Question

Parabolic cylinder.

Short answer

The required values are mentioned below.

$$\begin{gathered} \nabla U=\frac{1}{\sqrt{u^2+v^2}}\frac{\partial U}{\partial u}{\stackrel{⏜}{e}}_u+\frac{1}{\sqrt{u^2+v^2}}\frac{\partial U}{\partial v}{\stackrel{⏜}{e}}_v+\frac{\partial U}{\partial z}{\stackrel{⏜}{e}}_z \\ \nabla .V=\frac{1}{\sqrt{u^2+v^2}}\frac{\partial }{\partial u}\left(\sqrt{u^2+v^2}{V}_1\right)+\frac{1}{\sqrt{u^2+v^2}}\frac{\partial }{\partial u}\left(\sqrt{u^2+v^2}{V}_2\right)+\frac{\partial V_3}{\partial z} \\ {\nabla }^{2}U=\frac{1}{\sqrt{u^2+v^2}}\frac{{\partial }^{2}U}{\partial u^2}+\frac{1}{\sqrt{u^2+v^2}}\frac{{\partial }^{2}U}{\partial u^2}+\frac{{\partial }^{2}U}{\partial z^2} \\ \nabla \times V=\frac{1}{\sqrt{u^2+v^2}}\left\{\frac{\partial V_3}{\partial z}-\frac{\partial }{\partial z}\left(\sqrt{u^2+v^2}{V}_2\right)\right\}{\stackrel{⏜}{e}}_u+\frac{1}{\sqrt{u^2+v^2}}\left\{\frac{\partial }{\partial z}\left(\sqrt{u^2+v^2}{V}_1\right)-\frac{\partial V_3}{\partial z}\right\}{\stackrel{⏜}{e}}_v \\ +\frac{1}{u^2+v^2}\left\{\frac{\partial }{\partial u}\left(\sqrt{u^2+v^2}{V}_2\right)-\frac{\partial }{\partial v}\left(\sqrt{u^2+v^2}{V}_1\right)\right\}{\stackrel{⏜}{e}}_z \end{gathered}$$

Step-by-step solution

Given Information

The parabolic cylinder.

Definition of cylindrical coordinates.

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

Find the value.

The scale factors are given below.

$$\begin{gathered} h_u=\sqrt{u^2+v^2} \\ {h}_v=\sqrt{u^2+v^2} \\ {h}_z=1 \end{gathered}$$

Find $\nabla U$in Parabolic coordinates.

$$\begin{gathered} \nabla U=\frac{1}{h_u}\frac{\partial U}{\partial u}{\stackrel{⏜}{e}}_u+\frac{1}{h_v}\frac{\partial U}{\partial v}{\stackrel{⏜}{e}}_v+\frac{1}{h_z}\frac{\partial U}{\partial z}{\stackrel{⏜}{e}}_z \\ \nabla U=\frac{1}{\sqrt{u^2+v^2}}\frac{\partial U}{\partial u}{\stackrel{⏜}{e}}_u+\frac{1}{\sqrt{u^2+v^2}}\frac{\partial U}{\partial v}{\stackrel{⏜}{e}}_v+\frac{\partial U}{\partial z}{\stackrel{⏜}{e}}_z \end{gathered}$$

Find the divergence of V.

$$\begin{gathered} \nabla .V=\frac{1}{u^2+v^2}\left\{\frac{\partial }{\partial u}\left(\sqrt{u^2+v^2}{V}_1\right)+\frac{\partial }{\partial v}\left(\sqrt{u^2+v^2}{V}_2\right)+\frac{\partial }{\partial z}\left(\left(u^2+v^2\right)V_3\right)\right\} \\ \nabla .V=\frac{1}{u^2+v^2}\frac{\partial }{\partial u}\left(\sqrt{u^2+v^2}{V}_1\right)+\frac{1}{u^2+v^2}\frac{\partial }{\partial v}\left(\sqrt{u^2+v^2}{V}_2\right)+\frac{\partial V_3}{\partial z} \end{gathered}$$

