Step 1

The objective is to express the point \(\displaystyle{P}-{2},{6},{3}\) and vector \(\displaystyle{B}={y}{a}{x}+{x}+{z}{a}{y}\) in cylindrical and spherical coordinates.

Step 2

To convert Cartesian coordinates x,y,z to cylindrical coordinates \(\displaystyle{r},\theta,{z}\)

\(\displaystyle{r}={x}{2}+{y}{2}\theta={\tan{-}}{1}{y}{x}{z}={z}\)

The point P-2,6,3 in cylindrical coordinates is,

\(\displaystyle{r}={4}+{36}={40}={210}\)

And \(\displaystyle\theta={\tan{-}}{16}-{2}=-{1.25}\)

Cylindrical coordinates is \(\displaystyle{210},-{1.25},{3}\)

To convert Cartesian coordinates x,y,z to spherical coordinates \(\displaystyle\rho\theta\phi\)

\(\displaystyle\rho={x}{2}+{y}{2}+{z}{2}\theta={\tan{-}}{1}{y}{x}\phi={\tan{-}}{1}{x}{2}+{y}{2}{z}\)

The point \(\displaystyle{P}-{2},{6},{3}\) in spherical coordinates is,

\(\displaystyle\rho={x}{2}+{y}{2}+{z}{2}={4}+{36}+{9}={49}={7}\)

And \(\displaystyle\theta={\tan{-}}{16}-{2}=-{1.25}\)

And \(\displaystyle\phi={\tan{-}}{14}+{363}={\tan{-}}{1403}={1.3034}\)

Spherical coordinates is \(\displaystyle{7},-{1.25},{1.3034}\)

Step 3

The vector \(\displaystyle{B}={y}{a}{x}+{x}+{z}{a}{y}\) in cylindrical and spherical coordinates.

In the cartesian system B at P is

\(\displaystyle{B}={6}{a}{x}+{a}{y}\)

For vector B,

\(\displaystyle{B}{x}={y}{B}{y}={x}+{z}{B}{z}={0}\)

In cylindrical system

\(\displaystyle{A}{r}={y}{\cos{\theta}}+{x}+{z}{\sin{\theta}}{A}\theta=-{y}{\sin{\theta}}+{x}+{z}{\cos{\theta}}{A}{z}={0}\)

From step 2 we have substituting values of \(\displaystyle\theta\), we get

\(\displaystyle{A}=-{0.9487}{a}{r}-{6.008}{a}\theta\)

Similarly in spherical system

\(\displaystyle{A}\rho={y}{\sin{\theta}}{\cos{\phi}}+{x}+{z}{\sin{\theta}}{\sin{\phi}}{A}\theta={y}{\cos{\theta}}{\cos{\phi}}+{x}+{z}{\cos{\theta}}{\sin{\phi}}{A}\phi=-{y}{\sin{\phi}}+{\left({x}+{z}\right)}{\cos{\phi}}\)

Since,

\(\displaystyle{x}=\rho{\sin{\theta}}{\cos{\phi}}{y}=\rho{\sin{\theta}}{\sin{\phi}}{z}=\rho{\cos{\theta}}\)

From step 2, substitute \(\displaystyle\rho,\theta,\phi\)

we get,

\(\displaystyle{A}=-{0.8571}{a}\rho-{0.4066}{a}\theta-{6.008}{a}\phi\)

The objective is to express the point \(\displaystyle{P}-{2},{6},{3}\) and vector \(\displaystyle{B}={y}{a}{x}+{x}+{z}{a}{y}\) in cylindrical and spherical coordinates.

Step 2

To convert Cartesian coordinates x,y,z to cylindrical coordinates \(\displaystyle{r},\theta,{z}\)

\(\displaystyle{r}={x}{2}+{y}{2}\theta={\tan{-}}{1}{y}{x}{z}={z}\)

The point P-2,6,3 in cylindrical coordinates is,

\(\displaystyle{r}={4}+{36}={40}={210}\)

And \(\displaystyle\theta={\tan{-}}{16}-{2}=-{1.25}\)

Cylindrical coordinates is \(\displaystyle{210},-{1.25},{3}\)

To convert Cartesian coordinates x,y,z to spherical coordinates \(\displaystyle\rho\theta\phi\)

\(\displaystyle\rho={x}{2}+{y}{2}+{z}{2}\theta={\tan{-}}{1}{y}{x}\phi={\tan{-}}{1}{x}{2}+{y}{2}{z}\)

The point \(\displaystyle{P}-{2},{6},{3}\) in spherical coordinates is,

\(\displaystyle\rho={x}{2}+{y}{2}+{z}{2}={4}+{36}+{9}={49}={7}\)

And \(\displaystyle\theta={\tan{-}}{16}-{2}=-{1.25}\)

And \(\displaystyle\phi={\tan{-}}{14}+{363}={\tan{-}}{1403}={1.3034}\)

Spherical coordinates is \(\displaystyle{7},-{1.25},{1.3034}\)

Step 3

The vector \(\displaystyle{B}={y}{a}{x}+{x}+{z}{a}{y}\) in cylindrical and spherical coordinates.

In the cartesian system B at P is

\(\displaystyle{B}={6}{a}{x}+{a}{y}\)

For vector B,

\(\displaystyle{B}{x}={y}{B}{y}={x}+{z}{B}{z}={0}\)

In cylindrical system

\(\displaystyle{A}{r}={y}{\cos{\theta}}+{x}+{z}{\sin{\theta}}{A}\theta=-{y}{\sin{\theta}}+{x}+{z}{\cos{\theta}}{A}{z}={0}\)

From step 2 we have substituting values of \(\displaystyle\theta\), we get

\(\displaystyle{A}=-{0.9487}{a}{r}-{6.008}{a}\theta\)

Similarly in spherical system

\(\displaystyle{A}\rho={y}{\sin{\theta}}{\cos{\phi}}+{x}+{z}{\sin{\theta}}{\sin{\phi}}{A}\theta={y}{\cos{\theta}}{\cos{\phi}}+{x}+{z}{\cos{\theta}}{\sin{\phi}}{A}\phi=-{y}{\sin{\phi}}+{\left({x}+{z}\right)}{\cos{\phi}}\)

Since,

\(\displaystyle{x}=\rho{\sin{\theta}}{\cos{\phi}}{y}=\rho{\sin{\theta}}{\sin{\phi}}{z}=\rho{\cos{\theta}}\)

From step 2, substitute \(\displaystyle\rho,\theta,\phi\)

we get,

\(\displaystyle{A}=-{0.8571}{a}\rho-{0.4066}{a}\theta-{6.008}{a}\phi\)