Chapter 2, Problem 7P

a)

To determine

Convert the F vector to cylindrical and spherical system.

Explanation of Solution

Given:

$\text{F}=\frac{x{a}_x}{\sqrt{x^2+y^2+z^2}}+\frac{y{a}_y}{\sqrt{x^2+y^2+z^2}}+\frac{4a_z}{\sqrt{x^2+y^2+z^2}}$ . (I).

Calculation:

Write the expression for the vector in Cartesian coordinates system.

  $F=A_x{a}_x+A_y{a}_y+A_z{a}_z$        (II)

Compare the equation (I) and (II).

  $\begin{array}{l}{F}_x=\frac{x}{\sqrt{x^2+y^2+z^2}}\\ F_y=\frac{y}{\sqrt{x^2+y^2+z^2}}\\ F_z=\frac{z}{\sqrt{x^2+y^2+z^2}}\end{array}$

Write the expression for $A_{\rho }.A_{\varphi },\text{ and}{A}_z$ in cylindrical form.

  $\left[\begin{array}{c}{F}_{\rho }\\ F_{\varphi }\\ F_z\end{array}\right]=\left[\begin{array}{ccc}\cos \varphi & \sin \varphi & 0\\ -\sin \varphi & \cos \varphi & 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{c}{A}_x\\ A_y\\ A_z\end{array}\right]$

Substitute the obtained $A_x.A_y,\text{ and}{A}_z$ values in the above equation.

  $\left[\begin{array}{c}{F}_{\rho }\\ F_{*}\\ F_z\end{array}\right]=\left[\begin{array}{ccc}\cos \varphi & \sin \varphi & 0\\ -\sin \varphi & \cos \varphi & 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{c}\frac{x}{\sqrt{x^2+y^2+z^2}}\\ \frac{y}{\sqrt{x^2+y^2+z^2}}\\ \frac{4}{\sqrt{x^2+y^2+z^2}}\end{array}\right]$        (III)

Write the variable change of cylindrical to rectangular from table 2.1.

  $\begin{array}{l}\rho =\sqrt{x^2+y^2}\\ {\rho }^2=x^2+y^2\end{array}$

Substitute ${\rho }^2=x^2+y^2$        (III)

  $\left[\begin{array}{c}{F}_{\rho }\\ F_{*}\\ F_z\end{array}\right]=\left[\begin{array}{ccc}\cos \varphi & \sin \varphi & 0\\ -\sin \varphi & \cos \varphi & 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{c}\frac{x}{\sqrt{{\rho }^2+z^2}}\\ \frac{y}{\sqrt{{\rho }^2+z^2}}\\ \frac{4}{\sqrt{{\rho }^2+z^2}}\end{array}\right]$

Solve the above matrix to obtain the cylindrical coordinates.

  $\begin{array}{c}{F}_{\rho }=\frac{1}{\sqrt{{\rho }^2+z^2}}\left[\rho {\cos }^2\varphi +\rho {\sin }^2\varphi \right]\\ =\frac{\rho }{\sqrt{{\rho }^2+z^2}}\\ F_o=\frac{1}{\sqrt{{\rho }^2+z^2}}\left[-\rho \cos \varphi \sin \varphi +\rho \cos \varphi \sin \varphi \right]\\ =0\\ F_z=\frac{4}{\sqrt{{\rho }^2+z^2}}\end{array}$

Write the given F vector in cylindrical form.

  $\begin{array}{c}{F}_{\rho }=\frac{1}{\sqrt{{\rho }^2+z^2}}\left[\frac{\rho }{\sqrt{{\rho }^2+z^2}}{a}_{\rho }+0a_{\varphi }+\frac{4}{\sqrt{{\rho }^2+z^2}}{a}_z\right]\\ F=\frac{1}{\sqrt{{\rho }^2+z^2}}\left(\rho a_{\rho }+4a_z\right)\end{array}$

Thus, the given vector in cylindrical form is $F=\frac{1}{\sqrt{{\rho }^2+z^2}}\left(\rho a_{\rho }+4a_z\right)$.

Similarly write the matrix for the spherical form using Table 2.1.

  $\left[\begin{array}{c}{F}_r\\ F_{\theta }\\ F_{\varphi }\end{array}\right]=\left[\begin{array}{ccc}\sin \theta \cos \varphi & \sin \theta \sin \varphi & \cos \theta \\ \cos \theta \cos \varphi & \cos \theta \sin \varphi & -\sin \varphi \\ -\sin \varphi & \cos \varphi & 0\end{array}\right]\left[\begin{array}{c}\frac{x}{r}\\ \frac{y}{r}\\ \frac{4}{r}\end{array}\right]$

Solve the above matrix to obtain the spherical coordinates.

  $\begin{array}{c}{F}_r=\frac{r}{2}{\sin }^2\theta {\cos }^2\theta +\frac{r}{r}{\sin }^2\theta {\sin }^2\theta +\frac{4}{r}\cos \theta \\ ={\sin }^2\theta +\frac{4}{r}\cos \theta \\ F_{\theta }=\sin \theta \cos \theta {\cos }^2\theta +\sin \theta \cos \theta -\frac{4}{r}\cos \theta \\ =\sin \theta \cos \theta -\frac{4}{r}\sin \theta \\ F_{\varphi }=-\sin \theta \cos \varphi \sin \varphi +\sin \theta \sin \varphi \cos \varphi \\ =0\end{array}$

Write the given F vector in spherical form using the above equations.

