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

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you might want to use \cdot rather than \bullet hehe
I'll change the Function F in terms of t's in just a second
thanks @lgbasallote. Better? :P

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Other answers:

yes. yes it is.
oh my, I did I similar problem 28 days ago and I'm also looking at 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?
Hey i think I got it! \[x=t^3;y=-t^2;z=t\] YES?
That would give me: \[F(r(t))=sin(t^3)\hat i +cos(-t^2)\hat j+t^4 \hat k \]
\[\int_{t=0}^{t=1}\left(sin(t^3)\hat i +cos(-t^2)\hat j+t^4 \hat k\right)\cdot\left(3t^2\hat i -2t\hat j +\hat k \right) dt\]
and now I would just do the dot product correct?
well done
wait, I have to take the integral of several products?
separate them\[\int_{0}^{1} 3t^2 \sin t^3 dt+...+...\]
oh ok \[\int_0^1 3t^2sin(t^3)dt-\int_0^12tcos(-t^2)dt+\int_0^1t^4dt\] like this?
oh integration by parts! duh =D
\[-\cos t^3\] :-)
no integration by parts you can do it all with u-subs
do you get that?
Oh I see it now!!!!
cool :)
u=t^2 du=2t dt for the second integral
u=t^3 and du=3t^2 dt for the first integral...LOL that took me a while!
yeah, it's well set-up for the u-sub thing :)
u cooked the problem
\[-cos(1)+sin(1)+2\] My algebra is probably wrong...but Yeah @mukushla that was one long recipe!
\[\large -\cos t^3]_{0}^{1}+\sin -t^2]_{0}^{1}+t^5/5]_{0}^{1}\]
\[(-cos(1)+1)+(-sin(1)+1)+\frac 1 5\] \[-cos(1)-sin(1)+\frac{11}{5}\]
oops sine of 0 is zero
you have an extra +1 in there - -yep
-cos(1)-sin(1)+ 6/5
looks good to me :)
sigh...finally! Thanks guys!
welcome !

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