We seek the time dependence of the depth of a liquid in a hemispherical shell due to evaporation assuming the evaporation rate (change in volume w.r.t time) is proportional to the surface area and assuming constant pressure, temperature, humidity, and no wind. That is,
$$
\frac{dV}{dt}=\gamma A
$$
where ##V## is the volume,##\gamma## is the constant of proportionality, and ##A## is the surface area. We first compute the volume in terms of the depth of the liquid and construct the volume integral. From the diagram below,
From the diagram we see,
$$
h=\rho - H
$$
$$
\alpha = \sin^{-1}(\frac{h}{\rho})= \sin^{-1}(1 -\frac{H}{\rho})\\
$$
where ##H## is the liquid depth. The volume integral, in spherical coordinates, becomes
$$
V=\int_0^{2\pi} \int_{\frac{\pi}{2}+\sin^{-1}(1 -\frac{H}{\rho})}^{\pi}\int_{\rho -H}^{\rho}\rho^{'2}\sin(\theta^{'})d\rho{'}d\theta^{'}d\phi^{'}
$$
Evaluating the integral we find
$$
V=\frac{2\pi}{3}\rho^{3}y^2(3 -3y +y^2)
$$
$$
y=\frac{H}{\rho}
$$
We take the derivative of ##V## w.r.t time;
$$
\frac{dV}{dt}=\frac{2\pi}{3}\rho^{3}y(6-9y+4y^2)\frac{dy}{dt}
$$
The surface area is ##\pi R^2## with ##R^2=\rho^2-H^2## we have
$$
\frac{dV}{dt}=\gamma \pi \rho^2 y(2-y)
$$
$$
\frac{2\pi \rho^3}{3}y(6-9y+4y^2)\frac{dy}{dt}=\gamma \pi \rho^2 y(2-y)
$$
We integrate both sides of the equation
$$
\frac{2\rho}{3\gamma}\int \frac{(6-9y+4y^2)}{(2-y)}dy=\int dt
$$
These integrations evaluate to
$$
-y^2+\frac{y}{2}-\log((y-2)^2)+C=\frac{3\gamma}{4\rho}t
$$
where ##C## is the constant of integration. To determine ##C## we assume at time ##t=0##, ##y=1##, i.e. the hemisphere is filled to the brim. We find
$$
C=\frac{1}{2}
$$
To evaluate ##\gamma## take the measurement of depth (##H_0##) at time ##t_0##
$$
-(\frac{H_0}{\rho})^2 + \frac{H_0}{\rho}+\frac{1}{2}-\log((\frac{H_0}{\rho}-2)^2)=\frac{3\gamma}{4\rho}t_0
$$
and we find for the equation for normalized depth vs. time
$$
-y^2 + y + \frac{1}{2}-\log((y-2)^2)=\frac{t}{t_0}(-(\frac{H_0}{\rho})^2 +\frac{H_0}{\rho}+\frac{1}{2}-\log((\frac{H_0}{\rho}-2)^2))
$$
The equation of ##y## as a function of time can't be expressed in terms of elementary functions and must be evaluated by numerical techniques. I wrote a C++ routine to find ##y(t)## with accuracy ~one part in ##10^{-10}##. Taking ##\frac{H_0}{\rho}=\frac{1}{2}## and ##t_0=500## I obtained the following plot