2021-12-28

Transform the point P (-2, 6, 3) in spherical coordinate system.

karton

Expert

At point P: x = -2, y = 6, z = 3. Hence

$\rho =\sqrt{{x}^{2}+{y}^{2}}=\sqrt{4+36}=6.32$

$\varphi ={\mathrm{tan}}^{-1}\frac{y}{x}={\mathrm{tan}}^{-1}\frac{6}{-2}={108.43}^{\circ }$

$r=\sqrt{{x}^{2}+{y}^{2}+{z}^{2}}=\sqrt{4+36+9}=7$

$\theta ={\mathrm{tan}}^{-1}=\frac{\sqrt{{x}^{2}+{y}^{2}}}{z}={\mathrm{tan}}^{-1}\frac{\sqrt{40}}{3}={64.42}^{\circ }$

Thus

$P\left(-2,6,3\right)=P\left(6.32,{108.43}^{\circ },3\right)=P\left(7,{64.62}^{\circ },{108.43}^{\circ }\right)$

$\left[\begin{array}{c}{A}_{r}\\ {A}_{\theta }\\ {A}_{\varphi }\end{array}\right]=\left[\begin{array}{ccc}\mathrm{sin}\theta \mathrm{cos}\varphi & \mathrm{sin}\theta \mathrm{sin}\varphi & \mathrm{cos}\varphi \\ \mathrm{cos}\theta \mathrm{cos}\varphi & \mathrm{cos}\theta \mathrm{sin}\varphi & -\mathrm{sin}\theta \\ -\mathrm{sin}\theta & \mathrm{cos}\theta & 0\end{array}\right]\left[\begin{array}{c}y\\ x+z\\ 0\end{array}\right]$

or

${A}_{r}=y\mathrm{sin}\theta \mathrm{cos}\varphi +\left(x+z\right)\mathrm{sin}\theta \mathrm{sin}\varphi$

${A}_{\theta }=y\mathrm{cos}\theta \mathrm{cos}\varphi +\left(x+z\right)\mathrm{cos}\theta \mathrm{sin}\varphi$

${A}_{\varphi }=-y\mathrm{sin}\varphi +\left(x+z\right)\mathrm{cos}\varphi$

But $x=r\mathrm{sin}\theta \mathrm{cos}\varphi ,y=r\mathrm{sin}\theta \mathrm{sin}\varphi ,$ and $z=r\mathrm{cos}\theta$ . Substituting these yields

$A=\left({A}_{r},{A}_{\theta },{A}_{\varphi }\right)$

$=r\left[{\mathrm{sin}}^{2}\theta \mathrm{cos}\varphi \mathrm{sin}\varphi +\left(\mathrm{sin}\theta \mathrm{cos}\varphi +\mathrm{cos}\theta \right)\mathrm{sin}\theta \mathrm{sin}\varphi \right]{a}_{r}+r\left[\mathrm{sin}\theta \mathrm{cos}\theta \mathrm{sin}\varphi \mathrm{cos}\varphi +\left(\mathrm{sin}\theta \mathrm{cos}\varphi +\mathrm{cos}\theta \right)\mathrm{cos}\theta \mathrm{sin}\varphi \right]{a}_{\theta }+r\left[-\mathrm{sin}\theta {\mathrm{sin}}^{2}\varphi +\left(\mathrm{sin}\theta \mathrm{cos}\varphi +\mathrm{cos}\theta \right)\mathrm{cos}\varphi \right]{a}_{\varphi }$

At P

$r=7,\mathrm{tan}\varphi =\frac{6}{-2},\mathrm{tan}\theta =\frac{\sqrt{40}}{3}$

Hence, $\mathrm{cos}\varphi =\frac{-2}{\sqrt{40}},\mathrm{sin}\varphi =\frac{6}{\sqrt{40}},\mathrm{cos}\theta =\frac{3}{7},\mathrm{sin}\theta =\frac{\sqrt{40}}{7}$

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