${\partial }_x{G}_1$

$\=f\+x \left(f_r r_x\+f_{\mathrm{\theta }} {\mathrm{\theta }}_x\right)-y \left(g_r r_x\+g_{\mathrm{\theta }} {\mathrm{\theta }}_x\right)$

$\=f\+\left(x f_r-y g_r\right)r_x\+\left(x f_{\mathrm{\θ}}-y g_{\mathrm{\θ}}\right){\mathrm{\θ}}_x$

$\=f\+\left(r \cos \left(\mathrm{\θ}\right) f_r-r \sin \left(\mathrm{\θ}\right) g_r\right)\cos \left(\mathrm{\θ}\right)\+\left(r \cos \left(\mathrm{\θ}\right) f_{\mathrm{\θ}}-r \sin \left(\mathrm{\θ}\right) g_{\mathrm{\θ}}\right)\left(-\frac{\sin \left(\mathrm{\θ}\right)}{r}\right)$

$\=f\+r \left({\cos }^2\left(\mathrm{\θ}\right) f_r-\sin \left(\mathrm{\θ}\right)\cos \left(\mathrm{\θ}\right) g_r\right)-\left(\sin \left(\mathrm{\θ}\right)\cos \left(\mathrm{\θ}\right)f_{\mathrm{\θ}}-{\sin }^2\left(\mathrm{\θ}\right)g_{\mathrm{\θ}}\right)$

${\partial }_y{G}_2$

$\=f\+y \left(f_r r_y\+f_{\mathrm{\theta }} {\mathrm{\theta }}_y\right)\+x \left(g_r r_y\+g_{\mathrm{\theta }} {\mathrm{\theta }}_y\right)$

$\=f\+\left(y f_r\+x g_r\right)r_y\+\left(y f_{\mathrm{\θ}}\+x g_{\mathrm{\θ}}\right){\mathrm{\θ}}_y$

$\=f\+\left(r \sin \left(\mathrm{\θ}\right)f_r\+r \cos \left(\mathrm{\θ}\right)g_r\right) \sin \left(\mathrm{\θ}\right)\+\left(r \sin \left(\mathrm{\θ}\right)f_{\mathrm{\θ}}\+r \cos \left(\mathrm{\θ}\right)g_{\mathrm{\θ}}\right)\frac{\cos \left(\mathrm{\θ}\right)}{r}$

$\=f\+r \left({\sin }^2\left(\mathrm{\θ}\right)f_r\+\sin \left(\mathrm{\θ}\right)\cos \left(\mathrm{\θ}\right)g_r\right)\+\sin \left(\mathrm{\θ}\right)\cos \left(\mathrm{\θ}\right)f_{\mathrm{\θ}}\+{\cos }^2\left(\mathrm{\θ}\right)g_{\mathrm{\θ}}$

$\nabla ·\mathbf{G}$

$\=\left(r {\partial }_x{G}_1-G_1\cos \left(\mathrm{\θ}\right)\+r {\partial }_y{G}_2\+G_2\sin \left(\mathrm{\θ}\right)\right)\/r^2$

$\=\left({\partial }_x{G}_1\+{\partial }_y{G}_2\right)\/r-\left(G_1\cos \left(\mathrm{\θ}\right)\+G_2\sin \left(\mathrm{\θ}\right)\right)\/r^2$

$\=\frac{2 f\+r f_r\+g_{\mathrm{\θ}}}{r}-\frac{\left(r \cos \left(\mathrm{\theta }\right)f-r \sin \left(\mathrm{\theta }\right)g\right)\cos \left(\mathrm{\theta }\right)\+\left(r \sin \left(\mathrm{\theta }\right)f\+r \cos \left(\mathrm{\theta }\right)g\right)\sin \left(\mathrm{\theta }\right)}{r^2}$

