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)

\begin{align} \quad \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} \end{align}

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)

\begin{align} \quad \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} \quad , \quad \frac{\partial P}{\partial z} = \frac{\partial R}{\partial x} \quad , \quad \frac{\partial Q}{\partial z} = \frac{\partial R}{\partial y} \end{align}

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)

\begin{align} \quad \frac{\partial \phi}{\partial x} = P(x, y, z) = xy - \sin z \\ \quad \frac{\partial \phi}{\partial y} = Q(x, y, z) = \frac{1}{2}x^2 - \frac{e^y}{z} \\ \quad \frac{\partial \phi}{\partial z} = R(x, y, z) = \frac{e^y}{z^2} - x \cos z \end{align}

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)

\begin{align} \quad \int \frac{\partial \phi}{\partial x} \: dx = \int (xy - \sin z) \: dx \\ \quad \phi (x, y, z) = \frac{x^2y}{2} - x \sin z + h(y, z) \end{align}

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

(5)

\begin{align} \quad \frac{\partial}{\partial y} (\phi (x, y, z)) = \frac{\partial}{\partial y} \left ( \frac{x^2y}{2} - x \sin z + h(y, z) \right ) \\ \quad \frac{\partial \phi}{\partial y} = \frac{x^2}{2} + \frac{\partial h}{\partial y} \end{align}

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)

\begin{align} \quad \frac{1}{2}x^2 - \frac{e^y}{z} = \frac{x^2}{2} + \frac{\partial h}{\partial y} \end{align}

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)

\begin{align} \quad \int \frac{\partial h}{\partial y} \: dy = - \int \frac{e^y}{z} \: dy \\ \quad h(y, z) = -\frac{e^y}{z} + g(z) \end{align}

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

(8)

\begin{align} \quad \phi (x, y, z) = \frac{1}{2} x^2 y - x \sin z + h(y, z) \\ \quad \phi (x, y, z) = \frac{1}{2} x^2 y - x \sin z - \frac{e^y}{z} + g(z) \end{align}

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

(9)

\begin{align} \quad \frac{\partial}{\partial z} \left ( \phi (x, y, z) \right ) = \frac{\partial}{\partial z} \left ( \frac{1}{2} x^2 y - x \sin z - \frac{e^y}{z} + g(z) \right ) \\ \quad \frac{\partial \phi}{\partial z} = -x \cos z + \frac{e^y}{z^2} + g'(z) \end{align}

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)

\begin{align} \quad \frac{e^y}{z^2} - x \cos z = -x \cos z + \frac{e^y}{z^2} + g'(z) \end{align}

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)

\begin{align} \quad \phi (x, y, z) = \frac{1}{2} x^2 y - x \sin z - \frac{e^y}{z} + C \end{align}