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Question Number 60816 by ANTARES VY last updated on 26/May/19

S=4𝛑R^2    prove

$$\boldsymbol{\mathrm{S}}=\mathrm{4}\boldsymbol{\pi\mathrm{R}}^{\mathrm{2}} \:\:\:\boldsymbol{\mathrm{prove}} \\ $$

Answered by tanmay last updated on 26/May/19

ds=Rdθ×Rsinθdφ  S=R^2 ∫_0 ^π sinθdθ∫_0 ^(2π) dφ  =R^2 ×2×2π=4πR^2

$${ds}={Rd}\theta×{Rsin}\theta{d}\phi \\ $$$${S}={R}^{\mathrm{2}} \int_{\mathrm{0}} ^{\pi} {sin}\theta{d}\theta\int_{\mathrm{0}} ^{\mathrm{2}\pi} {d}\phi \\ $$$$={R}^{\mathrm{2}} ×\mathrm{2}×\mathrm{2}\pi=\mathrm{4}\pi{R}^{\mathrm{2}} \\ $$$$ \\ $$$$ \\ $$

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