$\=\frac{2 f\+r f_r\+g_{\mathrm{\theta }}}{r}-\frac{f}{r}$

$\=\frac{f}{r}\+f_r\+\frac{g_{\mathrm{\θ}}}{r}$

Initialize

Loading Student:-VectorCalculus

Define the polar field F

$⟨f\left(r\,\mathrm{\θ}\right)\,g\left(r\,\mathrm{\θ}\right)⟩$ = $\left[\begin{array}{c}f\left(r\,\mathrm{\theta }\right)\\ g\left(r\,\mathrm{\theta }\right)\end{array}\right]$$\stackrel{\text{apply coordinates}}{\to }$$\left[\begin{array}{c}f\left(r\,\mathrm{\theta }\right)\\ g\left(r\,\mathrm{\theta }\right)\end{array}\right]$$\stackrel{\text{to Vector Field}}{\to }$$\left[\begin{array}{c}f\left(r\,\mathrm{\theta }\right)\\ g\left(r\,\mathrm{\theta }\right)\end{array}\right]$$\stackrel{\text{assign to a name}}{\to }$$F$$$

Change F to Cartesian coordinates

$\mathbf{F}$ = $\left[\begin{array}{c}f\left(r\,\mathrm{\theta }\right)\\ g\left(r\,\mathrm{\theta }\right)\end{array}\right]$$\stackrel{\text{change coordinates}}{\to }$$\left[\begin{array}{c}\frac{f\left(\sqrt{x^2+y^2}\,\arctan \left(y\,x\right)\right)x}{\sqrt{x^2+y^2}}-\frac{g\left(\sqrt{x^2+y^2}\,\arctan \left(y\,x\right)\right)y}{\sqrt{x^2+y^2}}\\ \frac{f\left(\sqrt{x^2+y^2}\,\arctan \left(y\,x\right)\right)y}{\sqrt{x^2+y^2}}+\frac{g\left(\sqrt{x^2+y^2}\,\arctan \left(y\,x\right)\right)x}{\sqrt{x^2+y^2}}\end{array}\right]$$\stackrel{\text{assign to a name}}{\to }$$G$$$

Obtain the divergence in Cartesian coordinates as $\nabla ·\mathbf{G}\={\partial }_x{G}_1\+{\partial }_y{G}_2$ 

$\frac{\partial }{\partial x} G_1\+\frac{\partial }{\partial y} G_2$

$\frac{\left(\frac{{\mathrm{D}}_1\left(f\right)\left(\sqrt{x^2+y^2}\,\arctan \left(y\,x\right)\right)x}{\sqrt{x^2+y^2}}-\frac{{\mathrm{D}}_2\left(f\right)\left(\sqrt{x^2+y^2}\,\arctan \left(y\,x\right)\right)y}{x^2\left(1+\frac{y^2}{x^2}\right)}\right)x}{\sqrt{x^2+y^2}}+\frac{2f\left(\sqrt{x^2+y^2}\,\arctan \left(y\,x\right)\right)}{\sqrt{x^2+y^2}}-\frac{f\left(\sqrt{x^2+y^2}\,\arctan \left(y\,x\right)\right)x^2}{{\left(x^2+y^2\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}-\frac{\left(\frac{{\mathrm{D}}_1\left(g\right)\left(\sqrt{x^2+y^2}\,\arctan \left(y\,x\right)\right)x}{\sqrt{x^2+y^2}}-\frac{{\mathrm{D}}_2\left(g\right)\left(\sqrt{x^2+y^2}\,\arctan \left(y\,x\right)\right)y}{x^2\left(1+\frac{y^2}{x^2}\right)}\right)y}{\sqrt{x^2+y^2}}+\frac{\left(\frac{{\mathrm{D}}_1\left(f\right)\left(\sqrt{x^2+y^2}\,\arctan \left(y\,x\right)\right)y}{\sqrt{x^2+y^2}}+\frac{{\mathrm{D}}_2\left(f\right)\left(\sqrt{x^2+y^2}\,\arctan \left(y\,x\right)\right)}{x\left(1+\frac{y^2}{x^2}\right)}\right)y}{\sqrt{x^2+y^2}}-\frac{f\left(\sqrt{x^2+y^2}\,\arctan \left(y\,x\right)\right)y^2}{{\left(x^2+y^2\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}+\frac{\left(\frac{{\mathrm{D}}_1\left(g\right)\left(\sqrt{x^2+y^2}\,\arctan \left(y\,x\right)\right)y}{\sqrt{x^2+y^2}}+\frac{{\mathrm{D}}_2\left(g\right)\left(\sqrt{x^2+y^2}\,\arctan \left(y\,x\right)\right)}{x\left(1+\frac{y^2}{x^2}\right)}\right)x}{\sqrt{x^2+y^2}}$

