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Show-that-volume-of-a-region-of-space-bounded-by-a-boundary-surface-S-is-V-1-3-S-rcos-dA-being-the-angle-between-the-position-vector-of-a-point-P-on-the-surface-and-the-outer-normal-to




Question Number 23393 by ajfour last updated on 29/Oct/17
Show that volume of a region  of space bounded by a boundary  surface S is  V= (1/3)∫∫_(S ) rcos θdA .  θ being the angle between the  position vector of a point P  on  the surface, and the outer normal  to the surface at P.  r is the distance of point P from  origin.
$${Show}\:{that}\:{volume}\:{of}\:{a}\:{region} \\ $$$${of}\:{space}\:{bounded}\:{by}\:{a}\:{boundary} \\ $$$${surface}\:{S}\:{is}\:\:{V}=\:\frac{\mathrm{1}}{\mathrm{3}}\underset{{S}\:} {\int\int}{r}\mathrm{cos}\:\theta{dA}\:. \\ $$$$\theta\:{being}\:{the}\:{angle}\:{between}\:{the} \\ $$$${position}\:{vector}\:{of}\:{a}\:{point}\:{P}\:\:{on} \\ $$$${the}\:{surface},\:{and}\:{the}\:{outer}\:{normal} \\ $$$${to}\:{the}\:{surface}\:{at}\:{P}. \\ $$$${r}\:{is}\:{the}\:{distance}\:{of}\:{point}\:{P}\:{from} \\ $$$${origin}. \\ $$
Commented by ajfour last updated on 07/Nov/19
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