I can't seem to remember... What do i substitute in for x and y, do I pick which one I want t to be?

6. anonymous

Hey i think I got it! $x=t^3;y=-t^2;z=t$ YES?

7. anonymous

That would give me: $F(r(t))=sin(t^3)\hat i +cos(-t^2)\hat j+t^4 \hat k$

8. anonymous

$\int_{t=0}^{t=1}\left(sin(t^3)\hat i +cos(-t^2)\hat j+t^4 \hat k\right)\cdot\left(3t^2\hat i -2t\hat j +\hat k \right) dt$

9. anonymous

and now I would just do the dot product correct?

10. anonymous

exactly

11. anonymous

$\int_0^1(3t^2sin(t^3)-2tcos(-t^2)+t^4)dt$

12. anonymous

well done

13. anonymous

wait, I have to take the integral of several products?

14. anonymous

separate them$\int_{0}^{1} 3t^2 \sin t^3 dt+...+...$

15. anonymous

oh ok $\int_0^1 3t^2sin(t^3)dt-\int_0^12tcos(-t^2)dt+\int_0^1t^4dt$ like this?

16. anonymous

yep

17. anonymous

oh integration by parts! duh =D

18. anonymous

$-\cos t^3$ :-)

19. TuringTest

no integration by parts you can do it all with u-subs

20. TuringTest

do you get that?

21. anonymous

Oh I see it now!!!!

22. TuringTest

cool :)

23. anonymous

u=t^2 du=2t dt for the second integral

24. TuringTest

yep

25. anonymous

u=t^3 and du=3t^2 dt for the first integral...LOL that took me a while!

26. TuringTest

yeah, it's well set-up for the u-sub thing :)

27. anonymous

u cooked the problem

28. anonymous

$-cos(1)+sin(1)+2$ My algebra is probably wrong...but Yeah @mukushla that was one long recipe!

29. anonymous

$\large -\cos t^3]_{0}^{1}+\sin -t^2]_{0}^{1}+t^5/5]_{0}^{1}$

30. anonymous

$(-cos(1)+1)+(-sin(1)+1)+\frac 1 5$ $-cos(1)-sin(1)+\frac{11}{5}$

31. anonymous

oops sine of 0 is zero

32. TuringTest

you have an extra +1 in there - -yep

33. anonymous

-cos(1)-sin(1)+ 6/5

34. TuringTest

looks good to me :)

35. anonymous

sigh...finally! Thanks guys!

36. TuringTest

welcome !

37. anonymous

:)