# Math Insight

### Testing if three-dimensional vector fields are conservative

Here are two examples of testing whether or not three-dimensional vector fields are conservative (which is also called path-independent).

#### Example 1

Does the integral $$\int_\dlc (x^2-ze^y) dx + (y^3-xze^y) dy + (z^4-xe^y) dz$$ depend on the specific path $\dlc$ takes?

Solution: The. integral is of the vector field $\dlvf(x,y,z) = (x^2-ze^y, y^3-xze^y, z^4-xe^y)$. The vector field $\dlvf$ defined on $\R^3$, which is simply connected. We can show path-independence if the curl is zero.

The partial derivatives of $\dlvf$ are \begin{align*} \pdiff{\dlvfc_1}{y} &= \pdiff{\dlvfc_2}{x} = -ze^y\\ \pdiff{\dlvfc_1}{z} &= \pdiff{\dlvfc_3}{x} = -e^y\\ \pdiff{\dlvfc_2}{z} &= \pdiff{\dlvfc_3}{y} = -xe^y. \end{align*} Therefore, the curl is zero, and $\dlvf$ is conservative. The integral of $\dlvf$ does not depend on the particular path the $\dlc$ takes, but depends only on the endpoints of the curve.

Note that path-independence doesn't depend on $\pdiff{\dlvfc_1}{x}$, $\pdiff{\dlvfc_2}{y}$, or $\pdiff{\dlvfc_3}{z}$. You can also easily conclude from this example that \begin{align*} \dlvf(x,y,z) = (x^2-ze^y+x^{1000}, y^3-xze^y-\cos y^{372}, z^4-xe^y+e^{z+99}) \end{align*} is also conservative.

#### Example 2

Does the integral $$\int_\dlc (x^2-xe^y) dx + (y^3-xze^y) dy + (z^4-xe^y) dz$$ depend on the specific path $\dlc$ takes?

Solution: The integral is of the vector field $\dlvf(x,y,z) = (x^2-xe^y, y^3-xze^y, z^4-xe^y)$. The partial derivatives of $\dlvf$ are \begin{gather*} \pdiff{\dlvfc_1}{y} = -xe^y \neq \pdiff{\dlvfc_2}{x} = -ze^y\\ \pdiff{\dlvfc_1}{z} = 0 \neq \pdiff{\dlvfc_3}{x} = -e^y\\ \pdiff{\dlvfc_2}{z} = \pdiff{\dlvfc_3}{y} = -xe^y\\ \end{gather*} The curl of $\dlvf$ is not zero, so $\dlvf$ is path-dependent. The integral will, in general, depend on the specific path of $\dlc$.