Let C be the circle \(x^2+y^2 =1\) oriented counterclockwise in the xy-plane. What is the value of the line integral
\(\oint_C(2x-y)dx +(x+3y)dy \)

- Loser66

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- Loser66

@dan815

- ganeshie8

you don't wanto use green's thm ?

- Loser66

If it helps, why not?

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## More answers

- ganeshie8

use it then

- Loser66

ok, let me try. Actually, I didn't know what the notation \(\oint\) mean. I never see it before. :)

- ganeshie8

that just means the curve is a closed loop

- Loser66

I am working on it, will tag you to check it later, ok?

- ganeshie8

ok, just need to find the curl and setup double integral

- ganeshie8

knw how to find the curl ?

- Loser66

yes, I divide it into 2 parts, \(y = \pm \sqrt{1-x^2}\) ,hence the limit for the first part will go from 0 to 1, right?

- Loser66

oh, We talk about 2 different things. ha!!

- ganeshie8

lol yeah actually we don't need to do much work here, find the curl, you will know why :)

- Loser66

|dw:1434636122182:dw|

- Loser66

ok, give me your way, please. hehehe..

- ganeshie8

find the curl first

- ganeshie8

\(\large Mdx + Ndy\)
curl = \(N_x - M_y\)

- ganeshie8

\(\large (2x-y)dx +(x+3y)dy\)
curl = ?

- Loser66

It looks like differential equation part? finding exactness, right?

- ganeshie8

\(\large (2x-y)dx +(x+3y)dy\)
\(M = 2x-y\)
\(N = x+3y\)
\(N_x = 1\)
\(M_y = -1\)
curl = \(N_x - M_y = 1-(-1) = 2\)

- Loser66

I DO lost. :)

- ganeshie8

Easy.. just take partials and subtract

- Loser66

I know, but don't know why we have to do that.

- ganeshie8

because we want to use green's thm

- Loser66

I was taught that I have to find parametric equations for x, y and replace and take a loooooooooong steps to get the answer. This is somehow different.

- Loser66

But that is the reason i post the problem here to learn the shorter way. :)

- ganeshie8

\[\oint_C(2x-y)dx +(x+3y)dy ~~=~~ \iint_R~2 dxdy = 2\iint_R~1 dxdy = ?\]

- Loser66

You still use x, y , not r and theta?

- Loser66

oh, that is perimeter of the circle?

- ganeshie8

I just applied green's theorem to convert line integral into double integral

- Loser66

Ok, I got you. Thanks a lot. Need practice more. :)

- Loser66

One more question:

- Loser66

If the curve is not a circle, we must define the limits of x,y to put into the double integral, right?

- ganeshie8

green's theorem works only if the curve is a closed loop

- Loser66

Yes,
Again, don't we have to change to polar form?

- ganeshie8

for all other cases you need to work it by parameterizing

- Loser66

YES.

- ganeshie8

you can but its not really needed here if you recall the fact that \(\iint_R ~1 ~dxdy\) represents the area of the region.

- ganeshie8

\[\oint_C(2x-y)dx +(x+3y)dy ~~=~~ \iint_R~2 dxdy = 2\color{red}{\iint_R~1 dxdy }= ?\]
that red part represents the area of the circular unit disk

- Loser66

hey, on the previous comment (and you delete it), you stated the result is 4pi, ha!! now it turns to 2pi??

- ganeshie8

that red part is 2pi
final answer is 4pi

- Loser66

how?

- Loser66

area of unit circle is pi

- ganeshie8

Oops! you're right haha

- Loser66

hihihi... ok, got you now. Much appreciate for being patient to me.

- ganeshie8

np :) maybe for practice, work it by parameterizing also

- Loser66

Yes, Sir.

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