Question

find the integral of the vector field f (x, y) = yi + xj to move a particle on the curve L.

Answer 1

@satellite73
@myininaya
This guy needs more brain power than I have.

Answer 2

Is the curve L including y=0?

Answer 3

no

Answer 4

y is de value of the blue line. I mean the part green is the line segment along the x-axis stretching from -2,0) y 2,0

Answer 5

y is the value of the blue line. I mean the part green is the line segment along the x-axis stretching from -2,0) y 2,0

Answer 6

http://video.mit.edu/watch/lecture-4-introduction-to-line-integrals-10485/

for a refresher

Answer 7

F = <y,x> = <2sin(t), 2cos(t)> ; along the curve we want

r = <2cos(t), 2sin(t)>
dr = <-2sin(t), 2cos(t)>

\[\int_{0}^{pi}(F.dr)dt\]
\[\int_{0}^{pi}-4sin^2(t)+4cos^2(t)\ dt\]

is what im thinking for the curve part

Answer 8

^correct. that above can also be simplified to \[\int\limits_{0}^{pi} 4 \cos(2x)\], which leads to a peculiar result (wink wink)

Answer 9

^sorry t, not x

Answer 10

and what about if we try to solve it for double integral,,,,and like Green theorem said, the slution have to be same

Answer 11

(y,x)is conservative so you can choose best line x-axis
int(ydx+xdy) from (2,0) to (-2,0) =xy2-xy1=0-0=0

Answer 12

I have a feeling the answer is 0. And it should be cuzz the field seems to be conservative and it looks like a closed loop :)

Answer 13

yes but I need the process... I get confuse when I try to parametri