Chapter 2, Problem 10P

a)

To determine

Convert the vector A from cylindrical to rectangular coordinate system.

Explanation of Solution

Given:

  $A=\rho \sin \varphi a_p+\rho \cos a_{\varphi }+\rho \cos a_z$

Calculation:

Write the given vector A.

  $A=\rho \sin \varphi a_p+\rho \cos \varphi a_{\varphi }-2z{a}_z$

Here, $A_{\rho }=\rho \sin \varphi$, $A_{\varphi }=\rho \cos \varphi$, and $A_z=-2z$.

Write the expression for the transformation of vector A from $\left(A_x,A_y,A_z\right)\text{to}\left(A_{\rho },A_{\varphi },A_z\right)$.

  $\begin{array}{c}\left[\begin{array}{l}{A}_x\hfill \\ A_y\hfill \\ A_z\hfill \end{array}\right]={\left[\begin{array}{ccc}\cos \varphi & \sin \varphi & 0\\ -\sin \varphi & \cos \varphi & 0\\ 0& 0& 1\end{array}\right]}^{-1}\left[\begin{array}{c}{A}_{\rho }\\ A_{\varphi }\\ A_z\end{array}\right]\\ =\left[\begin{array}{c}\rho \sin \varphi \cos \varphi -\rho \cos \varphi \sin \varphi \\ \rho {\sin }^2\varphi +\rho {\cos }^2\varphi \\ -2z\end{array}\right]\\ =\left[\begin{array}{c}0\\ \rho \left({\sin }^2\varphi +{\cos }^2\varphi \right)\\ -2z\end{array}\right]\\ \left[\begin{array}{l}{A}_x\hfill \\ A_y\hfill \\ A_z\hfill \end{array}\right]=\left[\begin{array}{c}0\\ \rho \\ -2z\end{array}\right]\end{array}$        (I)

Write the relationship between cylindrical to rectangular.

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

Substitute $\rho =\sqrt{x^2+y^2}$ and $z=z$ in equation (I).

  $\begin{array}{l}\left[\begin{array}{l}{A}_x\hfill \\ A_y\hfill \\ A_z\hfill \end{array}\right]=\left[\begin{array}{c}0\\ \sqrt{x^2+y^2}\\ -2z\end{array}\right]\\ A=\sqrt{x^2+y^2}{a}_y-2z{a}_z\end{array}$

Thus, the vector A in rectangular coordinate system is $\underset{\_}{A=\sqrt{x^2+y^2}{a}_y-2z{a}_z}$.

b)

To determine

Convert the vector B from spherical to rectangular coordinate system.

Explanation of Solution

Given:

  $\text{B}=4r\cos \varphi {\text{a}}_r+r{\text{a}}_{\theta }$        (II)

Calculation:

Write the given vector B.

  $\text{B}=4r\cos \varphi {\text{a}}_r+r{\text{a}}_{\theta }$

Here, $B_r=4r\cos \varphi ,B_{\theta }=r,\text{ and}{B}_{\varphi }=0$.

  $\begin{array}{l}\left[\begin{array}{c}{B}_x\\ B_y\\ B_z\end{array}\right]=\left[\begin{array}{ccc}\sin \theta \cos \varphi & \cos \theta \cos \varphi & -\sin \varphi \\ \sin \theta \sin \varphi & \cos \theta \sin \varphi & \cos \varphi \\ \cos \theta & -\sin \theta & 0\end{array}\right]\left[\begin{array}{c}4r\cos \varphi \\ r\\ 0\end{array}\right]\\ \left[\begin{array}{c}{B}_x\\ B_y\\ B_z\end{array}\right]=\left[\begin{array}{c}4r\cos \varphi \cos \varphi \sin \theta +r\cos \theta \cos \varphi \\ 4r\cos \varphi \sin \varphi \sin \theta +r\cos \theta \sin \varphi \\ 4r\cos \varphi \cos \theta -r\sin \theta \end{array}\right]\\ \left[\begin{array}{c}{B}_x\\ B_y\\ B_z\end{array}\right]=\left[\begin{array}{c}4r{\cos }^2\varphi \sin \theta +r\cos \theta \cos \varphi \\ 4r\cos \varphi \sin \varphi \sin \theta +r\cos \theta \sin \varphi \\ 4r\cos \varphi \cos \theta -r\sin \theta \end{array}\right]\end{array}$        (III)

Write the relationship between spherical to rectangular.

