Loser66
  • Loser66
Let C be the circle \(x^2+y^2 =1\) oriented counterclockwise in the xy-plane. What is the value of the line integral \(\oint_C(2x-y)dx +(x+3y)dy \)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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Loser66
  • Loser66
@dan815
ganeshie8
  • ganeshie8
you don't wanto use green's thm ?
Loser66
  • Loser66
If it helps, why not?

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More answers

ganeshie8
  • ganeshie8
use it then
Loser66
  • Loser66
ok, let me try. Actually, I didn't know what the notation \(\oint\) mean. I never see it before. :)
ganeshie8
  • ganeshie8
that just means the curve is a closed loop
Loser66
  • Loser66
I am working on it, will tag you to check it later, ok?
ganeshie8
  • ganeshie8
ok, just need to find the curl and setup double integral
ganeshie8
  • ganeshie8
knw how to find the curl ?
Loser66
  • 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
  • Loser66
oh, We talk about 2 different things. ha!!
ganeshie8
  • ganeshie8
lol yeah actually we don't need to do much work here, find the curl, you will know why :)
Loser66
  • Loser66
|dw:1434636122182:dw|
Loser66
  • Loser66
ok, give me your way, please. hehehe..
ganeshie8
  • ganeshie8
find the curl first
ganeshie8
  • ganeshie8
\(\large Mdx + Ndy\) curl = \(N_x - M_y\)
ganeshie8
  • ganeshie8
\(\large (2x-y)dx +(x+3y)dy\) curl = ?
Loser66
  • Loser66
It looks like differential equation part? finding exactness, right?
ganeshie8
  • 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
  • Loser66
I DO lost. :)
ganeshie8
  • ganeshie8
Easy.. just take partials and subtract
Loser66
  • Loser66
I know, but don't know why we have to do that.
ganeshie8
  • ganeshie8
because we want to use green's thm
Loser66
  • 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
  • Loser66
But that is the reason i post the problem here to learn the shorter way. :)
ganeshie8
  • ganeshie8
\[\oint_C(2x-y)dx +(x+3y)dy ~~=~~ \iint_R~2 dxdy = 2\iint_R~1 dxdy = ?\]
Loser66
  • Loser66
You still use x, y , not r and theta?
Loser66
  • Loser66
oh, that is perimeter of the circle?
ganeshie8
  • ganeshie8
I just applied green's theorem to convert line integral into double integral
Loser66
  • Loser66
Ok, I got you. Thanks a lot. Need practice more. :)
Loser66
  • Loser66
One more question:
Loser66
  • Loser66
If the curve is not a circle, we must define the limits of x,y to put into the double integral, right?
ganeshie8
  • ganeshie8
green's theorem works only if the curve is a closed loop
Loser66
  • Loser66
Yes, Again, don't we have to change to polar form?
ganeshie8
  • ganeshie8
for all other cases you need to work it by parameterizing
Loser66
  • Loser66
YES.
ganeshie8
  • 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
  • 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
  • Loser66
hey, on the previous comment (and you delete it), you stated the result is 4pi, ha!! now it turns to 2pi??
ganeshie8
  • ganeshie8
that red part is 2pi final answer is 4pi
Loser66
  • Loser66
how?
Loser66
  • Loser66
area of unit circle is pi
ganeshie8
  • ganeshie8
Oops! you're right haha
Loser66
  • Loser66
hihihi... ok, got you now. Much appreciate for being patient to me.
ganeshie8
  • ganeshie8
np :) maybe for practice, work it by parameterizing also
Loser66
  • Loser66
Yes, Sir.

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