I have a mathematical question that I need help with. I have a conical funnel with a radius of $r$ and a height of $h = r\sqrt{3}$. The funnel is filled up to a height of $\frac{7}{8}h$, and then a sphere is completely placed inside the funnel. As a result, the water in the funnel rises exactly up to the rim.
I would like to calculate the radius of the sphere in terms of the radius of the funnel.
First I thought I could just take $1/8$ of the volume and solve than for $r_\text{sphere}$ but $7/8$ is the height and the $1/8$ at the bottom has a diffrent volume from the one on top. So my next thought was to calculate the volume vor $8/8$ of the height than form $7/8$ and sub the lower from the higher. But with this I have the problem that I think that the $r$ should be smaller since we dont use the full hight and the funnel gets wider as it gets higher. None of my friends know the answer. Can someone help me?
The 1/8 volume is a frustum
$$\begin{align} V &=\frac{\pi h\left(r^2+\left(\frac{7}{8}r\right)^2+\frac{7}{8}r\times r\right)}{3}\\ &=\frac{\pi \frac{1}{8}r^3\sqrt{3}\left(1+\left(\frac{7}{8}\right)^2+\frac{7}{8}\right)}{3}\\ &=\frac{169\pi r^3}{512\sqrt{3}} \end{align}$$
$$\begin{align}r_\text{sphere} &=\sqrt[3]{\frac{3\left(\frac{169\pi r^3}{512\sqrt{3}}\right)}{4\pi}}\\ &=r\sqrt[3]{\frac{169\sqrt{3}}{2048}}\\ &=\frac{6.5^{\frac{2}{3}}3^{\frac{1}{6}}r}{8} \end{align} $$