Spinnin' Curves

The tangent to the curve $y = \ln x$ at the point $(u,\ln u)$ has gradient $u^{-1}$ , so the normal at this point has gradient $-u$ , and hence has equation $y - \ln u \; = \; -u(x - u)$ which passes through the origin when $\begin{aligned} -\ln u & = \; u^2 \\ ue^{u^2} & = \; 1 \\ 2u^2e^{2u^2} & = \; 2 \\ 2u^2 & = \; W(2) \end{aligned}$ so the shortest distance $r$ from the origin to the curve $y = \ln x$ is given by $r^2 \; = \; u^2 + (\ln u)^2 \; = \; u^4 + u^2 \; = \; \tfrac14W(2)^2 + \tfrac12 W(2) \; = \; \tfrac14W(2)(W(2) +2)$ and so the area untouched by the rotation process is $\pi r^2 \; = \; \frac{\pi}{4}W(2)(W(2) +2)$ making the answer $4 \times 2 = \boxed{8}$ .