 Solution

The wind removes $2\pi rW(r)dr$ of mass per unit time from an annulus; $W$ is already the surface-density loss term in the supplied mass conservation equation, so there is no additional two-face factor. Integrating the photoevaporation profile givesThe convergence at infinity is important: the loss is concentrated near the photoevaporative gravitational radius. Using $r_g=1.4\times 10^{14}\mathrm{cm}$ yields ${\dot{M}}_w\simeq 2.46\times 10^{17}\mathrm{g}{\mathrm{s}}^{-1}$, or about $3.9\times 10^{-9}{M}_{\odot }\mathrm{y}{\mathrm{r}}^{-1}$. The initial disk mass is $2\times 10^{30}\mathrm{g}$, so the wind-only depletion time isThis estimate treats the heated area and wind normalization as fixed and neglects additional removal through stellar accretion. Once the disk shrinks, its wind rate and geometry need not remain constant.