>>199
∬_[x^2+y^2≦1-l^2]√{1+(∂z/∂x)^2+(∂z/∂y)^2}dxdy = (4/3)π

左辺= ∬_[x^2+y^2≦1-l^2]{1/√(1-x^2-y^2)}dxdy
=∫[θ=0,2π]dθ∫[r=0,√(1-l^2)]{r/√(1-r^2)}dr
=(2π)*(1-l)

l = 1/3