Question Number 97888 by M±th+et+s last updated on 10/Jun/20
$${how}\:{to}\:{prove}\:\left(\left({the}\:{volumn}\:{of}\right.\right. \\ $$$$\left.{d}\left.{imensional}\:{sphare}\right)\right)\:{formula} \\ $$$${V}_{{N}} \left({R}\right)=\frac{\pi^{{N}/\mathrm{2}} }{\Gamma\left(\frac{{N}}{\mathrm{2}}+\mathrm{1}\right)}\:{R}^{{N}} \\ $$$$ \\ $$
Commented by EmericGent last updated on 10/Jun/20
I think you can integrate N times, I'm sorry, I can't try it now
Commented by M±th+et+s last updated on 10/Jun/20
$${thanks}\:{for}\:{giving}\:{the}\:{idea}\:{mybe}\:{i}\:{will} \\ $$$${try}\:{later}\: \\ $$
Answered by smridha last updated on 10/Jun/20
![at first calculate v_N (1)[volume of unit circle in N dim] then multipy by R^N . you can use multidimentional calculus and constracted Jacobian for N dimention. you can see this... volume of sphere in 3D is (4/3)𝛑R^3 in 2D it became 𝛑R^2 just the area of circle in 1D it became 2R...... it′s very lengthy proof... I can solve this....thank you...](https://www.tinkutara.com/question/Q97899.png)
$$\boldsymbol{{at}}\:\boldsymbol{{first}}\:\boldsymbol{{calculate}}\:\boldsymbol{{v}}_{\boldsymbol{{N}}} \left(\mathrm{1}\right)\left[\boldsymbol{{volume}}\:\boldsymbol{{of}}\:\boldsymbol{{unit}}\:\boldsymbol{{circle}}\:\boldsymbol{{in}}\:\boldsymbol{{N}}\:\boldsymbol{{dim}}\right] \\ $$$$\boldsymbol{{then}}\:\boldsymbol{{multipy}}\:\boldsymbol{{by}}\:\boldsymbol{{R}}^{{N}} . \\ $$$$\boldsymbol{{you}}\:\boldsymbol{{can}}\:\boldsymbol{{use}}\:\boldsymbol{{multidimentional}} \\ $$$$\boldsymbol{{calculus}}\:\boldsymbol{{and}}\:\boldsymbol{{constracted}}\:\boldsymbol{{Jacobian}} \\ $$$$\boldsymbol{{for}}\:\boldsymbol{{N}}\:\boldsymbol{{dimention}}. \\ $$$$\boldsymbol{{you}}\:\boldsymbol{{can}}\:\boldsymbol{{see}}\:\boldsymbol{{this}}… \\ $$$$\boldsymbol{{volume}}\:\boldsymbol{{of}}\:\boldsymbol{{sphere}}\:\boldsymbol{{in}}\:\mathrm{3}\boldsymbol{{D}}\:{is} \\ $$$$\frac{\mathrm{4}}{\mathrm{3}}\boldsymbol{\pi{R}}^{\mathrm{3}} \:\boldsymbol{{in}}\:\mathrm{2}\boldsymbol{{D}}\:\boldsymbol{{it}}\:\boldsymbol{{became}}\:\boldsymbol{\pi{R}}^{\mathrm{2}} \:\boldsymbol{{just}}\:\boldsymbol{{the}} \\ $$$$\boldsymbol{{area}}\:\boldsymbol{{of}}\:\boldsymbol{{circle}}\:\boldsymbol{{in}}\:\mathrm{1}\boldsymbol{{D}}\:\boldsymbol{{it}}\:\boldsymbol{{became}} \\ $$$$\mathrm{2}\boldsymbol{{R}}…… \\ $$$$\boldsymbol{{it}}'\boldsymbol{{s}}\:\boldsymbol{{very}}\:\boldsymbol{{lengthy}}\:\boldsymbol{{proof}}… \\ $$$$\boldsymbol{{I}}\:\boldsymbol{{can}}\:\boldsymbol{{solve}}\:\boldsymbol{{this}}….\boldsymbol{{thank}}\:\boldsymbol{{you}}… \\ $$$$ \\ $$
Commented by M±th+et+s last updated on 10/Jun/20
$${thank}\:{you}\:{very}\:{much}\:{sir}\:{good}\:{explaining} \\ $$