Kierra Griffith

2022-11-28

Using spherical coordinates,
$x=p\mathrm{sin}\left(\varphi \right)\mathrm{cos}\left(\theta \right)$
$y=p\mathrm{sin}\left(\varphi \right)\mathrm{sin}\left(\theta \right)$
$z=p\mathrm{cos}\left(\theta \right)$
Show that $\left({x}^{2}+{y}^{2}+{z}^{2}{\right)}^{-3/2}={p}^{-3}$

polarnost0Yi

LHS $=\left({x}^{2}+{y}^{2}+{z}^{2}{\right)}^{-3/2}$
Put the value of $x,y,z$
$=\left(\left(p\mathrm{sin}\varphi \mathrm{cos}\theta {\right)}^{2}+\left(p\mathrm{sin}\varphi \mathrm{sin}\theta {\right)}^{2}+\left(p\mathrm{cos}\varphi {\right)}^{2}{\right)}^{-3/2}$
$=\left[{p}^{2}\left({\mathrm{sin}}^{2}\varphi {\mathrm{cos}}^{2}\theta +{\mathrm{sin}}^{2}\varphi {\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{2}\varphi \right){\right]}^{-3/2}$
$=\left({p}^{2}{\right)}^{-3/2}\left({\mathrm{sin}}^{2}\varphi \left({\mathrm{cos}}^{2}\theta +{\mathrm{sin}}^{2}\theta \right)+{\mathrm{cos}}^{2}\left(\varphi \right){\right)}^{-3/2}$
$={p}^{-3}\left({\mathrm{sin}}^{2}\varphi +{\mathrm{cos}}^{2}\left(\varphi \right){\right)}^{-3/2}$
$={p}^{-3}\left(1{\right)}^{-3/2}$
$={p}^{-3}=R.H.S.$

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