${\mathit{e}}_{\mathbf{\ρ}}\=\left[\begin{array}{c}\cos \left(\mathrm{\theta }\right)\sin \(\mathrm{\φ}\)\\ \sin \left(\mathrm{\theta }\right)\sin \left(\mathrm{\phi }\right)\\ \cos \left(\mathrm{\phi }\right)\end{array}\right]$

${\mathit{e}}_{\mathbf{\φ}}\=\left[\begin{array}{c}\cos \left(\mathrm{\theta }\right)\cos \(\mathrm{\φ}\)\\ \sin \left(\mathrm{\theta }\right)\cos \left(\mathrm{\phi }\right)\\ -\sin \left(\mathrm{\phi }\right)\end{array}\right]$

${\mathit{e}}_{\mathbf{\θ}}\=\left[\begin{array}{c}-\sin \left(\mathrm{\theta }\right)\\ \cos \left(\mathrm{\theta }\right)\\ 0\end{array}\right]$

$\mathrm{\ρ}\=\sqrt{x^2\+y^2\+z^2}$

$\cos \left(\mathrm{\φ}\right)\=z\/\mathrm{\ρ}\=z\/\sqrt{x^2\+y^2\+z^2}$

$\sin \left(\mathrm{\φ}\right)\=\sqrt{1-{\cos }^2\left(\mathrm{\φ}\right)}\=\sqrt{1-z^2\/{\mathrm{\ρ}}^2}\=\sqrt{{\mathrm{\ρ}}^2-z^2}\/\mathrm{\ρ}\=\frac{\sqrt{x^2\+y^2}}{\sqrt{x^2\+y^2\+z^2}}$

$\sin \left(\mathrm{\θ}\right)\=\frac{y}{\mathrm{\ρ} \sin \left(\mathrm{\φ}\right)}\=\frac{y}{\mathrm{\ρ} \frac{\sqrt{x^2\+y^2}}{\mathrm{\ρ}}}\=y\/\sqrt{x^2\+y^2}$

$\cos \left(\mathrm{\θ}\right)\=\frac{x}{\mathrm{\rho } \sin \left(\mathrm{\phi }\right)}\=\frac{x}{\mathrm{\rho } \frac{\sqrt{x^2\+y^2}}{\mathrm{\rho }}}\=x\/\sqrt{x^2\+y^2}$

${\mathit{e}}_{\mathbf{\rho }}\=\left[\begin{array}{c}\cos \left(\mathrm{\theta }\right)\sin \(\mathrm{\phi }\)\\ \sin \left(\mathrm{\theta }\right)\sin \left(\mathrm{\phi }\right)\\ \cos \left(\mathrm{\phi }\right)\end{array}\right]$ = $\left[\begin{array}{c}\frac{x}{\sqrt{x^2\+y^2}} \frac{\sqrt{x^2\+y^2}}{\mathrm{\ρ}}\\ \frac{y}{\sqrt{x^2\+y^2}}\frac{\sqrt{x^2\+y^2}}{\mathrm{\ρ}}\\ \frac{z}{\mathrm{\ρ}}\end{array}\right]$ = $\frac{1}{\sqrt{x^2\+y^2\+z^2}}\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$

${\mathit{e}}_{\mathbf{\phi }}\=\left[\begin{array}{c}\cos \left(\mathrm{\theta }\right)\cos \(\mathrm{\phi }\)\\ \sin \left(\mathrm{\theta }\right)\cos \left(\mathrm{\phi }\right)\\ -\sin \left(\mathrm{\phi }\right)\end{array}\right]$ = $\left[\begin{array}{c}\frac{x}{\sqrt{x^2\+y^2}} \frac{z}{\mathrm{\ρ}}\\ \frac{y}{\sqrt{x^2\+y^2}}\frac{z}{\mathrm{\ρ}}\\ -\frac{\sqrt{x^2\+y^2}}{\mathrm{\ρ}}\end{array}\right]$ = $\frac{1}{\sqrt{x^2\+y^2}\sqrt{x^2\+y^2\+z^2}}\left[\begin{array}{c}x z\\ y z\\ -\(x^2\+y^2\)\end{array}\right]$

