Find the volume that remains after the hole of radius 1 bored through the center of a solid sphere of radius 3. (a) 18π (b) ((28)/3)π (c) 36π (d) ((56π)/3)


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Question Number 124666 by benjo_mathlover last updated on 05/Dec/20

$F i n d t h e v o l u m e t h a t r e m a i n s$ $a f t e r t h e h o l e o f r a d i u s 1 b o r e d$ $t h r o u g h t h e c e n t e r o f a s o l i d$ $s p h e r e o f r a d i u s 3 .$ $(a) 18 π (b) \frac{28}{3} π (c) 36 π (d) \frac{56 π}{3}$
Commented by ajfour last updated on 05/Dec/20

Commented by ajfour last updated on 05/Dec/20

$c a p v o l u m e V .$ $V = \int_{h}^{R} π r^{2} d y$ $r = R \sin θ, y = R \cos θ$ $= \int_{0}^{α} π (R^{2} \sin^{2} θ) R \sin θ d θ$ $\sin α = \frac{1}{3}, \cos α = \frac{2 \sqrt{2}}{3}$ $= π R^{3} \int_{\frac{2 \sqrt{2}}{3}}^{1} (1 - t^{2}) d t; t = \cos θ$ $= π R^{3} (t - \frac{t^{3}}{3}) ∣_{\frac{2 \sqrt{2}}{3}}^{1}$ $= 27 π [\frac{2}{3} - (\frac{2 \sqrt{2}}{3} - \frac{16 \sqrt{2}}{81})]$ $= 27 π [\frac{2}{3} - \frac{38 \sqrt{2}}{81}]$ $r e m a i n i n g v o l u m e$ $= \frac{4}{3} π (27) - 2 π {(1)}^{2} (2 \sqrt{2}) - 2 V$ $= 36 π - 4 \sqrt{2} π - 36 π + \frac{76 \sqrt{2}}{3} π$ $V_{r e m a i n i n g} = \frac{64 \sqrt{2} π}{3} .$
Commented by benjo_mathlover last updated on 05/Dec/20

$n o t h i n g a v a i l a b l e c h o i c e a n s w e r$ $i s c o r r e c t ?$
Commented by mr W last updated on 05/Dec/20

$y e s . a l l a n s w e r s g i v e n a r e w r o n g .$ $i g o t a f o r m u l a f o r t h e r e m a i n i n g$ $v o l u m e w h i c h i s$ $V_{r e m a i n} = \frac{4 π {(R^{2} - r^{2})}^{\frac{3}{2}}}{3}$
	Answered by mr W last updated on 05/Dec/20

	$M e t h o d I$ $R = r a d i u s o f s p h e r e$ $r = r a d i u s o f h o l e$ $h = \sqrt{R^{2} - r^{2}} = \sqrt{3^{2} - 1^{2}} = 2 \sqrt{2}$ $v o l u m e o f s p h e r e$ $V = \frac{4 π R^{3}}{3}$ $v o l u m e o f c y l i n d e r w i t h h e i g h t 2 h$ $V_{1} = 2 h π r^{2}$ $v o l u m e o f e a c h c a p$ $V_{2} = π {(Δ h)}^{2} (R - \frac{Δ h}{3}) w i t h Δ h = R - h$ $v o l u m e o f r e m a i n i n g s p h e r e p a r t$ $V_{R e m a i n} = V - V_{1} - 2 V_{2}$ $= \frac{4 π R^{3}}{3} - 2 π {h r}^{2} - 2 π {(R - h)}^{2} \frac{(2 R + h)}{3}$ $= \frac{4 π 3^{3}}{3} - 2 π 2 \sqrt{2} \times 1^{2} - 2 π {(3 - 2 \sqrt{2})}^{2} \frac{(2 \times 3 + 2 \sqrt{2})}{3}$ $= \frac{108 - 12 \sqrt{2} - 4 \times 27 + 4 \times 19 \sqrt{2}}{3} π$ $= \frac{64 \sqrt{2} π}{3}$
	Commented by mr W last updated on 05/Dec/20

	Commented by mr W last updated on 05/Dec/20

	$M e t h o d I I$ $r ⩽ x ⩽ R$ $h (x) = \sqrt{R^{2} - x^{2}}$ $d V = 2 h (x) 2 π x = 4 π x \sqrt{R^{2} - x^{2}}$ $V_{R e m a i n} = \int_{r}^{R} 4 π x \sqrt{R^{2} - x^{2}} d x$ $= - 2 π \int_{r}^{R} \sqrt{R^{2} - x^{2}} d (R^{2} - x^{2})$ $= \frac{4 π}{3} {[{(R^{2} - x^{2})}^{\frac{3}{2}}]}_{R}^{r}$ $= \frac{4 π {(R^{2} - r^{2})}^{\frac{3}{2}}}{3}$ $= \frac{4 π {(3^{2} - 1^{2})}^{\frac{3}{2}}}{3}$ $= \frac{64 \sqrt{2} π}{3}$
	Commented by benjo_mathlover last updated on 05/Dec/20

	$t y p o s i r . \frac{4 π \times 8 \sqrt{8}}{3} = \frac{64 π \sqrt{2}}{3}$
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