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Question-125215




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.
$${A}\:{cylindrical}\:{hole}\:{of}\:{radius}\:{half} \\ $$$${that}\:{of}\:{the}\:{sphere}\:{is}\:{drilled}\:{out}\:, \\ $$$${touching}\:{the}\:{equatorial}\:{plane}\:{of} \\ $$$${sphere}.\:{Find}\:{fraction}\:{of}\:{volume} \\ $$$${drilled}\:{out}. \\ $$
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.
$${R}={radius}\:{of}\:{sphere} \\ $$$${x}={R}\:\mathrm{cos}\:\theta \\ $$$${x}=\mathrm{0}\:\rightarrow\:{R} \\ $$$$\theta=\frac{\pi}{\mathrm{2}}\:\rightarrow\:\mathrm{0} \\ $$$${height}\:{of}\:{shaded}\:{cylindrical}\:{shell}=\mathrm{2}{h} \\ $$$${h}=\sqrt{{R}^{\mathrm{2}} −{x}^{\mathrm{2}} }={R}\:\mathrm{sin}\:\theta \\ $$$${volume}\:{of}\:{shaded}\:{cylindrical}\:{shell}: \\ $$$${dV}=\mathrm{2}\left(\pi−\theta\right){x}×\mathrm{2}{hdx}=−\mathrm{4}{R}^{\mathrm{3}} \mathrm{sin}^{\mathrm{2}} \:\theta\mathrm{cos}\:\theta\left(\pi−\theta\right){d}\theta \\ $$$${V}=\mathrm{4}{R}^{\mathrm{3}} \int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}^{\mathrm{2}} \:\theta\mathrm{cos}\:\theta\left(\pi−\theta\right){d}\theta \\ $$$$=\mathrm{4}{R}^{\mathrm{3}} \left\{\pi\left[\frac{\mathrm{sin}^{\mathrm{3}} \:\theta}{\mathrm{3}}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} −\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}^{\mathrm{2}} \:\theta\:\mathrm{cos}\:\theta\:\theta{d}\theta\right\} \\ $$$$=\mathrm{4}{R}^{\mathrm{3}} \left\{\frac{\pi}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{9}}\left[\mathrm{3}\theta\:\mathrm{sin}^{\mathrm{3}} \:\theta−\mathrm{cos}^{\mathrm{3}} \:\theta+\mathrm{3cos}\:\theta\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \right\} \\ $$$$=\mathrm{4}{R}^{\mathrm{3}} \left\{\frac{\pi}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{9}}\left[\frac{\mathrm{3}\pi}{\mathrm{2}}−\mathrm{2}\right]\right\} \\ $$$$=\left(\frac{\mathrm{2}\pi}{\mathrm{3}}+\frac{\mathrm{8}}{\mathrm{9}}\right){R}^{\mathrm{3}} \\ $$$$ \\ $$$${volume}\:{of}\:{hole}:\:\frac{\mathrm{4}{R}^{\mathrm{3}} }{\mathrm{9}}\left(\frac{\mathrm{3}\pi}{\mathrm{2}}−\mathrm{2}\right)=\left(\frac{\mathrm{2}\pi}{\mathrm{3}}−\frac{\mathrm{8}}{\mathrm{9}}\right){R}^{\mathrm{3}} \\ $$$${fraction}=\frac{\frac{\mathrm{2}\pi}{\mathrm{3}}−\frac{\mathrm{8}}{\mathrm{9}}}{\frac{\mathrm{4}\pi}{\mathrm{3}}}=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{2}}{\mathrm{3}\pi}\approx\mathrm{28}.\mathrm{8\%} \\ $$$${i}.{e}.\:{more}\:{than}\:{a}\:{fourth}\:{but}\:{less}\:{than} \\ $$$${a}\:{third}\:{of}\:{the}\:{sphere}\:{is}\:{drilled}\:{out}. \\ $$
Commented by ajfour last updated on 09/Dec/20
Thanks Sir,  I absolutely agree with  and understand your solution!  Thanks for the effort and care.
