# Conservative Vector Fields Examples 3

Recall from the Conservative Vector Fields page that a vector field $\mathbf{F}$ is said to be conservative on the domain $D$ if there exists a function $\phi$ known as a potential function such that $\mathbf{F} = \nabla \phi$ on $D$.

We also proved a necessary condition for a vector field on $\mathbb{R}^2$ and a vector field on $\mathbb{R}^3$ to possess. Recall that if $\mathbf{F}(x, y) = P(x, y) \vec{i} + Q(x, y) \vec{j}$ is a conservative vector field on $D$ then we must have that for all points in $D$ that:

(1)

Furthermore, recall that if $\mathbf{F}(x, y, z) = P(x, y, z) \vec{i} + Q(x, y, z) \vec{j} + R(x, y, z) \vec{k}$ is a conservative vector field on $D$ then we must have that for all points in $D$ that:

(2)

We will now look at some examples of determining whether a vector field is conservative or not.

# Example 1

**Show that the vector field $\mathbf{F}(x, y, z) = (xy – \sin z) \vec{i} + \left ( \frac{1}{2} x^2 – \frac{e^y}{z} \right ) \vec{j} + \left ( \frac{e^y}{z^2} – x \cos z \right ) \vec{k}$ is a conservative vector field on $\mathbf{R}^3 \setminus \{ (0, 0, 0) \}$ by finding a potential function $\phi$.**

If $\mathbf{F}$ is indeed conservative, then for a potential function $\phi = \phi(x, y, z)$ we should have that:

(3)

Let’s use the first equation, $\frac{\partial \phi}{\partial x} = xy – \sin z$. We will integrate both sides of this equation with respect to $x$ to get that:

(4)

We will now partial differentiate both sides of the equation above with respect to $y$ to get:

(5)

We already know that $\frac{\partial \phi}{\partial y} = \frac{1}{2}x^2 – \frac{e^y}{z}$ though, and so setting this equal to what we got above and we get that:

(6)

Therefore we see that $\frac{\partial h}{\partial y} = – \frac{e^y}{z}$. If we integrate both sides with respect to $y$ then we get that:

(7)

Substituting this into our equation for the potential function $\phi$ and we have that:

(8)

We will now take the equation above and partial differentiate it with respect to $z$ to get that:

(9)

We already know that $\frac{\partial \phi}{\partial z} = \frac{e^y}{z^2} – x \cos z$ though, and so setting this equal to what we computed above and we have that:

(10)

This equation implies that $g'(z) = 0$. Thus if we integrate $g'(z) = 0$ with respect to $z$ then we have that $g(z) = C$ for some constant $C$, and so plugging this into the equation for our potential function and we get that:

(11)

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