${\mathit{e}}_{\mathbf{\theta }}\=\left[\begin{array}{c}-\sin \left(\mathrm{\theta }\right)\\ \cos \left(\mathrm{\theta }\right)\\ 0\end{array}\right]$ = $\left[\begin{array}{c}-\frac{y}{\sqrt{x^2\+y^2}}\\ \frac{x}{\sqrt{x^2\+y^2}}\\ 0\end{array}\right]$ = $\frac{1}{\sqrt{x^2\+y^2}}\left[\begin{array}{c}-y\\ x\\ 0\end{array}\right]$

$\mathbf{F}\=a {\mathit{e}}_{\mathbf{\ρ}}\+b {\mathit{e}}_{\mathbf{\φ}}\+c {\mathit{e}}_{\mathbf{\θ}}$

$\=\frac{a}{\mathrm{\ρ}} \left[\begin{array}{c}x\\ y\\ z\end{array}\right]\+\frac{b}{\mathrm{\ρ} \sqrt{x^2\+y^2}}\left[\begin{array}{c}x z\\ y z\\ -\(x^2\+y^2\)\end{array}\right]\+\frac{c}{\sqrt{x^2\+y^2}} \left[\begin{array}{c}-y\\ x\\ 0\end{array}\right]$

$\=\left[\begin{array}{c}\frac{ax}{\sqrt{x^2\+y^2\+z^2}}\+\frac{bxz}{\sqrt{x^2\+y^2}\sqrt{x^2\+y^2\+z^2}}-\frac{cy}{\sqrt{x^2\+y^2}}\\ \frac{ay}{\sqrt{x^2\+y^2\+z^2}}\+\frac{byz}{\sqrt{x^2\+y^2}\sqrt{x^2\+y^2\+z^2}}\+\frac{cx}{\sqrt{x^2\+y^2}}\\ \frac{az}{\sqrt{x^2\+y^2\+z^2}}\+\frac{b\left(-x^2-y^2\right)}{\sqrt{x^2\+y^2}\sqrt{x^2\+y^2\+z^2}}\end{array}\right]$

Initialize

Loading Student:-VectorCalculus

Define the spherical field F

$⟨a\,b\,c⟩$ = $\left[\begin{array}{c}a\\ b\\ c\end{array}\right]$$\stackrel{\text{apply coordinates}}{\to }$$\left[\begin{array}{c}a\\ b\\ c\end{array}\right]$$\stackrel{\text{to Vector Field}}{\to }$$\left[\begin{array}{c}a\\ b\\ c\end{array}\right]$$\stackrel{\text{assign to a name}}{\to }$$F$

Express F in Cartesian coordinates

$\mathbf{F}$ = $\left[\begin{array}{c}a\\ b\\ c\end{array}\right]$$\stackrel{\text{change coordinates}}{\to }$$\left[\begin{array}{c}\frac{ax}{\sqrt{x^2+y^2+z^2}}+\frac{bxz}{\sqrt{x^2+y^2}\sqrt{x^2+y^2+z^2}}-\frac{cy}{\sqrt{x^2+y^2}}\\ \frac{ay}{\sqrt{x^2+y^2+z^2}}+\frac{byz}{\sqrt{x^2+y^2}\sqrt{x^2+y^2+z^2}}+\frac{cx}{\sqrt{x^2+y^2}}\\ \frac{az}{\sqrt{x^2+y^2+z^2}}+\frac{b\left(-x^2-y^2\right)}{\sqrt{x^2+y^2}\sqrt{x^2+y^2+z^2}}\end{array}\right]$

Initialize

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

Define $\mathbf{F}\=a {\mathit{e}}_{\mathit{r}}\+b {\mathit{e}}_{\mathbf{\θ}}$ as a polar vector field

$\mathbf{F}≔\mathrm{VectorField}\left(⟨a\,b\,c⟩\,\mathrm{spherical}\left[\mathrm{\ρ}\,\mathrm{\φ}\,\mathrm{\θ}\right]\right)\:$

Change to Cartesian coordinates

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