If I interpreted your question correctly then you simply do as Américo Tavares does and substitute in the values. Given that $V_{cap}=0.4$ and $R=1$ you get: $V_{cap}=\frac{1}{3}\pi h^2(3R-h)\Longrightarrow0.4=\frac{1}{3}\pi h^2(3-h)$ Simplifying gives: $0=5\pi h^3-15h^2+6$ However, unfortunately it seems as though the zero's of the cubic seem to have rather nasty closed forms which can be found here.
An alternate method of finding the volume of a cap can be used using a little bit of calculus.
Say we have a circle with radius $r$ centered at $r$. Then we get a circle with equation $(x-r)^2+y^2=r^2$.
Here is an example with radius 3 with center $(3,0)$ 
We can simplify the general equation and get $y^2=r^2-(x^2-2xr+r^2)=2xr-x^2$. Since the volume of revolution of a circle is a sphere, the function for its volume is $\int\pi y^2dx=\int \pi(2xr-x^2)dx$ Now, if you want to find the volume of a cap with height $x$ you get: $V_{cap}(x)=\int_0^x\pi(2tr-t^2)dt=\pi r x^2-\frac{\pi x^3}{3}$
Edit: My "alternate method" is in fact the exact same method as that givien on the link you provided, just with a slightly different proof. I lieu of this, please consider only the first part of the answer.