Operator

Notation

Meaning

Gradient

$\nabla f$

Measure of change in the scalar function $f$

Divergence

$\nabla ·\mathbf{F}$

Local measure of "spread" of the vector field F

Curl

$\nabla \times \mathbf{F}$

Local measure of "rotation" in the vector field F

Laplacian

${\nabla }^{2}f\= \nabla ·\left(\nabla f\right)$

The divergence (i.e., spread) of the gradient field of the scalar $f$

Table 9.3.1   Interpreting the four basic differential operators of vector calculus

Gradient: $\nabla f$

Divergence: $\nabla ·\mathbf{F}$

Laplacian: ${\nabla }^{2}f\= \nabla ·\left(\nabla f\right)$

Cartesian

$f_x \mathbf{i}\+f_y \mathbf{j}\+f_z \mathbf{k}$

$u_x\+v_y\+w_z$

$f_{\mathrm{xx}}\+f_{\mathrm{yy}}\+f_{\mathrm{zz}}$

polar

$f_r {\mathit{e}}_{\mathit{r}}\+\left(f_{\mathrm{\θ}}\/r\right){\mathit{e}}_{\mathit{r}}$

$\frac{{\left(r u\right)}_r}{r}\+\frac{v_{\mathrm{\θ}}}{r}$

$\frac{{\left(r f_r\right)}_r}{r}\+\frac{f_{\mathrm{\theta \theta }}}{r^2}$

cylindrical

$f_r {\mathit{e}}_{\mathit{r}}\+\left(f_{\mathrm{\theta }}\/r\right){\mathit{e}}_{\mathit{r}}\+f_z \mathbf{k}$

$\frac{{\left(r u\right)}_r}{r}\+\frac{v_{\mathrm{\theta }}}{r}\+w_z$

$\frac{{\left(r f_r\right)}_r}{r}\+\frac{f_{\mathrm{\θ\θ}}}{r^2}\+f_{\mathrm{zz}}$

spherical

$f_{\mathrm{\ρ}} {\mathit{e}}_{\mathbf{\ρ}}\+\frac{f_{\mathrm{\φ}}}{\mathrm{\ρ}} {\mathit{e}}_{\mathbf{\φ}}\+\frac{f_{\mathrm{\θ}}}{\mathrm{\ρ} \sin \left(\mathrm{\φ}\right)} {\mathit{e}}_{\mathbf{\θ}}$

$\frac{{\left({\mathrm{\ρ}}^{2}u\right)}_{\mathrm{\ρ}}}{{\mathrm{\ρ}}^2}\+\frac{{\left(v \sin \left(\mathrm{\φ}\right)\right)}_{\mathrm{\φ}}}{\mathrm{\ρ} \sin \left(\mathrm{\φ}\right)}\+\frac{w_{\mathrm{\θ}}}{\mathrm{\ρ} \sin \left(\mathrm{\φ}\right)}$

$\frac{{\left({\mathrm{\rho }}^2{f}_{\mathrm{\rho }}\right)}_{\mathrm{\rho }}}{{\mathrm{\rho }}^2}\+\frac{{\left(f_{\mathrm{\phi }}\sin \left(\mathrm{\phi }\right)\right)}_{\mathrm{\phi }}}{{\mathrm{\rho }}^2\sin \left(\mathrm{\phi }\right)}\+\frac{f_{\mathrm{\theta \theta }}}{{\mathrm{\rho }}^2{\sin }^2\left(\mathrm{\phi }\right)}$

Table 9.3.2   Gradient, divergence, and Laplacian in different coordinate systems

Cartesian

cylindrical

spherical

$\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ {\partial }_x& {\partial }_y& {\partial }_z\\ u& v& w\end{array}\right|\=\left[\begin{array}{c}{w}_y-v_z\\ u_z-w_x\\ v_x-u_y\end{array}\right]$

$\left|\begin{array}{ccc}\frac{{\mathit{e}}_{\mathit{r}}}{r}& {\mathit{e}}_{\mathbf{\θ}}& \frac{\mathbf{k}}{r}\\ {\partial }_r& {\partial }_{\mathrm{\θ}}& {\partial }_z\\ u& r v& w\end{array}\right|\=\left[\begin{array}{c}\frac{w_{\mathrm{\theta }}-\(r v{\)}_z}{r}\\ u_z-w_r\\ \frac{\(r v{\)}_r-u_{\mathrm{\theta }}}{r}\end{array}\right]$

