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Question Number 85074 by ajfour last updated on 18/Mar/20

Commented by ajfour last updated on 19/Mar/20
$If a cylindrical cavity be drilled out of a hemisphere of radius R, in the place shown; with a driller of radius r, find fraction of hemisphere material drilled out.$
$I f a c y l i n d r i c a l c a v i t y b e d r i l l e d$ $o u t o f a h e m i s p h e r e o f r a d i u s R,$ $i n t h e p l a c e s h o w n; w i t h a d r i l l e r$ $o f r a d i u s r, f i n d f r a c t i o n o f$ $h e m i s p h e r e m a t e r i a l d r i l l e d o u t .$
	Answered by mr W last updated on 19/Mar/20

	Commented by mr W last updated on 19/Mar/20

	$a = R - r$ ${P M}^{2} = a^{2} + ρ^{2} - 2 a ρ \cos θ$ $l e n g t h o f c y l i n d e r a t p o i n t P = l$ $l = 2 \sqrt{R^{2} - {P M}^{2}} = 2 \sqrt{R^{2} - a^{2} - ρ^{2} + 2 a ρ \cos θ}$ $d V = l d A = l ρ d ρ d θ = 2 ρ \sqrt{R^{2} - a^{2} - ρ^{2} + 2 a ρ \cos θ} d ρ d θ$ $t h e v o l u m e o f i n t e r s e c t i o n f r o m$ $s p h e r e a n d c y l i n d e r w i t h d i s t a n c e a :$ $g e n e r a l l y :$ $V = 4 \int_{0}^{r} ρ \int_{0}^{π} \sqrt{R^{2} - a^{2} - ρ^{2} + 2 a ρ \cos θ} d θ d ρ$ $i n o u r c a s e :$ $V = 4 \int_{0}^{r} ρ \int_{0}^{π} \sqrt{r (2 R - r) - ρ^{2} + 2 (R - r) ρ \cos θ} d θ d ρ$ $. . . . m a y b e n o t i n t e g r a b l e ?$
	Commented by ajfour last updated on 19/Mar/20

	$T h a n k s S i r, i t w a s g r e a t f o l l o w i n g$ $y o u r s o l u t i o n!$
	Commented by mr W last updated on 19/Mar/20

	$t h a n k s t o y o u t o o s i r!$ $i h a v e p r o b l e m t o s o l v e t h e f i n a l$ $i n t e g r a l . d o y o u h a v e a n i d e a ?$
	Commented by mr W last updated on 22/Mar/20
	$an other attempt: V=4∫_(−r) ^( r) ∫_0 ^(√(r^2 −x^2 )) ∫_0 ^(√(R^2 −(a−x)^2 −y^2 )) dzdydx =4∫_(−r) ^( r) ∫_0 ^(√(r^2 −x^2 )) (√(R^2 −(a−x)^2 −y^2 ))dydx =2∫_(−r) ^( r) {[R^2 −(a−x)^2 ]sin^(−1) ((√(r^2 −x^2 ))/(√(R^2 −(a−x)^2 )))+(√((r^2 −x^2 )[R^2 −r^2 −(a−x)^2 +x^2 ]))}dx =2∫_(−r) ^( r) [(R^2 −a^2 +2ax−x^2 )sin^(−1) ((√(r^2 −x^2 ))/(√(R^2 −a^2 +2ax−x^2 )))+(√((r^2 −x^2 )(R^2 −r^2 −a^2 +2ax)))]dx$
	$a n o t h e r a t t e m p t :$ $V = 4 \int_{- r}^{r} \int_{0}^{\sqrt{r^{2} - x^{2}}} \int_{0}^{\sqrt{R^{2} - {(a - x)}^{2} - y^{2}}} d z d y d x$ $= 4 \int_{- r}^{r} \int_{0}^{\sqrt{r^{2} - x^{2}}} \sqrt{R^{2} - {(a - x)}^{2} - y^{2}} d y d x$ $= 2 \int_{- r}^{r} {[R^{2} - {(a - x)}^{2}] \sin^{- 1} \frac{\sqrt{r^{2} - x^{2}}}{\sqrt{R^{2} - {(a - x)}^{2}}} + \sqrt{(r^{2} - x^{2}) [R^{2} - r^{2} - {(a - x)}^{2} + x^{2}]}} d x$ $= 2 \int_{- r}^{r} [(R^{2} - a^{2} + 2 a x - x^{2}) \sin^{- 1} \frac{\sqrt{r^{2} - x^{2}}}{\sqrt{R^{2} - a^{2} + 2 a x - x^{2}}} + \sqrt{(r^{2} - x^{2}) (R^{2} - r^{2} - a^{2} + 2 a x)}] d x$
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