$\stackrel{\text{simplify}}{\=}$

$\frac{{\mathrm{D}}_1\left(f\right)\left(\sqrt{x^2+y^2}\,\arctan \left(y\,x\right)\right)\sqrt{x^2+y^2}+f\left(\sqrt{x^2+y^2}\,\arctan \left(y\,x\right)\right)+{\mathrm{D}}_2\left(g\right)\left(\sqrt{x^2+y^2}\,\arctan \left(y\,x\right)\right)}{\sqrt{x^2+y^2}}$

$\stackrel{\text{assign to a name}}{\to }$

$q$

Return to polar coordinates

$\genfrac{}{}{0ex}{}{q}{\phantom{x\=a}}|\genfrac{}{}{0ex}{}{\phantom{\mathrm{f(x)}}}{x\=r \cos \left(\mathrm{\θ}\right)\,y\=r \sin \left(\mathrm{\θ}\right)}$

$\frac{{\mathrm{D}}_1\left(f\right)\left(\sqrt{r^2{\cos \left(\mathrm{\theta }\right)}^2+r^2{\sin \left(\mathrm{\theta }\right)}^2}\,\arctan \left(r\sin \left(\mathrm{\theta }\right)\,r\cos \left(\mathrm{\theta }\right)\right)\right)\sqrt{r^2{\cos \left(\mathrm{\theta }\right)}^2+r^2{\sin \left(\mathrm{\theta }\right)}^2}+f\left(\sqrt{r^2{\cos \left(\mathrm{\theta }\right)}^2+r^2{\sin \left(\mathrm{\theta }\right)}^2}\,\arctan \left(r\sin \left(\mathrm{\theta }\right)\,r\cos \left(\mathrm{\theta }\right)\right)\right)+{\mathrm{D}}_2\left(g\right)\left(\sqrt{r^2{\cos \left(\mathrm{\theta }\right)}^2+r^2{\sin \left(\mathrm{\theta }\right)}^2}\,\arctan \left(r\sin \left(\mathrm{\theta }\right)\,r\cos \left(\mathrm{\theta }\right)\right)\right)}{\sqrt{r^2{\cos \left(\mathrm{\theta }\right)}^2+r^2{\sin \left(\mathrm{\theta }\right)}^2}}$

$\stackrel{\text{assuming positive}}{\to }$

$\frac{{\mathrm{D}}_1\left(f\right)\left(r\,\arctan \left(\sin \left(\mathrm{\theta }\right)\,\cos \left(\mathrm{\theta }\right)\right)\right)r+f\left(r\,\arctan \left(\sin \left(\mathrm{\theta }\right)\,\cos \left(\mathrm{\theta }\right)\right)\right)+{\mathrm{D}}_2\left(g\right)\left(r\,\arctan \left(\sin \left(\mathrm{\theta }\right)\,\cos \left(\mathrm{\theta }\right)\right)\right)}{r}$

$\stackrel{\text{assuming real range}}{\to }$

$\frac{{\mathrm{D}}_1\left(f\right)\left(r\,\mathrm{\theta }\right)r+f\left(r\,\mathrm{\theta }\right)+{\mathrm{D}}_2\left(g\right)\left(r\,\mathrm{\theta }\right)}{r}$

$\stackrel{\text{expand}}{\=}$

${\mathrm{D}}_1\left(f\right)\left(r\,\mathrm{\theta }\right)+\frac{f\left(r\,\mathrm{\theta }\right)}{r}+\frac{{\mathrm{D}}_2\left(g\right)\left(r\,\mathrm{\theta }\right)}{r}$

$\to$

$\frac{\partial }{\partial r}f\left(r\,\mathrm{\theta }\right)+\frac{f\left(r\,\mathrm{\theta }\right)}{r}+\frac{\frac{\partial }{\partial \mathrm{\theta }}g\left(r\,\mathrm{\theta }\right)}{r}$

Compare to Maple's built-in divergence operator

Common Symbols palette: Del and dot product operators
Context Panel: Evaluate and Display Inline

