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Question Number 3843 by Rasheed Soomro last updated on 22/Dec/15
Let the side of the following mentioned figures is s: The area of square is s^2 , the 3D area(volume) of a cube is s^3 ,the 4D area/volume of 4D hypercube can be said s^4 and so on. Now if radius is r, the area of circle is 𝛑r^2 , the 3D area(volume) of a sphere is(4/3)𝛑r^3 , what will be the 4D area/volume of 4D sphere and 5D area/volume of 5D sphere?
$$\mathcal{L}{et}\:{the}\:{side}\:{of}\:{the}\:{following}\:{mentioned} \\ $$$${figures}\:{is}\:\boldsymbol{\mathrm{s}}: \\ $$$${The}\:{area}\:{of}\:{square}\:{is}\:\boldsymbol{\mathrm{s}}^{\mathrm{2}} ,\:{the}\:\mathrm{3}{D}\:{area}\left({volume}\right) \\ $$$${of}\:{a}\:{cube}\:{is}\:\boldsymbol{\mathrm{s}}^{\mathrm{3}} ,{the}\:\mathrm{4}{D}\:{area}/{volume}\:{of}\:\mathrm{4}{D} \\ $$$${hypercube}\:{can}\:{be}\:{said}\:\boldsymbol{\mathrm{s}}^{\mathrm{4}} \:{and}\:{so}\:{on}. \\ $$$$ \\ $$$${Now}\:{if}\:{radius}\:{is}\:\boldsymbol{\mathrm{r}},\:{the}\:{area}\:{of}\:{circle}\:{is}\:\boldsymbol{\pi\mathrm{r}}^{\mathrm{2}} , \\ $$$${the}\:\mathrm{3}{D}\:{area}\left({volume}\right)\:{of}\:{a}\:{sphere}\:{is}\frac{\mathrm{4}}{\mathrm{3}}\boldsymbol{\pi\mathrm{r}}^{\mathrm{3}} , \\ $$$${what}\:{will}\:{be}\:{the}\:\mathrm{4}{D}\:{area}/{volume}\:{of}\:\mathrm{4}{D}\: \\ $$$${sphere}\:{and}\:\mathrm{5}{D}\:{area}/{volume}\:{of}\:\mathrm{5}{D}\:{sphere}? \\ $$
Commented by prakash jain last updated on 22/Dec/15
Volume V_n =((Ο€^(n/2) r^n )/(Ξ“(1+(n/2)))) Surface area S_(nβˆ’1) =((2Ο€^(n/2) r^(nβˆ’1) )/(Ξ“((n/2)))) (surface area)
$$\mathrm{Volume}\:{V}_{{n}} =\frac{\pi^{{n}/\mathrm{2}} {r}^{{n}} }{\Gamma\left(\mathrm{1}+\frac{{n}}{\mathrm{2}}\right)}\: \\ $$$$\mathrm{Surface}\:\mathrm{area}\:{S}_{{n}βˆ’\mathrm{1}} =\frac{\mathrm{2}\pi^{{n}/\mathrm{2}} {r}^{{n}βˆ’\mathrm{1}} }{\Gamma\left(\frac{{n}}{\mathrm{2}}\right)}\:\:\left({surface}\:{area}\right) \\ $$
Commented by Rasheed Soomro last updated on 22/Dec/15
Volume V_n =((Ο€^(n/2) r^n )/(Ξ“(1+(n/2)))) V_3 =((Ο€^(3/2) r^3 )/(Ξ“(1+(3/2)))) =^(?) (4/3)Ο€r^3 For 4D S_(4βˆ’1) =S_3 =((2Ο€^(4/2) r^(4βˆ’1) )/(Ξ“((4/2)))) [How many dimentions?] We consider surface as 2D
$$\mathrm{Volume}\:{V}_{{n}} =\frac{\pi^{{n}/\mathrm{2}} {r}^{{n}} }{\Gamma\left(\mathrm{1}+\frac{{n}}{\mathrm{2}}\right)}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{V}_{\mathrm{3}} =\frac{\pi^{\mathrm{3}/\mathrm{2}} {r}^{\mathrm{3}} }{\Gamma\left(\mathrm{1}+\frac{\mathrm{3}}{\mathrm{2}}\right)}\:\overset{?} {=}\frac{\mathrm{4}}{\mathrm{3}}\pi{r}^{\mathrm{3}} \\ $$$${For}\:\mathrm{4D}\:\:\mathrm{S}_{\mathrm{4}βˆ’\mathrm{1}} =\mathrm{S}_{\mathrm{3}} \:=\frac{\mathrm{2}\pi^{\mathrm{4}/\mathrm{2}} {r}^{\mathrm{4}βˆ’\mathrm{1}} }{\Gamma\left(\frac{\mathrm{4}}{\mathrm{2}}\right)}\:\left[{How}\:{many}\:{dimentions}?\right] \\ $$$$\mathrm{W}{e}\:{consider}\:{surface}\:{as}\:\mathrm{2D} \\ $$
Commented by prakash jain last updated on 22/Dec/15
For 3βˆ’D surface area r^2 For 4βˆ’D surface area r^3 S_(nβˆ’1) is surface area equivalent n dimentional sphere. Will be proportional to r^(nβˆ’1) . Ξ“((5/2))=(3/2)Ξ“((1/2))=((3(βˆšΟ€))/4)
$$\mathrm{For}\:\mathrm{3}βˆ’{D}\:{surface}\:{area}\:{r}^{\mathrm{2}} \\ $$$$\mathrm{For}\:\mathrm{4}βˆ’{D}\:{surface}\:{area}\:{r}^{\mathrm{3}} \\ $$$${S}_{{n}βˆ’\mathrm{1}} \:\mathrm{is}\:\mathrm{surface}\:\mathrm{area}\:\mathrm{equivalent}\:{n}\:{dimentional} \\ $$$${sphere}.\:{W}\mathrm{ill}\:\mathrm{be}\:\mathrm{proportional}\:\mathrm{to}\:{r}^{{n}βˆ’\mathrm{1}} . \\ $$$$\Gamma\left(\frac{\mathrm{5}}{\mathrm{2}}\right)=\frac{\mathrm{3}}{\mathrm{2}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\mathrm{3}\sqrt{\pi}}{\mathrm{4}} \\ $$
Commented by Rasheed Soomro last updated on 24/Dec/15
THΞ±nkS! If for n dimentions surface is of nβˆ’1 dimentions.This is consistant with 3D figures.Because in 3D, surface is 2D.