  $F=\left({\sin }^2\theta +\frac{4}{r}\sin \theta \right)a_r+\sin \theta \left(\cos \theta -\frac{4}{r}\right)a_{\theta }$

Thus, the given vector in spherical form is $F=\left({\sin }^2\theta +\frac{4}{r}\sin \theta \right)a_r+\sin \theta \left(\cos \theta -\frac{4}{r}\right)a_{\theta }$.

b)

To determine

Convert the G vector to cylindrical and spherical system.

Explanation of Solution

Given:

$G=\left(x^2+y^2\right)\left[\frac{x{a}_x}{\sqrt{x^2+y^2+z^2}}+\frac{y{a}_y}{\sqrt{x^2+y^2+z^2}}+\frac{z{a}_x}{\sqrt{x^2+y^2+z^2}}\right]$

Calculation:

Write the given vector G.

  $\begin{array}{l}G=\left(x^2+y^2\right)\left[\frac{x{a}_x}{\sqrt{x^2+y^2+z^2}}+\frac{y{a}_y}{\sqrt{x^2+y^2+z^2}}+\frac{z{a}_x}{\sqrt{x^2+y^2+z^2}}\right]\\ G=\left[\frac{x\left(x^2+y^2\right)a_x}{\sqrt{x^2+y^2+z^2}}+\frac{y\left(x^2+y^2\right)a_y}{\sqrt{x^2+y^2+z^2}}+\frac{z\left(x^2+y^2\right)a_x}{\sqrt{x^2+y^2+z^2}}\right]\end{array}$        (IV)

Write the variable change of spherical to rectangular from table 2.1.

  $\begin{array}{l}\rho =\sqrt{x^2+y^2}\\ {\rho }^2=x^2+y^2\end{array}$

Substitute ${\rho }^2=x^2+y^2$ in equation (IV).

  $G=\left[\frac{x\left({\rho }^2\right)a_x}{\sqrt{{\rho }^2+z^2}}+\frac{y\left({\rho }^2\right)a_y}{\sqrt{{\rho }^2+z^2}}+\frac{z\left({\rho }^2\right)a_x}{\sqrt{{\rho }^2+z^2}}\right]$

Write the matrix for the cylindrical form using Table 2.1.

  $\left[\begin{array}{c}{G}_{\rho }\\ G_{\varphi }\\ G_z\end{array}\right]=\left[\begin{array}{ccc}\cos \varphi & \sin \varphi & 0\\ -\sin \varphi & \cos \varphi & 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{c}\frac{x\left({\rho }^2\right)}{\sqrt{{\rho }^2+z^2}}\\ \frac{y\left({\rho }^2\right)}{\sqrt{{\rho }^2+z^2}}\\ \frac{z\left({\rho }^2\right)}{\sqrt{{\rho }^2+z^2}}\end{array}\right]$

Solve the above matrix to obtain the cylindrical coordinates.

  $\begin{array}{c}{G}_{\rho }=\frac{{\rho }^2}{\sqrt{{\rho }^2+z^2}}\left[\rho {\cos }^2\varphi +\rho {\sin }^2\varphi \right]\\ =\frac{{\rho }^3}{\sqrt{{\rho }^2+z^2}}\end{array}$

  $G_{\varphi }=0$

  $G_z=\frac{z{\rho }^2}{\sqrt{{\rho }^2+z^2}}$

Write the given F vector in cylindrical form using the above equations.

  $\begin{array}{l}{G}_z=\frac{{\rho }^3}{\sqrt{{\rho }^2+z^2}}{a}_x+0a_y+\frac{z{\rho }^2}{\sqrt{{\rho }^2+z^2}}{a}_z\\ G_z=\frac{{\rho }^2}{\sqrt{{\rho }^2+z^2}}\left(\rho a_{\rho }+z{a}_z\right)\end{array}$

Thus, the given vector in cylindrical form $G_z=\frac{{\rho }^2}{\sqrt{{\rho }^2+z^2}}\left(\rho a_{\rho }+z{a}_z\right)$.

Similarly write the matrix for the spherical form using Table 2.1.

  $\left[\begin{array}{c}{G}_r\\ G_{\theta }\\ G_{\varphi }\end{array}\right]=\left[\begin{array}{ccc}\sin \theta \cos \varphi & \sin \theta \sin \varphi & \cos \theta \\ \cos \theta \cos \varphi & \cos \theta \sin \varphi & -\sin \varphi \\ -\sin \varphi & \cos \varphi & 0\end{array}\right]\left[\begin{array}{c}\frac{xr\sin \theta }{r}\\ \frac{y\sin \theta }{r}\\ z\sin \theta \end{array}\right]$

Solve the above matrix to obtain the spherical coordinates.

  $\begin{array}{c}{G}_r=r{\sin }^2\theta {\cos }^2\varphi +r{\sin }^2\theta {\sin }^2\varphi +r{\cos }^2\theta \sin \theta \\ =r{\sin }^3\theta +r{\cos }^2\sin \theta \\ =r\sin \theta \\ G_{\theta }=r{\sin }^2\theta \cos \theta {\cos }^2\theta +r{\sin }^2\theta \cos \left({\sin }^3\varphi \right)-r{\sin }^3\theta \cos \theta \\ =r{\sin }^2\theta \cos \theta -r{\sin }^2\cos \theta \\ =0\\ G_{\varphi }=-r{\sin }^2\theta \sin \varphi \cos \varphi \sin \varphi +r{\sin }^2\theta \cos \varphi \sin \varphi \\ =0\end{array}$

Write the given F vector in spherical form using the above equations.

  $\begin{array}{c}G=r\sin \theta a_r+0+0\\ =r\sin \theta a_r\end{array}$

Thus, the given vector in spherical form is $G=r\sin \theta a_r$.