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Question-85074




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.
IfacylindricalcavitybedrilledoutofahemisphereofradiusR,intheplaceshown;withadrillerofradiusr,findfractionofhemispherematerialdrilledout.
Answered by mr W last updated on 19/Mar/20
Commented by mr W last updated on 19/Mar/20
a=R−r  PM^2 =a^2 +ρ^2 −2aρ cos θ    length of cylinder at point P =l  l=2(√(R^2 −PM^2 ))=2(√(R^2 −a^2 −ρ^2 +2aρ cos θ))    dV=l dA=lρdρdθ=2ρ(√(R^2 −a^2 −ρ^2 +2aρ cos θ)) dρdθ  the volume of intersection from  sphere and cylinder with distance a:  generally:  V=4∫_0 ^( r) ρ∫_0 ^( π) (√(R^2 −a^2 −ρ^2 +2aρ cos θ)) dθdρ  in our case:  V=4∫_0 ^( r) ρ∫_0 ^( π) (√(r(2R−r)−ρ^2 +2(R−r)ρ cos θ)) dθdρ  .... maybe not integrable ?
a=RrPM2=a2+ρ22aρcosθlengthofcylinderatpointP=ll=2R2PM2=2R2a2ρ2+2aρcosθdV=ldA=lρdρdθ=2ρR2a2ρ2+2aρcosθdρdθthevolumeofintersectionfromsphereandcylinderwithdistancea:generally:V=40rρ0πR2a2ρ2+2aρcosθdθdρinourcase:V=40rρ0πr(2Rr)ρ2+2(Rr)ρcosθdθdρ.maybenotintegrable?
Commented by ajfour last updated on 19/Mar/20
Thanks Sir, it was great following  your solution!
ThanksSir,itwasgreatfollowingyoursolution!
Commented by mr W last updated on 19/Mar/20
thanks to you too sir!  i have problem to solve the final  integral. do you have an idea?
thankstoyoutoosir!ihaveproblemtosolvethefinalintegral.doyouhaveanidea?
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
anotherattempt:V=4rr0r2x20R2(ax)2y2dzdydx=4rr0r2x2R2(ax)2y2dydx=2rr{[R2(ax)2]sin1r2x2R2(ax)2+(r2x2)[R2r2(ax)2+x2]}dx=2rr[(R2a2+2axx2)sin1r2x2R2a2+2axx2+(r2x2)(R2r2a2+2ax)]dx

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