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V-4-3-R-3-prove-




Question Number 60817 by ANTARES VY last updated on 26/May/19
V=(4/3)𝛑R^3    prove
\boldsymbolV=43\boldsymbolπR3\boldsymbolprove
Commented by Prithwish sen last updated on 26/May/19
http://mathcentral.uregina.ca/QQ/database/QQ.09.01/rahul1.html
Commented by Prithwish sen last updated on 26/May/19
This explanation is geometry based.
Thisexplanationisgeometrybased.
Answered by ajfour last updated on 26/May/19
Commented by ajfour last updated on 26/May/19
  An intermediate differential  layer has volume  dV = 4πr^2 dr  ⇒  V=∫_0 ^(  R) 4πr^2 dr = ((4πR^3 )/3) .
AnintermediatedifferentiallayerhasvolumedV=4πr2drV=0R4πr2dr=4πR33.
Commented by mr W last updated on 26/May/19
or  the surface can be divided into infinite  small pieces, each piece is the base of  a pyramid whose top is the center of  the sphere. the volume of the pyramid  is  ΔV=((hΔS)/3) with h=R  the volume of the sphere is  V=ΣΔV=(h/3)ΣΔS  ⇒V=(h/3)S=((h4πR^2 )/3)=((4πR^3 )/3)
orthesurfacecanbedividedintoinfinitesmallpieces,eachpieceisthebaseofapyramidwhosetopisthecenterofthesphere.thevolumeofthepyramidisΔV=hΔS3withh=RthevolumeofthesphereisV=ΣΔV=h3ΣΔSV=h3S=h4πR23=4πR33
Commented by mr W last updated on 26/May/19
Answered by tanmay last updated on 26/May/19
Commented by tanmay last updated on 26/May/19
dv=(rdθ)×(rsinθdφ)×(dr)  V=∫_0 ^R r^2 dr∫_0 ^π sinθdθ∫_0 ^(2π) dφ  =(R^3 /3)×2×2π  =(4/3)πR^3
dv=(rdθ)×(rsinθdϕ)×(dr)V=0Rr2dr0πsinθdθ02πdϕ=R33×2×2π=43πR3
Answered by Kunal12588 last updated on 26/May/19
there is a brilliant way I want to share
thereisabrilliantwayIwanttoshare
Commented by Kunal12588 last updated on 26/May/19
Commented by Kunal12588 last updated on 26/May/19
Commented by Prithwish sen last updated on 26/May/19
which book ? Sir.
whichbook?Sir.
Commented by Kunal12588 last updated on 26/May/19
Precalculus Mathematics
PrecalculusMathematics
Commented by Prithwish sen last updated on 26/May/19
thank you sir
thankyousir
Answered by tanmay last updated on 26/May/19
eqn of sphere centre origin(0,0,0) and radius R  (x^2 /R^2 )+(y^2 /R^2 )+(z^2 /R^2 )=1  u_1 =((x/R))^2 →x=R×u_1 ^(1/2) →dx=(R/2)×u^((−1)/2) du_1 =(R/2)×u_1 ^((1/2)−1) du_1   dy=(R/2)×u_2 ^((1/2)−1) du_2   dz=(R/2)×u_3 ^((1/2)−1) du_3   volume of sphere =8∫_0 ^1 ∫_0 ^1 ∫_0 ^1 ((R/2))^3 u_1 ^((1/2)−1) u_2 ^((1/2)−1) u_3 ^((1/2)−1) du_1 du_2 du_3   =8×(R^3 /8)×((⌈((1/2))×⌈((1/2))×⌈((1/2)))/(⌈(1+(1/2)+(1/2)+(1/2))))  =R^3 ×((((√π) )^3 )/(⌈((3/2)+1)))=((π(√π) )/((3/4)(√π)))×R^3 =((4πR^3 )/3)  using Dirchhilet theorem...attaching them    note below  ⌈((3/2)+1)=(3/2)⌈((1/2)+1)=(3/2)×(1/2)⌈((1/2))=(3/4)×(√π)
eqnofspherecentreorigin(0,0,0)andradiusRx2R2+y2R2+z2R2=1u1=(xR)2x=R×u112dx=R2×u12du1=R2×u1121du1dy=R2×u2121du2dz=R2×u3121du3volumeofsphere=8010101(R2)3u1121u2121u3121du1du2du3=8×R38×(12)×(12)×(12)(1+12+12+12)=R3×(π)3(32+1)=ππ34π×R3=4πR33usingDirchhilettheoremattachingthemnotebelow(32+1)=32(12+1)=32×12(12)=34×π
Commented by tanmay last updated on 26/May/19

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