Thread: Potential Function

F(x,y) is defined by the integral:



Express F(x,y) as a function of x and y, eliminating the integral sign.

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My approach:

I found the curl of the function to be zero, which means it is path independent and that a potential function exists... I'm now working towards building f from F and finally calculating f(x,y)-f(2,pi)

Which I did and I got:


which means the final value for the integral is:



Does my answer seem correct to you?

I just thought of a way to confirm that my answer is correct. Here's how:

If you differentiate with respect to 'u' you get the first integrand and if you differentiate with respect to 'v' you get the second integrand in F(x,y)...

So must been the potential function to begin with.

Anybody wanna tap me on the shoulder?