y=logx
y'=1/x
接点のx座標をαとして
接線は
y-logα=(1/α)*(x-α)
原点を通るから
logα=1
α=e

V = (π/3)*e*1^2 - π∫[x:1,e](logx)^2*dx
= (π/3)*e*1^2 - π(e-2)
= (2π/3)(3-e)


∫[x:1,e](logx)^2*dx
= [x*(logx)^2] - ∫[x:1,e]2*logx*dx
= e - 2*[x*logx] + 2*∫[x:1,e]1*dx
= e - 2e + 2e - 2
= e - 2