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

- schrodinger

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- Loser66

- ganeshie8

you don't wanto use green's thm ?

- Loser66

If it helps, why not?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## 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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.