# Given point P(-2, 6, 3) and vector B=ya_{x}+(x+z)a_{y}, express P and B in cylindrical and spherical coordinates. Evaluate A at P in the Cartesian, cylindrical and spherical systems.

Ava-May Nelson 2021-02-05 Answered
Given point P(-2, 6, 3) and vector $B=y{a}_{x}+\left(x+z\right){a}_{y}$, express P and B in cylindrical and spherical coordinates. Evaluate A at P in the Cartesian, cylindrical and spherical systems.
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dieseisB

Step 1 The objective is to express the point P(−2,6,3) and vector $B=y{a}_{x}+\left(x+z\right){a}_{y}$ in cylindrical and spherical coordinates. Step 2 To convert Cartesian coordinates $\left(x,y,z\right)$ to cylindrical coordinates $\left(r,\theta ,z\right)$ $r=\sqrt{{x}^{2}+{y}^{2}}$
$\theta ={\mathrm{tan}}^{-1}\left(\frac{y}{x}\right)$
$z=z$ The point $P\left(-2,6,3\right)$ in cylindrical coordinates is, $r=\sqrt{4+36}$
$=\sqrt{40}$
$=2\sqrt{10}$ And $\theta ={\mathrm{tan}}^{-1}\left(\frac{6}{-2}\right)$
$=-1.25$ Cylindrical coordinates is $\left(2\sqrt{10},-1.25,3\right)$ To convert Cartesian coordinates $\left(x,y,z\right)$ to spherical coordinates $\left(\rho ,\theta ,\varphi \right)$ $p=\sqrt{{x}^{2}+{y}^{2}+{z}^{2}}$
$\theta ={\mathrm{tan}}^{-1}\left(\frac{y}{x}\right)$
$\varphi ={\mathrm{tan}}^{-1}\left(\frac{\sqrt{{x}^{2}+{y}^{2}}}{z}\right)$ The point $P\left(-2,6,3\right)$ in spherical coordinates is, $p=\sqrt{{x}^{2}+{y}^{2}+{z}^{2}}$
$\sqrt{4+36+9}$
$=\sqrt{49}$
$=7$ And $\theta ={\mathrm{tan}}^{-1}\left(\frac{6}{-2}\right)$
$=-1.25$ And $\varphi ={\mathrm{tan}}^{-1}\left(\frac{\sqrt{4+36}}{3}\right)$
$={\mathrm{tan}}^{-1}\left(\frac{\sqrt{40}}{3}\right)$
$=1.3034$ Spherical coordinates is $\left(7,-1.25,1.3034\right)$ Step 3 The vector $B=y{a}_{x}+\left(x+z\right){a}_{y}$ in cylindrical and spherical coordinates. In the cartesian system B at P is $B=6{a}_{x}+{a}_{y}$ For vector B, ${B}_{x}=y$
${B}_{y}=x+z$
${B}_{z}=0$ In cylindrical system ${A}_{r}=y\mathrm{cos}\theta +\left(x+z\right)\mathrm{sin}\theta$
${A}_{\theta }=-y\mathrm{sin}\theta +\left(x+z\right)\mathrm{cos}\theta$