Show that S=4πr^2


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Question Number 153409 by SOMEDAVONG last updated on 07/Sep/21

$Show that S = 4 π r^{2}$
	Answered by puissant last updated on 07/Sep/21

	$S i s d e r i v a t e o f V . n o w,$ $V = \frac{4}{3} π R^{3} .$ $\frac{\partial}{\partial R} V = S \Rightarrow S = \frac{4}{3} \times 3 π R^{2}$ $\Rightarrow S = 4 π R^{2} . .$
	Commented by puissant last updated on 07/Sep/21

	$d i s k A r e a i s S = π (R^{2} - z^{2})$ $n o w V = \int_{- R}^{R} S (z) d z$ $= \int_{- R^{}}^{R} π (R^{2} - z^{2}) d z = \frac{4}{3} π R^{3} . .$ $S = \frac{\partial (\frac{4}{3} π R^{3})}{\partial R} = 4 π R^{2} i s a a r e a o f s p h e r e . .$
	Commented by MJS_new last updated on 07/Sep/21

	$great! now show that S = \frac{d^{2} \sqrt{10 (5 - \sqrt{5})}}{2}$
	Commented by puissant last updated on 07/Sep/21

	${i t}^{'} s w e i r d$
	Commented by MJS_new last updated on 07/Sep/21

	$sorry . for me it seems weird to answer this$ $question and similar ones . can you be sure$ $S is the surface of a sphere ? the question is$ $not complete .$
	Commented by puissant last updated on 07/Sep/21

	$y e s s s i r t h e s u r f a c e o f a s p h e r e i s S = 4 π R^{2} .$
	Commented by MJS_new last updated on 07/Sep/21

	$yes indeed . and S = \frac{d^{2} \sqrt{10 (5 - \sqrt{5})}}{2} is the$ $surface of a dodecahedron with diagonal d .$
	Commented by puissant last updated on 07/Sep/21

	$S = \frac{12 d^{2}}{2} c o t a n (\frac{π}{12}) = \frac{6 d^{2}}{t a n (\frac{π}{12})} . .$
	Commented by prakash jain last updated on 07/Sep/21

	$S is derivate of V ? How ?$
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