Find the Laplacian of U.

role="math" localid="1659269426501" $$\begin{gathered} {\nabla }^{2}U=\frac{1}{u^2+v^2}\left\{\frac{\partial }{\partial u}\left(\frac{\partial U}{\partial u}\right)+\frac{\partial }{\partial v}\left(\frac{\partial U}{\partial u}\right)+\frac{\partial }{\partial z}\left(\left(u^2+v^2\right)\frac{\partial U}{\partial z}\right)\right\} \\ {\nabla }^{2}U=\frac{1}{u^2+v^2}\frac{{\partial }^{2}U}{\partial u^2}+\frac{1}{u^2+v^2}\frac{{\partial }^{2}U}{\partial v^2}+\frac{{\partial }^{2}U}{\partial z^2} \end{gathered}$$

Find curl of V.

$$\begin{gathered} \nabla \times V=\frac{1}{\sqrt{u^2+v^2}}\left\{\frac{\partial V_3}{\partial v}-\frac{\partial }{\partial z}\left(\sqrt{u^2+v^2}{V}_2\right)\right\}{\stackrel{⏜}{e}}_u+\frac{1}{\sqrt{u^2+v^2}}\left\{\frac{\partial }{\partial z}\left(\sqrt{u^2+v^2}{V}_1\right)-\frac{\partial V_3}{\partial u}\right\}{\stackrel{⏜}{e}}_v \\ +\frac{1}{u^2+v^2}\left\{\frac{\partial }{\partial u}\left(\sqrt{u^2+v^2}{V}_2\right)-\frac{\partial }{\partial v}\left(\sqrt{u^2+v^2}{V}_1\right)\right\}{\stackrel{⏜}{e}}_z \end{gathered}$$

Hence, the required values are mentioned below.

$$\begin{gathered} \nabla U=\frac{1}{\sqrt{u^2+v^2}}\frac{\partial U}{\partial u}{\stackrel{⏜}{e}}_u+\frac{1}{\sqrt{u^2+v^2}}\frac{\partial U}{\partial v}{\stackrel{⏜}{e}}_v+\frac{\partial U}{\partial z}{\stackrel{⏜}{e}}_z \\ \nabla .V=\frac{1}{\sqrt{u^2+v^2}}\frac{\partial }{\partial u}\left(\sqrt{u^2+v^2}{V}_1\right)+\frac{1}{u^2+v^2}\frac{\partial }{\partial v}\left(\sqrt{u^2+v^2}{V}_2\right)+\frac{\partial V_3}{\partial z} \\ {\nabla }^{2}U=\frac{1}{\sqrt{u^2+v^2}}\frac{{\partial }^{2}U}{\partial u^2}+\frac{1}{u^2+v^2}\frac{{\partial }^{2}U}{\partial v^2}+\frac{{\partial }^{2}U}{\partial z^2} \\ \nabla \times V=\left[\begin{array}{c}\frac{1}{\sqrt{u^2+v^2}}\left\{\frac{\partial V_3}{\partial v}-\frac{\partial }{\partial z}\left(\sqrt{u^2+v^2}{V}_2\right)\right\}{\stackrel{⏜}{e}}_u+\frac{1}{\sqrt{u^2+v^2}}\left\{\frac{\partial }{\partial z}\left(\sqrt{u^2+v^2}{V}_1\right)\right\}{\stackrel{⏜}{e}}_v\\ +\frac{1}{\sqrt{u^2+v^2}}\left\{\frac{\partial }{\partial u}\left(\sqrt{u^2+v^2}{V}_2\right)-\frac{\partial }{\partial v}\left(\sqrt{u^2+v^2}{V}_1\right)\right\}{\stackrel{⏜}{e}}_z\end{array}\right] \end{gathered}$$