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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.

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    ... from which you get the cubic equation $5\pi h^{3}-15\pi h^{2}+6=0.$2012-03-03

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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 $(3,0)$ enter image description here

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.

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    @Nuthinking: In [this Quartic/Cubic/Quadratic Equation Calculator](http://www.freewebs.com/brianjs/ultimateequationsolver.htm) I have computed \begin{eqnarray*} h &\approx &2.956294511506 \\ h &\approx &-0.338263678856 \\ h &\approx &0.38196916735. \end{eqnarray*}2012-03-03