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.


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Question Number 23393 by ajfour last updated on 29/Oct/17

$S h o w t h a t v o l u m e o f a r e g i o n$ $o f s p a c e b o u n d e d b y a b o u n d a r y$ $s u r f a c e S i s V = \frac{1}{3} \underset{S}{\int \int} r \cos θ d A .$ $θ b e i n g t h e a n g l e b e t w e e n t h e$ $p o s i t i o n v e c t o r o f a p o i n t P o n$ $t h e s u r f a c e, a n d t h e o u t e r n o r m a l$ $t o t h e s u r f a c e a t P .$ $r i s t h e d i s t a n c e o f p o i n t P f r o m$ $o r i g i n .$
Commented by ajfour last updated on 07/Nov/19

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