$\nabla ·\mathbf{G}$

$\=\frac{r}{\sin \left(\mathrm{\theta }\right)}-\frac{r{\cos }^2\left(\mathrm{\theta }\right)}{\sin \left(\mathrm{\theta }\right)}-\frac{\cos \left(\mathrm{\theta }\right)}{r{\sin }^2\left(\mathrm{\theta }\right)}$

$\=\frac{r}{\sin \left(\mathrm{\theta }\right)}\left(1-{\cos }^2\left(\mathrm{\theta }\right)\right)-\frac{\cos \left(\mathrm{\θ}\right)}{r {\sin }^2\left(\mathrm{\θ}\right)}$

$\=r \sin \left(\mathrm{\θ}\right)-\frac{\cos \left(\mathrm{\θ}\right)}{r {\sin }^2\left(\mathrm{\θ}\right)}$

Initialize

Loading Student:-VectorCalculus

Define the transformation between polar and Cartesian coordinates

$\left[r\=\sqrt{x^2\+y^2}\,\mathrm{\θ}\=\arctan \left(y\,x\right)\right]$$\stackrel{\text{assign to a name}}{\to }$$T$$$

Define the vector field F and change to polar coordinates

$⟨x y\,x\/y⟩$ = $\left[\begin{array}{c}xy\\ \frac{x}{y}\end{array}\right]$$\stackrel{\text{to Vector Field}}{\to }$$\left[\begin{array}{c}xy\\ \frac{x}{y}\end{array}\right]$$\stackrel{\text{change coordinates}}{\to }$$\left[\begin{array}{c}{r}^2{\cos \left(\mathrm{\theta }\right)}^2\sin \left(\mathrm{\theta }\right)+\cos \left(\mathrm{\theta }\right)\\ -r^2\cos \left(\mathrm{\theta }\right){\sin \left(\mathrm{\theta }\right)}^2+\frac{{\cos \left(\mathrm{\theta }\right)}^2}{\sin \left(\mathrm{\theta }\right)}\end{array}\right]$$\stackrel{\text{assign to a name}}{\to }$$G$$$

Obtain the divergence in polar coordinates

$\mathbf{G}$

$\stackrel{\text{divergence}}{\to }$

$\frac{r^2{\cos \left(\mathrm{\θ}\right)}^2\sin \left(\mathrm{\θ}\right)-\cos \left(\mathrm{\θ}\right)\+r^2{\sin \left(\mathrm{\θ}\right)}^3-\frac{{\cos \left(\mathrm{\θ}\right)}^3}{{\sin \left(\mathrm{\θ}\right)}^2}}{r}$

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

$-\frac{r^2{\cos \left(\mathrm{\θ}\right)}^2\sin \left(\mathrm{\θ}\right)-\sin \left(\mathrm{\θ}\right)r^2\+\cos \left(\mathrm{\θ}\right)}{r{\sin \left(\mathrm{\θ}\right)}^2}$

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

$q$

Convert the divergence to Cartesian coordinates

$\genfrac{}{}{0ex}{}{q}{\phantom{x\=a}}|\genfrac{}{}{0ex}{}{\phantom{\mathrm{f(x)}}}{T}$ = $-\frac{\sqrt{x^2\+y^2}\left(\frac{x^{2}y}{\sqrt{x^2\+y^2}}-y\sqrt{x^2\+y^2}\+\frac{x}{\sqrt{x^2\+y^2}}\right)}{y^2}$$\stackrel{\text{expand}}{\=}$$y-\frac{x}{y^2}$$$

Initialize

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

Define the vector field F

$\mathbf{F}≔\mathrm{VectorField}\left(⟨x y\,x\/y⟩\right)\:$

Change F to polar coordinates

$\mathbf{G}≔\mathrm{MapToBasis}\left(\mathbf{F}\,\mathrm{polar}\left[r\,\mathrm{\θ}\right]\right)$

Obtain the divergence in polar coordinates

$q≔\mathrm{simplify}\left(\mathrm{Divergence}\left(G\right)\right)$

$-\frac{r^2{\cos \left(\mathrm{\θ}\right)}^2\sin \left(\mathrm{\θ}\right)-\sin \left(\mathrm{\θ}\right)r^2\+\cos \left(\mathrm{\θ}\right)}{r{\sin \left(\mathrm{\θ}\right)}^2}$

Change the divergence from polar to Cartesian coordinates

$\mathrm{expand}\left(\mathrm{eval}\left(q\,\left[r\=\sqrt{x^2\+y^2}\,\mathrm{\θ}\=\arctan \left(y\,x\right)\right]\right)\right)$ = $y-\frac{x}{y^2}$$$