91Ó°ÊÓ

Given are the spherical coordinates.

The coordinate system primarily utilized in three-dimensional systems is the spherical coordinates of the system. The spherical coordinate system is used to find the surface area in three-dimensional space.

The scale factors are given below.

$\begin{array}{c}{h}_{\mathrm{ϕ}}=r\sin \mathrm{θ}\\ h_r=1\\ h_{\mathrm{θ}}=r\end{array}$

Formula states that $∇U=\frac{∂U}{∂r}{\widehat{e}}_r+\frac{1}{r}\frac{∂U}{∂\mathrm{θ}}{\widehat{e}}_{\mathrm{θ}}+\frac{1}{r\sin \mathrm{θ}}\frac{∂U}{∂\mathrm{ϕ}}{\widehat{e}}_{\mathrm{ϕ}}$.

Find the divergence of V.

$\begin{array}{c}∇â‹\dots \mathbf{V}=\frac{1}{r^2\sin \mathrm{θ}}\{\frac{∂}{∂r}\left(r^2\sin \mathrm{θ}{V}_1\right)+\frac{∂}{∂\mathrm{θ}}\left(r\sin \mathrm{θ}{V}_2\right)+\frac{∂}{∂\mathrm{Ï•}}\left(r{V}_3\right)\}\\ ∇â‹\dots \mathbf{V}=\frac{1}{r^2}\frac{∂}{∂r}\left(r^2{V}_1\right)+\frac{1}{r\sin \mathrm{θ}}\frac{∂}{∂\mathrm{θ}}\left(\sin \mathrm{θ}{V}_2\right)+\frac{1}{r\sin \mathrm{θ}}\frac{∂V_3}{∂\mathrm{Ï•}}\end{array}$

Calculate the Laplacian of U

$\begin{array}{c}{∇}^{2}U=\frac{1}{r^2\sin \mathrm{θ}}\{\frac{∂}{∂r}\left(r^2\sin \mathrm{θ}\frac{∂U}{∂r}\right)+\frac{∂}{∂\mathrm{θ}}\left(\frac{r\sin \mathrm{θ}}{r}\frac{∂U}{∂\mathrm{θ}}\right)+\frac{∂}{∂\mathrm{ϕ}}\left(\frac{r}{r\sin \mathrm{θ}}\frac{∂U}{∂\mathrm{ϕ}}\right)\}\\ {∇}^{2}U=\frac{1}{r^2}\frac{∂}{∂r}\left(r^2\frac{∂U}{∂r}\right)+\frac{1}{r^2\sin \mathrm{θ}}\frac{∂}{∂\mathrm{θ}}\left(\sin \mathrm{θ}\frac{∂U}{∂\mathrm{θ}}\right)+\frac{1}{r^2{\sin }^2\mathrm{θ}}\frac{{∂}^{2}U}{∂{\mathrm{ϕ}}^2}\end{array}$

The curl of V is mentioned below.

$∇×\mathbf{V}=\frac{1}{r^2\sin \mathrm{θ}}\{\frac{∂}{∂\mathrm{θ}}\left(r\sin \mathrm{θ}{V}_3\right)−\frac{∂}{∂\mathrm{ϕ}}\left(r{V}_2\right)\}{\widehat{e}}_r+\frac{1}{r\sin \mathrm{θ}}\{\frac{∂}{∂\mathrm{ϕ}}\left(V_1\right)−\frac{∂}{∂r}\left(r\sin \mathrm{θ}{V}_3\right)\}{\widehat{e}}_{\mathrm{θ}}+\frac{1}{r}\{\frac{∂}{∂r}\left(r{V}_2\right)−\frac{∂}{∂\mathrm{θ}}\left(V_1\right)\}{\widehat{e}}_{\mathrm{ϕ}}$

The values are mentioned below.

role="math" localid="1664343316129" $\begin{array}{c}∇â‹\dots \mathbf{V}=\frac{1}{r^2}\frac{∂}{∂r}\left(r^2{V}_1\right)+\frac{1}{r\sin \mathrm{θ}}\frac{∂}{∂\mathrm{θ}}\left(\sin \mathrm{θ}{V}_2\right)+\frac{1}{r\sin \mathrm{θ}}\frac{∂V_3}{∂\mathrm{Ï•}}\\ {∇}^{2}U=\frac{1}{r^2}\frac{∂}{∂r}\left(r^2\frac{∂U}{∂r}\right)+\frac{1}{r^2\sin \mathrm{θ}}\frac{∂}{∂\mathrm{θ}}\left(\sin \mathrm{θ}\frac{∂U}{∂\mathrm{θ}}\right)+\frac{1}{r^2{\sin }^2\mathrm{θ}}\frac{{∂}^{2}U}{∂{\mathrm{Ï•}}^2}\\ ∇×\mathbf{V}=\frac{1}{r^2\sin \mathrm{θ}}\{\frac{∂}{∂\mathrm{θ}}\left(r\sin \mathrm{θ}{V}_3\right)−\frac{∂}{∂\mathrm{Ï•}}\left(r{V}_2\right)\}{\widehat{e}}_r+\frac{1}{r\sin \mathrm{θ}}\{\frac{∂}{∂\mathrm{Ï•}}\left(V_1\right)−\frac{∂}{∂r}\left(r\sin \mathrm{θ}{V}_3\right)\}{\widehat{e}}_{\mathrm{θ}}+\frac{1}{r}\{\frac{∂}{∂r}\left(r{V}_2\right)−\frac{∂}{∂\mathrm{θ}}\left(V_1\right)\}{\widehat{e}}_{\mathrm{Ï•}}\end{array}$