$\left|\begin{array}{ccc}\frac{{\mathit{e}}_{\mathbf{\ρ}}}{{\mathrm{\ρ}}^2\sin \left(\mathrm{\φ}\right)}& \frac{{\mathit{e}}_{\mathbf{\φ}}}{\mathrm{\ρ} \sin \left(\mathrm{\φ}\right)}& \frac{{\mathit{e}}_{\mathbf{\θ}}}{\mathrm{\ρ}}\\ {\partial }_{\mathrm{\ρ}}& {\partial }_{\mathrm{\φ}}& {\partial }_{\mathrm{\θ}}\\ u& \mathrm{\ρ} v& \mathrm{\ρ} w \sin \left(\mathrm{\θ}\right)\end{array}\right|\=\left[\begin{array}{c}\frac{w \cos \(\mathrm{\phi }\)}{\mathrm{\rho } \sin \(\mathrm{\phi }\)}\+\frac{w_{\mathrm{\phi }}}{\mathrm{\rho }}-\frac{v_{\mathrm{\θ}}}{\mathrm{\rho } \sin \(\mathrm{\phi }\)}\\ \frac{u_{\mathrm{\theta }}}{\mathrm{\rho } \sin \(\mathrm{\phi }\)}-\frac{w}{\mathrm{\rho }}-w_{\mathrm{\rho }}\\ \frac{v}{\mathrm{\rho }}\+v_{\mathrm{\rho }}-\frac{u_{\mathrm{\phi }}}{\mathrm{\rho }}\end{array}\right]$

Table 9.3.3   $\nabla \times \mathbf{F}$ in different coordinate systems

Example 9.3.1

For $f\left(x\,y\right)\=2 x^2\+3 y^2$, obtain $\nabla f$ at the point $\left(x\,y\right)\=\left(1\,1\right)$. Show that at this point, the gradient vector is orthogonal to a vector tangent to the level curve through this point.

Example 9.3.2

If $r$ and $\mathrm{\θ}$ are polar coordinates, obtain $\nabla f$ for $f\left(r\,\mathrm{\θ}\right)\=\ln \left(r\right) \tan \left(\mathrm{\θ}\right)$.

Example 9.3.3

Derive the expression for the gradient in polar coordinates.

Example 9.3.4

Integrate the gradient field for the scalar function $f\left(x\,y\right)\=x y$.

Example 9.3.5

Obtain the divergence of the Cartesian vector field $\mathbf{F}\=x y \mathbf{i}\+x\/y \mathbf{j}$.

Example 9.3.6

Change the Cartesian vector field $\mathbf{F}\=x y \mathbf{i}\+x\/y \mathbf{j}$ to polar coordinates, and obtain its divergence in those coordinates. Then express the result in Cartesian coordinates and compare to the result in Example 9.3.5.

Example 9.3.7

Derive the expression for the divergence in polar coordinates.

Example 9.3.8

Compute the curl of the Cartesian vector field $\mathbf{F}\=x y \mathbf{i}-y z \mathbf{j}-x z \mathbf{k}$.

Example 9.3.9

Change the Cartesian vector field $\mathbf{F}\=x y \mathbf{i}-y z \mathbf{j}-x z \mathbf{k}$ to cylindrical coordinates and obtain its curl in those coordinates. Then express the result in Cartesian coordinates and compare to the result in Example 9.3.8.

Example 9.3.10

Derive the expression for the curl in cylindrical coordinates.

Example 9.3.11

Compute the curl of the Cartesian vector field $\mathbf{F}\=y \mathbf{i}-z \mathbf{j}-x \mathbf{k}$, then change to spherical coordinates and again obtain the curl. Transform the result back to Cartesian coordinates and compare to the original results.

Example 9.3.12

Derive the expression for the curl in spherical coordinates.

Example 9.3.13

Obtain the Laplacian of the scalar function $f\left(x\,y\right)\=x\/y\+y\/x$. Show that it is equivalent to the divergence of the gradient.

Example 9.3.14

Change $f\left(x\,y\right)\=x\/y\+y\/x$ to polar coordinates, and in those coordinates obtain the Laplacian. Then show the results are equivalent to those in Example 9.3.13.

Example 9.3.15

Derive the expression for the Laplacian in polar coordinates.

Example 9.3.16

Graph the vector fields ${\mathbf{F}}_1\=x \mathbf{i}\+y \mathbf{j}$ and ${\mathbf{F}}_2\={\mathbf{F}}_1\/\left(x^2\+y^2\right)$. Show that $\nabla ·{\mathbf{F}}_1\=2$, but $\nabla ·{\mathbf{F}}_2\=0$, even though the arrows of both fields are radially outward, suggesting "divergence" for both.

Example 9.3.17

Graph the vector fields ${\mathbf{F}}_1\=y \mathbf{i}-x \mathbf{j}$ and ${\mathbf{F}}_2\={\mathbf{F}}_1\/\left(x^2\+y^2\right)$. Show that $\nabla \times {\mathbf{F}}_1\=-2 \mathbf{k}$, but $\nabla \times {\mathbf{F}}_2\=\mathbf{0}$, even though the arrows of both fields are tangent to concentric circles, suggesting "rotation" for both.