Multivariable calculus and vector analysis

Path-independence example in three dimensions

In three dimensions, we can write a vector field as

$$\begin{array}{lll}\hfill \mathbf{F}\left(x,y,z\right)=\left(F_1\left(x,y,z\right),F_2\left(x,y,z\right),F_3\left(x,y,z\right)\right).& & \hfill \end{array}$$

Assume that $\mathbf{F}$ is defined everywhere in ${\mathbf{R}}^3$, except for possibly a finite number of points. Then, $\mathbf{F}$ is path-independent if and only if the curl of $\mathbf{F}$ is zero, i.e., if

$$\begin{array}{lll}\hfill \nabla \times \mathbf{F}=\left(\frac{\partial F_3}{\partial y}-\frac{\partial F_2}{\partial z},\frac{\partial F_1}{\partial z}-\frac{\partial F_3}{\partial x},\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}\right)=\left(0,0,0\right)& & \hfill \end{array}$$

We can write this condition as

$$\begin{array}{lll}\hfill \frac{\partial F_1}{\partial y}=\frac{\partial F_2}{\partial x},\frac{\partial F_1}{\partial z}=\frac{\partial F_3}{\partial x},\text{ and }\frac{\partial F_2}{\partial z}=\frac{\partial F_3}{\partial y}.& & \hfill \end{array}$$

It’s the same thing.

Here are two examples.

Example 1: ${\int \; <!--nolimits\; -->}_C\left(x^2-z{e}^y\right)dx+\left(y^3-xz{e}^y\right)dy+\left(z^4-x{e}^y\right)dz$.
Is the integral path-independent?

This corresponds to $\mathbf{F}\left(x,y,z\right)=\left(x^2-z{e}^y,y^3-xz{e}^y,z^4-x{e}^y\right)$. The vector field $\mathbf{F}$ defined on ${\mathbf{R}}^3$. So we can use the above condition.

$$\begin{array}{lll}\hfill \frac{\partial F_1}{\partial y}=\frac{\partial F_2}{\partial x}=-z{e}^y,\frac{\partial F_1}{\partial z}=\frac{\partial F_3}{\partial x}=-e^y& & \hfill \\ \hfill \frac{\partial F_2}{\partial z}=\frac{\partial F_3}{\partial y}=-x{e}^y& & \hfill \end{array}$$

The curl is zero, so the integral is path-independent.

Example 2: ${\int \; <!--nolimits\; -->}_C\left(x^2-x{e}^y\right)dx+\left(y^3-xz{e}^y\right)dy+\left(z^4-x{e}^y\right)dz$.
Is the integral path-independent?

This corresponds to $\mathbf{F}\left(x,y,z\right)=\left(x^2-x{e}^y,y^3-xz{e}^y,z^4-x{e}^y\right)$.

$$\begin{array}{lll}\hfill \frac{\partial F_1}{\partial y}=-x{e}^y\ne \frac{\partial F_2}{\partial x}=-z{e}^y,\frac{\partial F_1}{\partial z}=0\ne \frac{\partial F_3}{\partial x}=-e^y& & \hfill \\ \hfill \frac{\partial F_2}{\partial z}=\frac{\partial F_3}{\partial y}=-x{e}^y& & \hfill \end{array}$$

The curl is not zero, so the integral is path-dependent.

Note: since path-independence doesn’t depend on $\frac{\partial F_1}{\partial x}$, $\frac{\partial F_2}{\partial y}$, $\frac{\partial F_z}{\partial z}$, you can easily conclude from Example 1 that

$$\begin{array}{lll}\hfill \mathbf{F}\left(x,y,z\right)=\left(x^2-z{e}^y+x^{1000},y^3-xz{e}^y-cos{y}^{372},z^4-x{e}^y+e^{z+99}\right)& & \hfill \end{array}$$

is also path-independent.

Contents by topic

1. Geometry in higher dimensions

2. Differential Calculus

3. Integral calculus

4. Vector calculus

5. The fundamental theorems of vector calculus

5a. Green's theorem

5b. The theorem for conservative vector fields

• Path-independent or conservative vector fields
• An example of a conservative vector field
• Path-independence implies no circulation
• Understanding the conditions for path-independence
• Finding the potential function
• Path-independence example in three dimensions

5c. Stokes' theorem

5d. The divergence theorem

6. Review material