Given $F(x,y,z) = \langle\,2xy, x^2 + z, y\,\rangle$ be a conservative vector field. What would the potential function of $F$ be?

Question

I know that a conservative vector field means that the field vector, $F$, is equal to $\nabla\,f.$

So this means that $\nabla$f would be the potential function of $F$ if I'm correct.

So I tried to take the partial derivative of each component with respect to $x, y$ and $z$ and got:

$\nabla$f = $\langle\,2y, 0, 0\,\rangle$

I believe I messed up in the above step but have no idea how to get the answer from this.

The options for the answers are:

i) $x^2 yz$

ii) $x^2 y + yz - 87$

iii) $x^6 yz$

iv) $x^2 y +yz + z$

v) $\langle\,x^2 y, yz, z\,\rangle$

Answer 1

Your second sentence is wrong: $f$ would be the potential function, not $\nabla f$.

So you need to ask "Is there a function $f$ whose partial derivative with respect to $x$ is $2xy$, and whose partial with respect to $y$ is $x^2 + z$, and whose partial with respect to $z$ is $y$?" Better still, your multiple choice options give you just five things to try as candidates for $f$!

Answer 2

Hint: You can compute the gradients of the options. Alternatively, if $F = \nabla f$, then you want to solve $$\frac{\partial f}{\partial x} = 2xy, \quad \frac{\partial f}{\partial y} = x^2+z, \quad \frac{\partial f}{\partial z} = y.$$The condition ${\rm curl}\,F = \vec{0}$ is a necessary condition for the above system having a solution. Since the domain of $F$ is simply-connected, this condition is also sufficient.

Answer 3

You are seeking the function $f$ for which $\nabla f = F$ (this is the 3-dimensional analogue of an antiderivative for vector fields). So you want to assume that $$f_x = 2xy \quad\quad f_y = x^2+z \quad\quad f_z = y$$ and then integrate each to find what $f$ should be. This should get you started toward the solution.