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December 21, 2014

December 21, 2014

Posted by **Donny** on Monday, November 28, 2011 at 9:26pm.

- vector calculus -
**Steve**, Tuesday, November 29, 2011 at 12:20amIf F is a conservative field, then there will be a G such that F = ∇G.

If we let G(x,y) = 1/2 (x^2 + 2xy - y^2) then

∇G = 1/2 (2x dx + 2y dx + 2x dy - 2y dy)

= (x+y)dx + (x-y)dy

So, F is a conservative field and the line integral is path-independent.

So, just evaluate G at the limits of integration

G(1,2) = 1/2 (1+4-4) = 1/2

G(0,0) = 0

So, the integral evaluates to just 1/2

If I recall my vector calculus correctly...

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