# Thread:

1. ## Potential Function

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

$\\<br /> \int\nolimits_{(2,\pi )}^{(x,y)} {[-2uv^{2}\sin (u^{2}v)]du+[\cos (u^{2}v)-u^{2}v\sin (u^{2}v)]dv}$

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

_______________________

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:
$f(u,v)=v[\cos (u^{2}v)]$

which means the final value for the integral is:

$y\cos \left ({x^{2}y} \right )-\pi$

Does my answer seem correct to you?

2. ## Confirmed!

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

If you differentiate $f(u,v)=v[\cos (u^{2}v)]$ 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 $f(u,v)=v[\cos (u^{2}v)]$ must been the potential function to begin with.

Anybody wanna tap me on the shoulder?

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