$${Thanks}\:{Sir},\:\:{I}\:{absolutely}\:{agree}\:{with} \\ $$$${and}\:{understand}\:{your}\:{solution}! \\ $$$${Thanks}\:{for}\:{the}\:{effort}\:{and}\:{care}. \\ $$
Commented by mr W last updated on 09/Dec/20
h^2 +x^2 =R^2   ⇒h=(√(R^2 −x^2 ))
$${h}^{\mathrm{2}} +{x}^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$$\Rightarrow{h}=\sqrt{{R}^{\mathrm{2}} −{x}^{\mathrm{2}} } \\ $$
Commented by mr W last updated on 09/Dec/20
Commented by mr W last updated on 09/Dec/20
let′s look at the solid sphere without  hole. it can be seen as a lot of   thin walled cylindrical shells with  radius x, thickness dx and height H  (H=2h=2(√(R^2 −x^2 ))).  the volume of this cylindrical shell  is dV=2πx×H×dx=4πx(√(R^2 −x^2 ))dx  the volume of the sphere is then  V=∫dV=4π∫_0 ^R x(√(R^2 −x^2 ))dx=((4πR^3 )/3)
$${let}'{s}\:{look}\:{at}\:{the}\:{solid}\:{sphere}\:{without} \\ $$$${hole}.\:{it}\:{can}\:{be}\:{seen}\:{as}\:{a}\:{lot}\:{of}\: \\ $$$${thin}\:{walled}\:{cylindrical}\:{shells}\:{with} \\ $$$${radius}\:{x},\:{thickness}\:{dx}\:{and}\:{height}\:{H} \\ $$$$\left({H}=\mathrm{2}{h}=\mathrm{2}\sqrt{{R}^{\mathrm{2}} −{x}^{\mathrm{2}} }\right). \\ $$$${the}\:{volume}\:{of}\:{this}\:{cylindrical}\:{shell} \\ $$$${is}\:{dV}=\mathrm{2}\pi{x}×{H}×{dx}=\mathrm{4}\pi{x}\sqrt{{R}^{\mathrm{2}} −{x}^{\mathrm{2}} }{dx} \\ $$$${the}\:{volume}\:{of}\:{the}\:{sphere}\:{is}\:{then} \\ $$$${V}=\int{dV}=\mathrm{4}\pi\int_{\mathrm{0}} ^{{R}} {x}\sqrt{{R}^{\mathrm{2}} −{x}^{\mathrm{2}} }{dx}=\frac{\mathrm{4}\pi{R}^{\mathrm{3}} }{\mathrm{3}} \\ $$
Commented by mr W last updated on 09/Dec/20
Commented by mr W last updated on 09/Dec/20
now we look at the situation with hole.  we can proceed in the same way. the  only difference is that the cylindrical  shell is no longer completely closed,  but cut by the hole. the perimeter of  the shell is (2π−2θ)x with cos θ=(x/R).  dV=(2π−2θ)x×H×dx=4(π−θ)x(√(R^2 −x^2 ))dx  V=∫_0 ^R 4(π−θ)x(√(R^2 −x^2 ))dx  the rest is clear, i think.
$${now}\:{we}\:{look}\:{at}\:{the}\:{situation}\:{with}\:{hole}. \\ $$$${we}\:{can}\:{proceed}\:{in}\:{the}\:{same}\:{way}.\:{the} \\ $$$${only}\:{difference}\:{is}\:{that}\:{the}\:{cylindrical} \\ $$$${shell}\:{is}\:{no}\:{longer}\:{completely}\:{closed}, \\ $$$${but}\:{cut}\:{by}\:{the}\:{hole}.\:{the}\:{perimeter}\:{of} \\ $$$${the}\:{shell}\:{is}\:\left(\mathrm{2}\pi−\mathrm{2}\theta\right){x}\:{with}\:\mathrm{cos}\:\theta=\frac{{x}}{{R}}. \\ $$$${dV}=\left(\mathrm{2}\pi−\mathrm{2}\theta\right){x}×{H}×{dx}=\mathrm{4}\left(\pi−\theta\right){x}\sqrt{{R}^{\mathrm{2}} −{x}^{\mathrm{2}} }{dx} \\ $$$${V}=\int_{\mathrm{0}} ^{{R}} \mathrm{4}\left(\pi−\theta\right){x}\sqrt{{R}^{\mathrm{2}} −{x}^{\mathrm{2}} }{dx} \\ $$$${the}\:{rest}\:{is}\:{clear},\:{i}\:{think}. \\ $$
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?
$$\:\mathrm{did}\:\mathrm{you}\:\mathrm{make}\:\mathrm{this}\:\mathrm{diagram}\:\mathrm{on}\:\mathrm{here}? \\ $$$$ \\ $$
Commented by mr W last updated on 10/Dec/20
no. i made the diagrams in other  apps.
$${no}.\:{i}\:{made}\:{the}\:{diagrams}\:{in}\:{other} \\ $$$${apps}. \\ $$
Answered by ajfour last updated on 09/Dec/20

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