Rate of Evaporation of a Vase 
The base of a vase is created by rotating the curve
$$f(x)=2.393794315((1.25916975182872)\left(\frac{x}{2.24581}\right)^4+(-4.29578020022745)(\frac{x}{2.24581})^3+(4.37766951188496)(\frac{x}{2.24581})^2+(-0.553610482668319)(\frac{x}{2.24581})+(-1.6343566433531))$$
...defined in the domain {${0.224581\le x\le 2.919553}$}, rotated $2\pi$ radians about the x-axis.

The units of the coordinate axes are in centimetres.


My goal is to find the time it takes when the base of the vase is completely filled with water to evaporate.

Through some basic evaporation equations from thermodynamics, I have discovered that:
$$\frac{dV}{dt}=-2.55\cdot Exposed\ Surface\ Area$$
As the curve above is being revolved around the x-axis, it is obvious that:
$$Exposed\ Surface\ Area=\pi \cdot (f(x))^2$$
However, this is as far as I've been able to go. In a worked example I saw, the rate of change in volume ($\frac{dV}{dt}$) was linked to the volume in the container remaining, and then the total volume of the vase was plugged in to solve the differential equation for time. Here it is:

SOLUTION



Please help me proceed. Any help will be greatly appreciated, thank you in advance.

 A: There are far more numbers in this question than I'd like so I'm going to try to remove everything not relevant to solving a slightly more general problem and then you should be able to finish off the problem.
Assume we have a cylindrical symmetrical vase and represent it by its radius $f$ as a function of the height above the ground $x$.
If the vase is filled water to a height $h$ then the exposed surface area of the water is $A(h)$ where we define: $$A(x)=\pi(f(x))^2.$$
The volume contained is the volume of revolution:
$$V(h) = \int_0^h A\mathrm d x.$$
Now we have evaporation given by some rate and the exposed surface area:
$$\frac{\mathrm d V}{\mathrm d t} = -k A(h),$$
where $h$ is the instantaneous water level and $k>0$ is some constant for the evaporation rate.
We want to know what the water level $h$ will be at various times so we want to consider $\frac{\mathrm d h}{\mathrm d t}.$
Ok. Now we're going to finish this off in a few lines. Look out for:


*

*The chain rule

*The fundamental theorem of calculus

*A bit of rearranging and substitution

*The assumption that $A\ne0$

*Solving a simple differential equation


Here we go:
\begin{align}
\frac{\mathrm d V}{\mathrm d t}&=\frac{\mathrm d h}{\mathrm d t}\frac{\mathrm d V}{\mathrm d h}\\
\frac{\mathrm d V}{\mathrm d h} &=
\frac{\mathrm d }{\mathrm d h}
\int_0^h A(x) \mathrm d x\\
&= A(h)\\
A(h) &= -\frac1k\frac{\mathrm d V}{\mathrm d t} \\
&=-\frac1k\frac{\mathrm d h}{\mathrm d t} A(h)\\
-k&=\frac{\mathrm d h}{\mathrm d t}\\
h&=h_0-k t
\end{align}
Given the initial height $h_0$ you can solve the final equation for the time at which $h=0.$
Perhaps this result may be surprising but if you accept that it is reasonable to work out a volume of revolution by integrating surface area (summing up a bunch of lamina) then it seems reasonable that removing them at a constant rate by evaporation would reduce the water level at a constant rate.
Another way to think of this is by applying some dimensional analysis to the evaporation rate: The evaporation rate is some volume (length cubed) that evaporates per a given surface area (length squared) in a given time. Thus it has dimension $\frac{L^3}{L^2T}=LT^{-1}$ which is a velocity so it is quite natural to suppose that evaporation rate is the same as the rate that the depth of fluid decreases, and indeed it is. 
To finish off your problem, you have $h_0=2.919553-0.224581=2.694972\text{ cm}$ and $k=2.55$ (unknown units; dimension $L^3T^{-1}L^{-2}=LT^{-1}$) so the time for the vase to empty would be $\frac{h_0}{k}\approx 1.06$ (unknown time units)
