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 $

Question

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 · Answer

@dan815

ganeshie8 · Answer

you don't wanto use green's thm ?

loser66 · Answer

If it helps, why not?

ganeshie8 · Answer

use it then

loser66 · Answer

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

ganeshie8 · Answer

that just means the curve is a closed loop

loser66 · Answer

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

ganeshie8 · Answer

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

ganeshie8 · Answer

knw how to find the curl ?

loser66 · Answer

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 · Answer

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

ganeshie8 · Answer

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

loser66 · Answer

|dw:1434636122182:dw|

loser66 · Answer

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

ganeshie8 · Answer

find the curl first

ganeshie8 · Answer

$\large Mdx + Ndy$

curl =  $N_x - M_y$

ganeshie8 · Answer

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

loser66 · Answer

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

ganeshie8 · Answer

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

loser66 · Answer

I DO lost. :)

ganeshie8 · Answer

Easy.. just take partials and subtract

loser66 · Answer

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

ganeshie8 · Answer

because we want to use green's thm

loser66 · Answer

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 · Answer

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

ganeshie8 · Answer

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

loser66 · Answer

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

loser66 · Answer

oh, that is perimeter of the circle?

ganeshie8 · Answer

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

loser66 · Answer

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

loser66 · Answer

One more question:

loser66 · Answer

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

ganeshie8 · Answer

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

loser66 · Answer

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

ganeshie8 · Answer

for all other cases you need to work it by parameterizing

loser66 · Answer

YES.

ganeshie8 · Answer

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 · Answer

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

loser66 · Answer

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

ganeshie8 · Answer

that red part is 2pi
final answer is 4pi

loser66 · Answer

how?

loser66 · Answer

area of unit circle is pi

ganeshie8 · Answer

Oops! you're right haha

loser66 · Answer

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

ganeshie8 · Answer

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

loser66 · Answer

Yes, Sir.