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Question Number 214498 by universe last updated on 10/Dec/24

Answered by mr W last updated on 10/Dec/24

$$=4\pi {\int }_0^{\mathrm{\infty }}{re}^{-r^2}{r}^{2}dr$$$$=2\pi {\int }_0^{\mathrm{\infty }}{pe}^{-p}dp$$$$=2\pi (-{pe}^{-p}+{\int }_0^{\mathrm{\infty }}{e}^{-p}dp)$$$$=2\pi {[-{pe}^{-p}-e^{-p}]}_0^{\mathrm{\infty }}$$$$=2\pi$$

Commented by universe last updated on 11/Dec/24

$$sorrysiridontgetit$$$$sircanyouexplainlittlebit$$

Commented by mr W last updated on 11/Dec/24

$$\sqrt{x^2+y^2+z^2}=r$$$$dxdydz=dV=Sdr=4\pi r^{2}dr$$$$-\mathrm{\infty }<x,y,z<+\mathrm{\infty }\Rightarrow 0⩽r<+\mathrm{\infty }$$

Commented by universe last updated on 11/Dec/24

$$thankusomuchsir$$

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