Kierra Griffith

2022-11-28

Using spherical coordinates,

$x=p\mathrm{sin}(\varphi )\mathrm{cos}(\theta )$

$y=p\mathrm{sin}(\varphi )\mathrm{sin}(\theta )$

$z=p\mathrm{cos}(\theta )$

Show that $({x}^{2}+{y}^{2}+{z}^{2}{)}^{-3/2}={p}^{-3}$

$x=p\mathrm{sin}(\varphi )\mathrm{cos}(\theta )$

$y=p\mathrm{sin}(\varphi )\mathrm{sin}(\theta )$

$z=p\mathrm{cos}(\theta )$

Show that $({x}^{2}+{y}^{2}+{z}^{2}{)}^{-3/2}={p}^{-3}$

polarnost0Yi

Beginner2022-11-29Added 7 answers

LHS $=({x}^{2}+{y}^{2}+{z}^{2}{)}^{-3/2}$

Put the value of $x,y,z$

$=((p\mathrm{sin}\varphi \mathrm{cos}\theta {)}^{2}+(p\mathrm{sin}\varphi \mathrm{sin}\theta {)}^{2}+(p\mathrm{cos}\varphi {)}^{2}{)}^{-3/2}$

$=[{p}^{2}({\mathrm{sin}}^{2}\varphi {\mathrm{cos}}^{2}\theta +{\mathrm{sin}}^{2}\varphi {\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{2}\varphi ){]}^{-3/2}$

$=({p}^{2}{)}^{-3/2}({\mathrm{sin}}^{2}\varphi ({\mathrm{cos}}^{2}\theta +{\mathrm{sin}}^{2}\theta )+{\mathrm{cos}}^{2}(\varphi ){)}^{-3/2}$

$={p}^{-3}({\mathrm{sin}}^{2}\varphi +{\mathrm{cos}}^{2}(\varphi ){)}^{-3/2}$

$={p}^{-3}(1{)}^{-3/2}$

$={p}^{-3}=R.H.S.$

Put the value of $x,y,z$

$=((p\mathrm{sin}\varphi \mathrm{cos}\theta {)}^{2}+(p\mathrm{sin}\varphi \mathrm{sin}\theta {)}^{2}+(p\mathrm{cos}\varphi {)}^{2}{)}^{-3/2}$

$=[{p}^{2}({\mathrm{sin}}^{2}\varphi {\mathrm{cos}}^{2}\theta +{\mathrm{sin}}^{2}\varphi {\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{2}\varphi ){]}^{-3/2}$

$=({p}^{2}{)}^{-3/2}({\mathrm{sin}}^{2}\varphi ({\mathrm{cos}}^{2}\theta +{\mathrm{sin}}^{2}\theta )+{\mathrm{cos}}^{2}(\varphi ){)}^{-3/2}$

$={p}^{-3}({\mathrm{sin}}^{2}\varphi +{\mathrm{cos}}^{2}(\varphi ){)}^{-3/2}$

$={p}^{-3}(1{)}^{-3/2}$

$={p}^{-3}=R.H.S.$

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