  $\begin{array}{l}r=\sqrt{x^2+y^2+z^2}\\ \theta ={\tan }^{-1}\frac{\sqrt{x^2+y^2}}{z}\\ \varphi ={\tan }^{-1}\frac{y}{x}\end{array}$

Calculate the $\sin \theta$ components.

  $\begin{array}{c}\sin \theta =\frac{\rho }{r}\\ =\frac{\sqrt{x^2+y^2}}{\sqrt{x^2+y^2+z^2}}\end{array}$

Calculate the $\cos \theta$ components.

  $\begin{array}{c}\cos \theta =\frac{z}{r}\\ =\frac{\sqrt{x^2+y^2}}{\sqrt{x^2+y^2+z^2}}\end{array}$

Calculate the $\sin \varphi$ components.

  $\begin{array}{c}\sin \varphi =\frac{y}{\rho }\\ =\frac{y}{\sqrt{x^2+y^2}}\end{array}$

Calculate the $\cos \varphi$ components.

  $\begin{array}{c}\cos \varphi =\frac{x}{\rho }\\ =\frac{x}{\sqrt{x^2+y^2}}\end{array}$

Calculate the x component of B vector from the equation (II).

  $\begin{array}{c}{B}_x=4r{\cos }^2\varphi \sin \theta +r\cos \theta \cos \varphi \\ =\left\{\begin{array}{l}4\left(\sqrt{x^2+y^2+z^2}\right){\left(\frac{x}{\sqrt{x^2+y^2}}\right)}^2\left(\frac{\sqrt{x^2+y^2}}{\sqrt{x^2+y^2+z^2}}\right)+\hfill \\ \left(\sqrt{x^2+y^2+z^2}\right)\left(\frac{z}{\sqrt{x^2+y^2+z^2}}\right)\left(\frac{x}{\sqrt{x^2+y^2}}\right)+\hfill \end{array}\right\}\\ B_x=\frac{4x^2}{\sqrt{x^2+y^2}}+\frac{zx}{\sqrt{x^2+y^2}}\end{array}$

Calculate the y component of B vector from the equation (II).

  $\begin{array}{c}{B}_y=4r\cos \varphi \sin \varphi \sin \theta +r\cos \theta \sin \varphi \\ =\left\{\begin{array}{l}4\left(\sqrt{x^2+y^2+z^2}\right)\left(\frac{x}{\sqrt{x^2+y^2}}\right)\left(\frac{y}{\sqrt{x^2+y^2}}\right)\left(\frac{\sqrt{x^2+y^2}}{\sqrt{x^2+y^2+z^2}}\right)+\hfill \\ \left(\sqrt{x^2+y^2+z^2}\right)\left(\frac{z}{\sqrt{x^2+y^2+z^2}}\right)\left(\frac{y}{\sqrt{x^2+y^2}}\right)\hfill \end{array}\right\}\\ B_y=\frac{4xy}{\sqrt{x^2+y^2}}+\frac{zy}{\sqrt{x^2+y^2}}\end{array}$

Calculate the z component of B vector from the equation (II).

  $\begin{array}{c}{B}_z=4r\cos \varphi \cos \theta -r\sin \theta \\ =\left\{\begin{array}{l}4\left(\sqrt{x^2+y^2+z^2}\right)\left(\frac{x}{\sqrt{x^2+y^2}}\right)\left(\frac{z}{\sqrt{x^2+y^2+z^2}}\right)\hfill \\ -\left(\sqrt{x^2+y^2+z^2}\right)\left(\frac{\sqrt{x^2+y^2}}{\sqrt{x^2+y^2+z^2}}\right)\hfill \end{array}\right\}\\ B_z=\frac{4xz}{\sqrt{x^2+y^2}}-\sqrt{x^2+y^2}\end{array}$

Substitute the component of Vector B in the equation (II).

  $B=\frac{1}{\sqrt{x^2+y^2}}\left(\left(4x^2+zx\right)a_x+(4xy+zy)a_y+\left(4xz-x^2-y^2\right)a_z\right)$.

Thus, the vector B in rectangular coordinates system is $\underset{\_}{B=\frac{1}{\sqrt{x^2+y^2}}\left(\left(4x^2+zx\right)a_x+(4xy+zy)a_y+\left(4xz-x^2-y^2\right)a_z\right)}$.