Question

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:
\[\hat{r}'(t)=3t^2\hat i-2t \hat j+ \hat k\]

Answer 1

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

Answer 2

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

Answer 3

thanks @lgbasallote. Better? :P

Answer 4

yes. yes it is.

Answer 5

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?

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

and now I would just do the dot product correct?

Answer 10

exactly

Answer 11

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

Answer 12

well done

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

yep

Answer 17

oh integration by parts! duh =D

Answer 18

\[-\cos t^3\] :-)

Answer 19

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

Answer 20

do you get that?

Answer 21

Oh I see it now!!!!

Answer 22

cool :)

Answer 23

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

Answer 24

yep

Answer 25

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

Answer 26

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

Answer 27

u cooked the problem

Answer 28

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

Answer 29

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

Answer 30

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

Answer 31

oops sine of 0 is zero

Answer 32

you have an extra +1 in there - -yep

Answer 33

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

Answer 34

looks good to me :)

Answer 35

sigh...finally! Thanks guys!

Answer 36

welcome !

Answer 37

:)