Evaporation of spherical droplet of liquid Suppose a spherical droplet of liquid evaporates at a rate proportional to its surface area.
$${dV\over dt} = -kA$$
where $V=volume (mm^3)$, $t=time(min)$, $k=$ evaporation rate $ (mm/min)$, and $A=$ surface area $(mm^2)$. Use Euler's method to compute the volume of the droplet from t=0 to 10 min using a step size of 0.25 min. Assume k = 0.1, and that the initial droplet has a radius of 3mm. Assess the validity of your results by determining the radius of your final computed volume and verifying that it is consistent with the evaporation rate.
I think I have a handle of the Euler's method aspect, so now I'm trying to determine the radius of the liquid mathematically and compare the two.
I tried converting the equation into terms of radius and time and integrating:
$${d {4\over 3} \pi r^3 \over dt} = -k4\pi r^2 $$
Simplifying...
$$ {d {1\over 3} r \over dt} = -k $$
$$ \int d r = \int -3k dt$$
$$ r = -3kt + C$$
Using r(t=0) to solve for C...
$$3 = -3k(0) + C$$
$$r = -3kt + 3$$
but the result isn't right. I don't think I can convert the volume derivative to radius and integrate like that, but I don't know how else to go about it.
I would appreciate, in addition to help on the question, if anyone could point me to similar examples. Thanks!

What you did to solve the ODE by substituting the formula for $V$ and $A$ was correct.
The error you made is when you dropped the $r^2$ term while $r^3$ was still in differntial form. So you should have differentiated $r^3$ first then you can factorise. Anyway here's the correct approach:
$$
\frac{dV}{dt}=-kA\\
\frac{d\frac{4\pi}{3}r^3}{dt}=-k4\pi r^2\\
\frac{4\pi}{3}\frac{d}{dt}r^3=-4\pi k r^2\\
\frac{1}{3}\times 3r^2\frac{dr}{dt}=-kr^2\\
r^2\frac{dr}{dt}=-kr^2\\
\frac{dr}{dt}=-k\\
r(t)=-kt+C
$$
And since $r(t=0)=3$ then:
$$
C=3
$$
So:
$$
r(t)=-kt+3
$$
