91Ó°ÊÓ

The Elliptical cylinder.

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.

The velocity and scale factors are given below.

$$\begin{gathered} h_u=a\sqrt{\sin h^{2}u+\sin h^{2}v} \\ {h}_v=a\sqrt{\sin h^{2}u+\sin h^{2}v} \\ {h}_z=1 \\ \stackrel{Ë™}{r}=\stackrel{Ë™}{u}a\sqrt{\sin h^{2}u+\sin h^{2}v}{\stackrel{Áåœ}{e}}_u+\stackrel{Ë™}{v}a\sqrt{\sin h^{2}u+\sin h^{2}v}{\stackrel{Áåœ}{e}}_v+\stackrel{Ë™}{z}{\stackrel{Áåœ}{e}}_z \end{gathered}$$

The lagrangian in the parabolic coordinates are given below.

$$\begin{gathered} L=\frac{1}{2}m{\stackrel{Ë™}{r}}^2-V\left(u,v,\mathrm{Ï•}\right) \\ L=\frac{1}{2}m\left[a^2\left(\sin h^{2}u+\sin h^{2}v\right){\stackrel{Ë™}{u}}^2+a^2\left(\sin h^{2}u+\sin h^{2}v\right){\stackrel{Ë™}{v}}^2+{\mathrm{Ï•}}^2\right]-V\left(u,v,\mathrm{Ï•}\right) \end{gathered}$$

Apply Euler-lagrange equation, the above equation becomes as follows.

$$\begin{gathered} \frac{d}{dt}\left(\frac{∂L}{∂\stackrel{˙}{u}}\right)=\frac{∂L}{∂u}m{a}^2\left[\left(\sin h^{2}u+{\sin }^{2}v\right)\stackrel{¨}{u}+\sin h\cos hu{\stackrel{˙}{u}}^2\right] \\ =m{a}^2\left[-2\sin v\cos v\stackrel{˙}{u}\stackrel{˙}{v}+\sin h\cos hu{\stackrel{˙}{v}}^2\right]-\frac{∂V}{∂u} \end{gathered}$$

Solve further.

$$\begin{gathered} \frac{d}{dt}\left(\frac{∂L}{∂\stackrel{˙}{v}}\right)=\frac{∂L}{∂v}m{a}^2\left[\left(\sin h^{2}u+{\sin }^{2}v\right)\stackrel{¨}{v}+\sin v\cos v{\stackrel{˙}{v}}^2\right] \\ =m{a}^2\left[\mathrm{sincos}v{\stackrel{˙}{u}}^2-2\stackrel{˙}{u}\stackrel{˙}{v}\sin h\cos hu\right]-\frac{∂V}{∂v} \\ \frac{d}{dt}\left(\frac{∂L}{∂\stackrel{˙}{z}}\right)=\frac{∂L}{∂z}\$\$m\stackrel{¨}{z} \\ =-\frac{∂V}{∂z} \end{gathered}$$

Divide the above equation by respective scalar factors.

$$\begin{gathered} ma\left[\sqrt{\sin h^{2}u+{\sin }^{2}v}\stackrel{¨}{u}+\frac{\sin hu\cos hu}{\sqrt{\sin h^{2}u+{\sin }^{2}v}}{\stackrel{˙}{u}}^2\right]+ma\left[\frac{2\sin v\cos v}{\sqrt{\sin h^{2}u+{\sin }^{2}v}}\stackrel{˙}{u}\stackrel{˙}{v}-\frac{\sin hu\cos hu}{\sqrt{\sin h^{2}u+{\sin }^{2}v}}{\stackrel{˙}{v}}^2\right]=-\frac{1}{h_u}\frac{∂V}{∂u} \\ ma\left[\sqrt{\sin h^{2}u+{\sin }^{2}v}\stackrel{¨}{v}+\frac{\sin v\cos v}{\sqrt{\sin h^{2}u+{\sin }^{2}v}}{\stackrel{˙}{u}}^2\right]+ma\left[\frac{2\sin hv\cos hv}{\sqrt{\sin h^{2}u+{\sin }^{2}v}}\stackrel{˙}{u}\stackrel{˙}{v}+\frac{\sin v\cos hv}{\sqrt{\sin h^{2}u+{\sin }^{2}v}}{\stackrel{˙}{v}}^2\right]=-\frac{1}{h_v}\frac{∂V}{∂v} \\ m\stackrel{¨}{z}=-\frac{∂V}{∂v} \end{gathered}$$

The components of acceleration are mentioned below.

$$\begin{gathered} a_u=a\left[\sqrt{\sin h^{2}u+{\sin }^{2}v}\stackrel{¨}{u}+\frac{\sin hu\cos hu}{\sqrt{\sin h^{2}u+{\sin }^{2}v}}{\stackrel{˙}{u}}^2\right]+a\left[\frac{2\sin v\cos v}{\sqrt{\sin h^{2}u+{\sin }^{2}v}}\stackrel{˙}{u}\stackrel{˙}{v}-\frac{\sin hu\cos hu}{\sqrt{\sin h^{2}u+{\sin }^{2}v}}{\stackrel{˙}{v}}^2\right] \\ {a}_v=a\left[\sqrt{\sin h^{2}u+{\sin }^{2}v}\stackrel{¨}{v}+\frac{\sin v\cos v}{\sqrt{\sin h^{2}u+{\sin }^{2}v}}{\stackrel{˙}{u}}^2\right]+a\left[\frac{2\sin hu\cos hv}{\sqrt{\sin h^{2}u+{\sin }^{2}v}}\stackrel{˙}{u}\stackrel{˙}{v}+\frac{\sin v\cos hv}{\sqrt{\sin h^{2}u+{\sin }^{2}v}}{\stackrel{˙}{v}}^2\right] \\ {a}_z=\stackrel{¨}{z} \end{gathered}$$

The values of components of acceleration are mentioned below.

$$\begin{gathered} a_u=\stackrel{¨}{u}\sqrt{u^2+v^2}+\frac{u}{\sqrt{u^2+v^2}}{\stackrel{˙}{u}}^2+\frac{2v}{\sqrt{u^2+v^2}}\stackrel{˙}{u}\stackrel{˙}{v}-\frac{u}{\sqrt{u^2+v^2}}{\stackrel{˙}{v}}^2 \\ {a}_v=\stackrel{¨}{v}\sqrt{u^2+v^2}-\frac{u}{\sqrt{u^2+v^2}}{\stackrel{˙}{u}}^2+\frac{2u}{\sqrt{u^2+v^2}}\stackrel{˙}{u}\stackrel{˙}{v}-\frac{v}{\sqrt{u^2+v^2}}{\stackrel{˙}{v}}^2 \\ {a}_z=\stackrel{¨}{z} \end{gathered}$$