$$\mathcal{TH}\alpha{nk}\mathcal{S}! \\ $$$$\mathcal{I}{f}\:{for}\:{n}\:{dimentions}\:{surface}\:{is}\:{of}\:\:{n}βˆ’\mathrm{1} \\ $$$${dimentions}.{This}\:{is}\:{consistant}\:{with}\:\mathrm{3D} \\ $$$${figures}.{Because}\:{in}\:\mathrm{3D},\:{surface}\:{is}\:\mathrm{2D}. \\ $$
Answered by Filup last updated on 22/Dec/15
Volume is area rotated around axis. ∴4D area/volume is the volume rotated around an axis. Where you rotate the whole 3D solid around the z axis (I think z axis)
$$\mathrm{Volume}\:\mathrm{is}\:\mathrm{area}\:\mathrm{rotated}\:\mathrm{around}\:\mathrm{axis}. \\ $$$$\therefore\mathrm{4}{D}\:{area}/{volume}\:{is}\:{the}\:{volume} \\ $$$${rotated}\:{around}\:{an}\:{axis}.\:\mathrm{Where}\:\mathrm{you}\:\mathrm{rotate} \\ $$$$\mathrm{the}\:\mathrm{whole}\:\mathrm{3D}\:\mathrm{solid}\:\mathrm{around}\:\mathrm{the}\:{z}\:\mathrm{axis}\: \\ $$$$\left(\mathrm{I}\:\mathrm{think}\:{z}\:\mathrm{axis}\right) \\ $$
Commented by Filup last updated on 22/Dec/15
this was meant to be a comment
$${this}\:{was}\:{meant}\:{to}\:{be}\:\:{a}\:{comment} \\ $$
Commented by Rasheed Soomro last updated on 22/Dec/15
Formula of volume of 4D sphere,in terms of r is required. Formulas of square,cube,hypercube etc form a GP: s^2 ,s^3 ,s^4 ,... Formulas of circle,sphere,etc form a sequence 𝛑r^2 ,(4/3)𝛑r^3 ,.... What is next term? I f we add one term at the begining ,the first sequence should be: s,s^2 ,s^3 ,... s may be considered length of line segment. The second sequence,I think should start with 0 (0 can be considered as β€²lengthβ€² of a point): 0,𝛑r^2 ,(4/3)𝛑r^3 ,... Point,Circle,Sphere,...
$${Formula}\:{of}\:{volume}\:{of}\:\mathrm{4}{D}\:{sphere},{in}\:{terms}\: \\ $$$${of}\:{r}\:{is}\:{required}. \\ $$$${Formulas}\:{of}\:{square},{cube},{hypercube}\:{etc} \\ $$$${form}\:{a}\:{GP}:\:\:\boldsymbol{\mathrm{s}}^{\mathrm{2}} ,\boldsymbol{\mathrm{s}}^{\mathrm{3}} ,\boldsymbol{\mathrm{s}}^{\mathrm{4}} ,… \\ $$$${Formulas}\:{of}\:{circle},{sphere},{etc}\:{form}\:{a}\:{sequence} \\ $$$$\boldsymbol{\pi\mathrm{r}}^{\mathrm{2}} ,\frac{\mathrm{4}}{\mathrm{3}}\boldsymbol{\pi\mathrm{r}}^{\mathrm{3}} ,…. \\ $$$$\mathcal{W}{hat}\:{is}\:{next}\:{term}? \\ $$$$\mathcal{I}\:{f}\:\:{we}\:{add}\:{one}\:{term}\:{at}\:{the}\:{begining}\:,{the}\:{first}\:{sequence} \\ $$$${should}\:{be}:\:\boldsymbol{\mathrm{s}},\boldsymbol{\mathrm{s}}^{\mathrm{2}} ,\boldsymbol{\mathrm{s}}^{\mathrm{3}} ,… \\ $$$$\boldsymbol{\mathrm{s}}\:{may}\:{be}\:{considered}\:{length}\:{of}\:{line}\:{segment}. \\ $$$${The}\:{second}\:{sequence},{I}\:{think}\:{should}\:{start}\:{with}\:\mathrm{0} \\ $$$$\left(\mathrm{0}\:{can}\:{be}\:{considered}\:{as}\:'{length}'\:{of}\:{a}\:{point}\right): \\ $$$$\mathrm{0},\boldsymbol{\pi\mathrm{r}}^{\mathrm{2}} ,\frac{\mathrm{4}}{\mathrm{3}}\boldsymbol{\pi\mathrm{r}}^{\mathrm{3}} ,… \\ $$$${Point},{Circle},{Sphere},… \\ $$

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