Calculation of volume of sphere using gauss diversion theorem
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This theorem says that if we have a solid limited by a closed surface and is a vector field, then
I have a A⊆RnA⊆Rn compact with smooth boundary and x=(0,...0)T∈Ax=(0,...0)T∈A. Now v(x)v(x) is the normal unit surface of AA and α(x)=∢(x,v(x))α(x)=∢(x,v(x)) of the position vector xx and the normal unit surface v(x)v(x). I need to show that for wdwd, the volume of Sn−1⊆RnSn−1⊆Rn follows:
∫∂Acos(α(x))∥x∥n−1dS(x)=wd∫∂Acos(α(x))‖x‖n−1dS(x)=wd
The hint is stated by observing:
Aϵ={x∈A|∥x∥>ϵ},ϵ>0Aϵ={x∈A|‖x‖>ϵ},ϵ>0
Is Aϵ={x∈A:∥x−0∥>ϵ}Aϵ={x∈A:‖x−0‖>ϵ} the outside of Sn−1={x∈Rn:∥x∥=1}Sn−1={x∈Rn:‖x‖=1}? I hope you can help me, because I don't even have a clue.
I have a A⊆RnA⊆Rn compact with smooth boundary and x=(0,...0)T∈Ax=(0,...0)T∈A. Now v(x)v(x) is the normal unit surface of AA and α(x)=∢(x,v(x))α(x)=∢(x,v(x)) of the position vector xx and the normal unit surface v(x)v(x). I need to show that for wdwd, the volume of Sn−1⊆RnSn−1⊆Rn follows:
∫∂Acos(α(x))∥x∥n−1dS(x)=wd∫∂Acos(α(x))‖x‖n−1dS(x)=wd
The hint is stated by observing:
Aϵ={x∈A|∥x∥>ϵ},ϵ>0Aϵ={x∈A|‖x‖>ϵ},ϵ>0
Is Aϵ={x∈A:∥x−0∥>ϵ}Aϵ={x∈A:‖x−0‖>ϵ} the outside of Sn−1={x∈Rn:∥x∥=1}Sn−1={x∈Rn:‖x‖=1}? I hope you can help me, because I don't even have a clue.
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