green's theorem.

[alamps3]

ok, so I dont know how to draw integration signs on this thing if it is possible, but let me try to ask the question anyway.

So Green's Theorem is :
line integral on closed curve C (M dx + N dy) = double integral on region R [(dN/dx - dM/dy) dA]

So, based on this, we can show that:
double integral on region R (dA) = 1/2 * line integral on closed curve C (-y dx + x dy)

Ok. I get that part, but then they say that it follows obviously that:
double integral on region R (dA) =
line integral on closed curve C (-y dx) =
line integral on closed curve C (x dy)

I dont see why this has to be true. Can someone explain it to me please. Thanks.

If anyone has the book, what I'm talking abt is in the Princeton Review Book on page 153 in the 3rd Ed and I'm not sure if its on a diff page or what in any other edition.

[kdilks]

The integral of a sum is the sum of the integrals.

[alamps3]

thanks for helping me. i think there is something that i still dont get though. what you are saying basically explains that...

(1/2)*integral(-y dx + x dy)
=(1/2)*integral(-y dx) + (1/2)*integral(x dy)

but i still don't see how it follows that ...
=(1/2)*integral(-y dx) + (1/2)*integral(x dy)
=integral(-y dx)
=integral(x dy)

why must each of the summands be equal? maybe there is something abt line integrals that i don't understand i guess. if you can point it out then that might help me out i guess. ok. thanks again.

[kdilks]

Didn't see the last two lines, that happens because of integration by parts.

For normal integration by parts, you have int from a to b u dv =uv evaluated from a to b -integral from a to b v du. You can do the same thing here, except the xy evaluated from a to b term becomes 0 since the range of integration is a loop (meaning your starting point and end point will be the same).

[alamps3]

oh ok. i think i get it now. thanks a lot