 Given point (-2,6,3) and vector B=ya_{x}+(x+z)a_{y}, Express P and B in cylindrical and spherical coordinates. Dolly Robinson 2021-08-02 Answered
Given point $$\displaystyle{P}{\left(-{2},{6},{3}\right)}$$ and vector $$\displaystyle{B}={y}{a}_{{{x}}}+{\left({x}+{z}\right)}{a}_{{{y}}}$$, Express P and B in cylindrical and spherical coordinates. Evaluate A at P in the Cartesian, cylindrical and spherical systems.

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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,
$$\displaystyle{A}=-{0.8571}{a}\rho-{0.4066}{a}\theta-{6.008}{a}\phi$$