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Question Number 20157 by ajfour last updated on 23/Aug/17

Commented by ajfour last updated on 23/Aug/17

$C o m p u t e t h e v o l u m e c o m m o n t o t h e$ $s p h e r e x^{2} + y^{2} + z^{2} = 4 a n d t h e i n s i d e o f$ $t h e p a r a b o l o i d x^{2} + y^{2} = 3 z .$
	Answered by ajfour last updated on 23/Aug/17

	Commented by ajfour last updated on 23/Aug/17

	$L e t x^{2} + y^{2} = r^{2}$ $\Rightarrow e q u a t i o n o f s p h e r e b e c o m e s$ $r^{2} + z^{2} = 4$ $e q u a t i o n o f p a r a b o l o i d i s$ $r^{2} = 3 z$ $i n t e r s e c t i o n i s a c i r c l e o f$ $r a d i u s r_{0} .$ $r_{0}^{2} + z_{0}^{2} = 4 & r_{0}^{2} = 3 z_{0}$ $\Rightarrow z_{0}^{2} + 3 z_{0} = 4 \Rightarrow z_{0} = 1, r_{0} = \sqrt{3} .$ $I f e n c l o s e d v o l u m e i s V,$ $V = \int_{0}^{r_{0}} (2 π r) d r (Δ z)$ $= 2 π \int_{0}^{\sqrt{3}} (\sqrt{4 - r^{2}} - \frac{r^{2}}{3}) r d r$ $= 2 π [- \frac{1}{3} {(4 - r^{2})}^{3 / 2} - \frac{r^{4}}{12}] ∣_{0}^{\sqrt{3}}$ $= 2 π [- \frac{1}{3} + \frac{8}{3} - \frac{3}{4}] = 2 π (\frac{28 - 9}{12})$ $V = \frac{19 π}{6} .$
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