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$\because ~\int (-\ln y)^2dy=2y-2y\ln y-y \ln^2 y\\ ~~~~~~~~~~~\lim_{y \to 0}y\ln y=0\\ ~~~~~~~~~~~\lim_{y \to 0}y\ln^2 y=0\\ \therefore V=\pi \int_0^1\varphi(y)^2dy=\pi\int_0^1(-\ln y)^2dy=2\pi \\$

[ Last edited by soliton923 on 2011-5-21 at 20:43 ]

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