Find a potential f for the vector function v = [ye^x, e^x, z^2].
Let F = [ye^x, e^x, z^2]
For F(x,y,z) to be conservative, we require that curl F = 0.
Curl F = (δF3/δy – δF2/δz, δF1/δz – δF3/δx, δF2/δx – δF1/δy)
Curl F = ((0 – 0), (0 – 0), (e^x – e^x)) = (0, 0, 0)
Curl F = zero
Hence F is conservative.
Let f(x,y,z) be a potential function for F s.t.
Div f = (δf/δx).i + (δf/δy).j + (δf/δz).k = F
Equating the vector components of F with the partial derivatives of f,
δf/δx = ye^x, δf/δy = e^x, δf/δz = z^2
Carrying out partial integration,
f = ʃ ye^x dx, wrt x, y,z constant
f = ye^x + g(y,z) ------------------------------------------------- (1)
f = ʃ e^x dy, wrt y, x,z constant
f = ye^x + h(x,z) ------------------------------------------------- (2)
f = ʃ z^2 dz, wrt z, x,y constant
f = z^3/3 + k(x,y) ----------------------------------------------- (3)
The expressions in (1), (2) and (3) must all be equal.
Comparing (1) with (2), g(y,z) = h(x,z) => g() = h() = G(z)
Comparing (1) and (2) now with (3), we conclude that G(z) = z^3/3 and k(x,y) = ye^x.
Our potential function then is: f(x,y,z) = ye^x + z^3/3 + const of integration.