Question

How would I do the line integral for Fdr where F=2xyi+x^2j and the line is described by r(t)=(t^2-2t)i+(tsin(t-2))j over 0 13 years ago

Answer 1

\[\int\limits Fdr, F=<2xy,x^2>, C: r(t)+<t^2-2t, tsin(t-2)>, 0<t<2\]

Answer 2

actually, is this green's theorem, since F is only 2 dimensions?

Answer 3

Would it come out to zero?

Answer 4

I guess in general I'm having trouble understanding how to construct the surface integral for Stokes. I know how to find the curl, but I'm not sure what I'm supposed to be dotting it with (dS? a normal vector? what is dS?)

Answer 5

F(r(t)).r'(t) sounds familiar

Answer 6

im not remembering the specific for stokes to be sure what to do on that end of it

Answer 7

Thanks for the reply. I believe F(r(t).r'(t)) is if i want to integrate it directly.

Answer 8

F= (2xy,x^2)

x=t^2-2t ; x' = 2t - 2
y=t sin(t-2) ; y' = t cos(t-2) + sin(t-2)

F(r(t))= 2 t sin(t-2) (t^2-2t) ; (t^2-2t)^2
2t - 2 ; t cos(t-2) + sin(t-2)
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whatever the dot is .............................

Answer 9

correct, if you just want to turn it all into one big sum of a quazihoots yes

Answer 10

if you want a readable explanation and example of Greens, this site is always good for me

http://tutorial.math.lamar.edu/Classes/CalcIII/GreensTheorem.aspx

Answer 11

:) yeah, it gets really ugly. I'm pretty sure the problem is engineered for Stokes, just haven't gotten a good grasp on how to apply it.

Answer 12

hey, i'll check it out. thank you.

Answer 13

that should have stokes on the site somewhere as well; if not theres a patrickJMT or another on youtube that is also pretty easy to follow for these things

Answer 14

its just been to long past and too few practices to recollect these for me :)

Answer 15

hey, that's some good stuff. I appreciate the help.

Answer 16

good luck :)