Showing that a three-dimensional vector field is conservative

[CactuarEnigma]

Alright, so the field is \mathbf{F} = (z^2 + 2xy,x^2,2xz)
it's a gradient only when f_x = z^2 + 2xy, f_y = x^2 and f_z = 2xz
integrate the first equation with respect to x to get f(x,y,z) = \int z^2 +2xy\,dx = xz^2 + x^2y + g(y,z)
now, f_z(x,y,z) = g_z(y,x) which is 2xz
integrate that with respect to z, g(y,z) = \int 2xz\,dz = xz^2 + h(y)
plug that into our previous expression f(x,y,z) = xz^2 + x^2y + g(y,z) = 2xz^2 + x^2y + h(y)
derive that with respect to y for f_y(x,y,z) = x^2 + h'(y) = x^2
so h'(y) = 0 and h(y) = c and we can set c = 0 and now the potential function is f(x,y,z) = 2xz^2 + x^2y which is wrong. It should be f(x,y,z) = xz^2 + x^2y. Help me please. Sorry if my LaTeX is wonky.

 

[nrqed]

Alright, so the field is \mathbf{F} = (z^2 + 2xy,x^2,2xz)
it's a gradient only when f_x = z^2 + 2xy, f_y = x^2 and f_z = 2xz
integrate the first equation with respect to x to get f(x,y,z) = \int z^2 +2xy\,dx = xz^2 + z^2y + g(y,z)

no...the second term is x^2y

 

[nrqed]

Alright, so the field is \mathbf{F} = (z^2 + 2xy,x^2,2xz)
it's a gradient only when f_x = z^2 + 2xy, f_y = x^2 and f_z = 2xz
integrate the first equation with respect to x to get f(x,y,z) = \int z^2 +2xy\,dx = xz^2 + z^2y + g(y,z)
now, f_z(x,y,z) = g_z(y,x) which is 2xz
integrate that with respect to z, g(y,z) = \int 2xz\,dz = xz^2 + h(y)
plug that into our previous expression f(x,y,z) = xz^2 + z^2y + g(y,z) = 2xz^2 + z^2y + h(y)
derive that with respect to y for F_y(x,y,z) = x^2 + h'(y) = x^2

No, your derivative is wrong. should be z^2 + h'

 

[CactuarEnigma]

Hold on, I'm trying to fix what I typed, as it doesn't match my paper here... alright, fixing the first thing fixed the second thing, so now this reflects how I did the problem.

 

[nrqed]

Alright, so the field is \mathbf{F} = (z^2 + 2xy,x^2,2xz)
it's a gradient only when f_x = z^2 + 2xy, f_y = x^2 and f_z = 2xz
integrate the first equation with respect to x to get f(x,y,z) = \int z^2 +2xy\,dx = xz^2 + x^2y + g(y,z)
now, f_z(x,y,z) = g_z(y,x) which is 2xz

no. Look at your f(x,y,z). f_z is 2xz + g_z

 

[CactuarEnigma]

no. Look at your f(x,y,z). f_z is 2xz + g_z

So then what do you do from there? f_z = 2xz + g_z and we know that f_z = 2xz so that g_z = 0 and g = h(y), is that right? That would give the right answer.

 

[nrqed]

So then what do you do from there? f_z = 2xz + g_z and we know that f_z = 2xz so that g_z = 0 and g = h(y), is that right? That would give the right answer.

yes..
You are done. the derivative with respect to y of f must give x^2. But now you know that f= xz^2 + x^2 y + h(y) so this shows that h is a constant.
You are done.