Context Panel: Expand≻Expand

$\nabla ·\mathbf{F}$ = $\frac{f\left(r\,\mathrm{\θ}\right)\+r\left(\frac{\partial }{\partial r}f\left(r\,\mathrm{\θ}\right)\right)\+\frac{\partial }{\partial \mathrm{\θ}}g\left(r\,\mathrm{\θ}\right)}{r}$$\stackrel{\text{expand}}{\=}$$\frac{f\left(r\,\mathrm{\θ}\right)}{r}\+\frac{\partial }{\partial r}f\left(r\,\mathrm{\θ}\right)\+\frac{\frac{\partial }{\partial \mathrm{\θ}}g\left(r\,\mathrm{\θ}\right)}{r}$$$

Alternative access to divergence operator

$\mathbf{F}$ = $\stackrel{\text{divergence}}{\to }$$\frac{f\left(r\,\mathrm{\θ}\right)\+r\left(\frac{\partial }{\partial r}f\left(r\,\mathrm{\θ}\right)\right)\+\frac{\partial }{\partial \mathrm{\θ}}g\left(r\,\mathrm{\θ}\right)}{r}$

Initialize

$$\begin{gathered} \mathrm{with}\left(\mathrm{Student}:-\mathrm{VectorCalculus}\right)\: \\ \mathrm{BasisFormat}\left(\mathrm{false}\right)\: \end{gathered}$$

Define the general polar field F

$\mathbf{F}≔\mathrm{VectorField}\left(⟨f\left(r\,\mathrm{\θ}\right)\,g\left(r\,\mathrm{\θ}\right)⟩\,\mathrm{polar}\left[r\,\mathrm{\θ}\right]\right)\:$

Change coordinates in F to Cartesian

$\mathbf{G}≔\mathrm{MapToBasis}\left(\mathbf{F}\,\mathrm{cartesian}\left[x\,y\right]\right)$

Take G as $\mathbf{G}\=G_1 \mathbf{i}\+G_2 \mathbf{j}$, and obtain its divergence: $\nabla ·\mathbf{G}\={\partial }_x{G}_1\+{\partial }_y{G}_2$ 

$q≔\mathrm{simplify}\left(\mathrm{diff}\left(G_1\,x\right)\+\mathrm{diff}\left(G_2\,y\right)\right)$

$\frac{{\mathrm{D}}_1\left(f\right)\left(\sqrt{x^2\+y^2}\,\arctan \left(y\,x\right)\right)\sqrt{x^2\+y^2}\+f\left(\sqrt{x^2\+y^2}\,\arctan \left(y\,x\right)\right)\+{\mathrm{D}}_2\left(g\right)\left(\sqrt{x^2\+y^2}\,\arctan \left(y\,x\right)\right)}{\sqrt{x^2\+y^2}}$

Change back to polar coordinates

$$\begin{gathered} \mathrm{temp}≔\mathrm{eval}\left(q\,\left[x\=r \cos \left(\mathrm{\theta }\right)\,y\=r \sin \left(\mathrm{\theta }\right)\right]\right)\: \\ \mathrm{Temp}≔\mathrm{simplify}\left(\mathrm{temp}\right) assuming r\>0\,\mathrm{\theta }\colon\colon \mathrm{RealRange}\left(\mathrm{Open}\left(-\mathrm{\pi }\right)\,\mathrm{\pi }\right)\: \\ \mathrm{expand}\left(\mathrm{convert}\left(\mathrm{Temp}\,\mathrm{diff}\right)\right) \end{gathered}$$

$\frac{\partial }{\partial r}f\left(r\,\mathrm{\θ}\right)\+\frac{f\left(r\,\mathrm{\θ}\right)}{r}\+\frac{\frac{\partial }{\partial \mathrm{\θ}}g\left(r\,\mathrm{\θ}\right)}{r}$

Compare to Maple's expression for the divergence in polar coordinates

$\mathrm{expand}\left(\mathrm{Divergence}\left(\mathbf{F}\right)\right)$

$\frac{\partial }{\partial r}f\left(r\,\mathrm{\θ}\right)\+\frac{f\left(r\,\mathrm{\θ}\right)}{r}\+\frac{\frac{\partial }{\partial \mathrm{\θ}}g\left(r\,\mathrm{\θ}\right)}{r}$