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Question Number 60817 by ANTARES VY last updated on 26/May/19

V=(4/3)𝛑R^3 prove

$$\boldsymbol{\mathrm{V}}=\frac{\mathrm{4}}{\mathrm{3}}\boldsymbol{\pi\mathrm{R}}^{\mathrm{3}} \:\:\:\boldsymbol{\mathrm{prove}} \\ $$

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.

$$\mathrm{This}\:\mathrm{explanation}\:\mathrm{is}\:\mathrm{geometry}\:\mathrm{based}. \\ $$

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) .

$$\:\:{An}\:{intermediate}\:{differential} \\ $$$${layer}\:{has}\:{volume}\:\:{dV}\:=\:\mathrm{4}\pi{r}^{\mathrm{2}} {dr} \\ $$$$\Rightarrow\:\:{V}=\int_{\mathrm{0}} ^{\:\:{R}} \mathrm{4}\pi{r}^{\mathrm{2}} {dr}\:=\:\frac{\mathrm{4}\pi{R}^{\mathrm{3}} }{\mathrm{3}}\:. \\ $$

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)

$${or} \\ $$$${the}\:{surface}\:{can}\:{be}\:{divided}\:{into}\:{infinite} \\ $$$${small}\:{pieces},\:{each}\:{piece}\:{is}\:{the}\:{base}\:{of} \\ $$$${a}\:{pyramid}\:{whose}\:{top}\:{is}\:{the}\:{center}\:{of} \\ $$$${the}\:{sphere}.\:{the}\:{volume}\:{of}\:{the}\:{pyramid} \\ $$$${is} \\ $$$$\Delta{V}=\frac{{h}\Delta{S}}{\mathrm{3}}\:{with}\:{h}={R} \\ $$$${the}\:{volume}\:{of}\:{the}\:{sphere}\:{is} \\ $$$${V}=\Sigma\Delta{V}=\frac{{h}}{\mathrm{3}}\Sigma\Delta{S} \\ $$$$\Rightarrow{V}=\frac{{h}}{\mathrm{3}}{S}=\frac{{h}\mathrm{4}\pi{R}^{\mathrm{2}} }{\mathrm{3}}=\frac{\mathrm{4}\pi{R}^{\mathrm{3}} }{\mathrm{3}} \\ $$

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

$${dv}=\left({rd}\theta\right)×\left({rsin}\theta{d}\phi\right)×\left({dr}\right) \\ $$$${V}=\int_{\mathrm{0}} ^{{R}} {r}^{\mathrm{2}} {dr}\int_{\mathrm{0}} ^{\pi} {sin}\theta{d}\theta\int_{\mathrm{0}} ^{\mathrm{2}\pi} {d}\phi \\ $$$$=\frac{{R}^{\mathrm{3}} }{\mathrm{3}}×\mathrm{2}×\mathrm{2}\pi \\ $$$$=\frac{\mathrm{4}}{\mathrm{3}}\pi{R}^{\mathrm{3}} \\ $$

Answered by Kunal12588 last updated on 26/May/19

there is a brilliant way I want to share

$${there}\:{is}\:{a}\:{brilliant}\:{way}\:{I}\:{want}\:{to}\:{share} \\ $$

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.

$$\mathrm{which}\:\mathrm{book}\:?\:\mathrm{Sir}. \\ $$

Commented by Kunal12588 last updated on 26/May/19

Precalculus Mathematics

$${Precalculus}\:{Mathematics} \\ $$

Commented by Prithwish sen last updated on 26/May/19

thank you sir

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

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)

$${eqn}\:{of}\:{sphere}\:{centre}\:{origin}\left(\mathrm{0},\mathrm{0},\mathrm{0}\right)\:{and}\:{radius}\:{R} \\ $$$$\frac{{x}^{\mathrm{2}} }{{R}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{R}^{\mathrm{2}} }+\frac{{z}^{\mathrm{2}} }{{R}^{\mathrm{2}} }=\mathrm{1} \\ $$$${u}_{\mathrm{1}} =\left(\frac{{x}}{{R}}\right)^{\mathrm{2}} \rightarrow{x}={R}×{u}_{\mathrm{1}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \rightarrow{dx}=\frac{{R}}{\mathrm{2}}×{u}^{\frac{−\mathrm{1}}{\mathrm{2}}} {du}_{\mathrm{1}} =\frac{{R}}{\mathrm{2}}×{u}_{\mathrm{1}} ^{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} {du}_{\mathrm{1}} \\ $$$${dy}=\frac{{R}}{\mathrm{2}}×{u}_{\mathrm{2}} ^{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} {du}_{\mathrm{2}} \\ $$$${dz}=\frac{{R}}{\mathrm{2}}×{u}_{\mathrm{3}} ^{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} {du}_{\mathrm{3}} \\ $$$${volume}\:{of}\:{sphere}\:=\mathrm{8}\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{{R}}{\mathrm{2}}\right)^{\mathrm{3}} {u}_{\mathrm{1}} ^{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} {u}_{\mathrm{2}} ^{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} {u}_{\mathrm{3}} ^{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} {du}_{\mathrm{1}} {du}_{\mathrm{2}} {du}_{\mathrm{3}} \\ $$$$=\mathrm{8}×\frac{{R}^{\mathrm{3}} }{\mathrm{8}}×\frac{\lceil\left(\frac{\mathrm{1}}{\mathrm{2}}\right)×\lceil\left(\frac{\mathrm{1}}{\mathrm{2}}\right)×\lceil\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\lceil\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right)} \\ $$$$={R}^{\mathrm{3}} ×\frac{\left(\sqrt{\pi}\:\right)^{\mathrm{3}} }{\lceil\left(\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{1}\right)}=\frac{\pi\sqrt{\pi}\:}{\frac{\mathrm{3}}{\mathrm{4}}\sqrt{\pi}}×{R}^{\mathrm{3}} =\frac{\mathrm{4}\pi{R}^{\mathrm{3}} }{\mathrm{3}} \\ $$$${using}\:{Dirchhilet}\:{theorem}...{attaching}\:{them} \\ $$$$ \\ $$$${note}\:{below} \\ $$$$\lceil\left(\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{1}\right)=\frac{\mathrm{3}}{\mathrm{2}}\lceil\left(\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{1}\right)=\frac{\mathrm{3}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{2}}\lceil\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\mathrm{3}}{\mathrm{4}}×\sqrt{\pi}\: \\ $$$$ \\ $$

Commented by tanmay last updated on 26/May/19

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