Calculate volume of sphere cap I found few equations here: http://mathworld.wolfram.com/SphericalCap.html
Developing one ($V_{\text{cap}}=\frac{1}{3}\pi h^{2}(3R-h)$) and considering that $R$ in my case is equals to $1$ and $V_{\text{cap}}$ is $0.4$ I can't figure out $h$.
I am stuck here: $0.4 = h^2 (3-h)$
Tried also to use various equation solvers but they all return 0.
Thanks a lot!
P.S.: I hope this doesn't sound as silly as it feels.
 A: 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.
