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.