Math Question and Answers


Question and Answers Forum

All Questions Topic List
Mensuration Questions
Previous in All Question Next in All Question
Previous in Mensuration Next in Mensuration
Question Number 125215 by ajfour last updated on 09/Dec/20

Commented by ajfour last updated on 09/Dec/20
$A cylindrical hole of radius half that of the sphere is drilled out , touching the equatorial plane of sphere. Find fraction of volume drilled out.$
$A c y l i n d r i c a l h o l e o f r a d i u s h a l f$ $t h a t o f t h e s p h e r e i s d r i l l e d o u t,$ $t o u c h i n g t h e e q u a t o r i a l p l a n e o f$ $s p h e r e . F i n d f r a c t i o n o f v o l u m e$ $d r i l l e d o u t .$
Commented by mr W last updated on 09/Dec/20
I found something interesting about this problem: https://diynovice.wordpress.com/2012/12/02/hole-in-sphere/
	Answered by mr W last updated on 09/Dec/20

	Commented by mr W last updated on 09/Dec/20
	$R=radius of sphere x=R cos θ x=0 → R θ=(π/2) → 0 height of shaded cylindrical shell=2h h=(√(R^2 −x^2 ))=R sin θ volume of shaded cylindrical shell: dV=2(π−θ)x×2hdx=−4R^3 sin^2 θcos θ(π−θ)dθ V=4R^3 ∫_0 ^(π/2) sin^2 θcos θ(π−θ)dθ =4R^3 {π[((sin^3 θ)/3)]_0 ^(π/2) −∫_0 ^(π/2) sin^2 θ cos θ θdθ} =4R^3 {(π/3)−(1/9)[3θ sin^3 θ−cos^3 θ+3cos θ]_0 ^(π/2) } =4R^3 {(π/3)−(1/9)[((3π)/2)−2]} =(((2π)/3)+(8/9))R^3 volume of hole: ((4R^3 )/9)(((3π)/2)−2)=(((2π)/3)−(8/9))R^3 fraction=((((2π)/3)−(8/9))/((4π)/3))=(1/2)−(2/(3π))≈28.8% i.e. more than a fourth but less than a third of the sphere is drilled out.$
	$R = r a d i u s o f s p h e r e$ $x = R \cos θ$ $x = 0 \to R$ $θ = \frac{π}{2} \to 0$ $h e i g h t o f s h a d e d c y l i n d r i c a l s h e l l = 2 h$ $h = \sqrt{R^{2} - x^{2}} = R \sin θ$ $v o l u m e o f s h a d e d c y l i n d r i c a l s h e l l :$ $d V = 2 (π - θ) x \times 2 h d x = - 4 R^{3} \sin^{2} θ \cos θ (π - θ) d θ$ $V = 4 R^{3} \int_{0}^{\frac{π}{2}} \sin^{2} θ \cos θ (π - θ) d θ$ $= 4 R^{3} {π {[\frac{\sin^{3} θ}{3}]}_{0}^{\frac{π}{2}} - \int_{0}^{\frac{π}{2}} \sin^{2} θ \cos θ θ d θ}$ $= 4 R^{3} {\frac{π}{3} - \frac{1}{9} {[3 θ \sin^{3} θ - \cos^{3} θ + 3 \cos θ]}_{0}^{\frac{π}{2}}}$ $= 4 R^{3} {\frac{π}{3} - \frac{1}{9} [\frac{3 π}{2} - 2]}$ $= (\frac{2 π}{3} + \frac{8}{9}) R^{3}$ $v o l u m e o f h o l e : \frac{4 R^{3}}{9} (\frac{3 π}{2} - 2) = (\frac{2 π}{3} - \frac{8}{9}) R^{3}$ $f r a c t i o n = \frac{\frac{2 π}{3} - \frac{8}{9}}{\frac{4 π}{3}} = \frac{1}{2} - \frac{2}{3 π} \approx 28 . 8 %$ $i . e . m o r e t h a n a f o u r t h b u t l e s s t h a n$ $a t h i r d o f t h e s p h e r e i s d r i l l e d o u t .$
	Commented by ajfour last updated on 09/Dec/20

	$T h a n k s S i r, I a b s o l u t e l y a g r e e w i t h$ $a n d u n d e r s t a n d y o u r s o l u t i o n!$ $T h a n k s f o r t h e e f f o r t a n d c a r e .$
	Commented by mr W last updated on 09/Dec/20

	$h^{2} + x^{2} = R^{2}$ $\Rightarrow h = \sqrt{R^{2} - x^{2}}$
	Commented by mr W last updated on 09/Dec/20

	Commented by mr W last updated on 09/Dec/20

	${l e t}^{'} s l o o k a t t h e s o l i d s p h e r e w i t h o u t$ $h o l e . i t c a n b e s e e n a s a l o t o f$ $t h i n w a l l e d c y l i n d r i c a l s h e l l s w i t h$ $r a d i u s x, t h i c k n e s s d x a n d h e i g h t H$ $(H = 2 h = 2 \sqrt{R^{2} - x^{2}}) .$ $t h e v o l u m e o f t h i s c y l i n d r i c a l s h e l l$ $i s d V = 2 π x \times H \times d x = 4 π x \sqrt{R^{2} - x^{2}} d x$ $t h e v o l u m e o f t h e s p h e r e i s t h e n$ $V = \int d V = 4 π \int_{0}^{R} x \sqrt{R^{2} - x^{2}} d x = \frac{4 π R^{3}}{3}$
	Commented by mr W last updated on 09/Dec/20

	Commented by mr W last updated on 09/Dec/20

	$n o w w e l o o k a t t h e s i t u a t i o n w i t h h o l e .$ $w e c a n p r o c e e d i n t h e s a m e w a y . t h e$ $o n l y d i f f e r e n c e i s t h a t t h e c y l i n d r i c a l$ $s h e l l i s n o l o n g e r c o m p l e t e l y c l o s e d,$ $b u t c u t b y t h e h o l e . t h e p e r i m e t e r o f$ $t h e s h e l l i s (2 π - 2 θ) x w i t h \cos θ = \frac{x}{R} .$ $d V = (2 π - 2 θ) x \times H \times d x = 4 (π - θ) x \sqrt{R^{2} - x^{2}} d x$ $V = \int_{0}^{R} 4 (π - θ) x \sqrt{R^{2} - x^{2}} d x$ $t h e r e s t i s c l e a r, i t h i n k .$
	Commented by mr W last updated on 09/Dec/20

	Commented by talminator2856791 last updated on 10/Dec/20

	$did you make this diagram on here ?$
	Commented by mr W last updated on 10/Dec/20

	$n o . i m a d e t h e d i a g r a m s i n o t h e r$ $a p p s .$
	Answered by ajfour last updated on 09/Dec/20

Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum