Question

Does anybody know how to use Green's Theorem to evaluate the line integral.......

Answer 1

\[\int\limits_{?}^{?}xy dx + y dy \]
where C is the unit circle.

Answer 2

Do you know the statement of green's theorem?

Answer 3

yes i found P=xy and Q=y
with dp/dy=x and dq/dx=0

so dq/dy - dp/dy =-x

Answer 4

so I got the double integral to be
\[\int\limits_{0}^{2\pi}\int\limits_{0}^{1}-\cos \theta r drd \theta \]

after i converted to polar coordinates

Answer 5

when i go through the integration i get 0 which i don't think is correct

Answer 6

Shouldn't x=rcos(theta)

Answer 7

yes i just wrote it as \[\cos \theta r dr d \theta \]

and since it is a -x it comes out to \[-rcos \theta dr d \theta \]

Answer 8

There should also be an r from the change of variables I believe.
Anyway have you tried evaluating the line integral normally to check. Unit circle can be parametrized easily.

Answer 9

i thought since the parametrization of the unit circle is \[x= \cos \theta\] and \[y=\sin \theta \]
with the r in front = to 1 since it is the unit circle you just have to include 1 r for the change of variables. I haven't tried evaluating the line integral normally. Not sure I remember how....lol

Answer 10

You arn't parametrizing when doing an integral over D. You should do that when actually evaluating the line integral. You need rcos(theta) because you are varying r from 0 to 1 to cover the unit disk.

Answer 11

ok i'm confused. how do you set up the integral using Green's Theorem?

Answer 12

Just like you did. It's \[-\int_{R} x\ dydx\]
where R is the unit disk. So you can forget that you are doing a line integral and just evaluate the double integral as you normally would.
\[(x,y)=(r\cos(\theta), r\sin(\theta)); dydx=rdrd\theta\]

Answer 13

so in this case what would the limits of integration be?
\[-\int\limits_{0}^{2\pi}\int\limits_{0}^{1}x dydx\] ??

Answer 14

Just like you have it. Except do the change of variables correctly.

Answer 15

so it should be \[-\int\limits\limits_{0}^{2\pi}\int\limits\limits_{0}^{1} r ^{2}\cos \theta drd \theta \]

???

Answer 16

Yes.

Answer 17

when i evaluate the integral i still get 0 which i don't think is correct

Answer 18

Why isn't it correct?

Answer 19

what does the integral represent? area? volume? and if so how can an area or volume be 0?

Answer 20

Remember that integrals evaluated signed area and volumes. There are negative areas and volumes.
You can check your answer by evaluating the original line integral.

Answer 21

Ok thanks, I will need to evaluate the line integral later, I have to run some errands. Thanks for you help.