Take $k \in \Bbb N$ intersecting American footballs and configure them inside a unit sphere such that each football touches two opposite ends of the sphere. Each of the shapes are spaced evenly apart.
In the case $k\to \infty$ the volume should be equal to the volume of a sphere.
Related Problem (intersection of $n$ congruent cylinders).
I want to tackle the case $k=2$ first. So I'm using "Cavalieri's Lemon" which is the surface of revolution of a parabolic arc, to model a football.
The equation for the parabolic arc to be revolved about the $x-$axis is: $f(x)=x(1-x).$
What is the volume in the region where the (two) footballs overlap?
I've calculated that the volume of one football is $\frac{\pi}{30}.$ If the interior shape is a well-known shape I can proabably search for the volume of that shape, but I'm having trouble visualizing the interior shape with the enclosed volume.
Let's assume the axes of the footballs are perpendicular to each other, e.g., $$(y^2 + z^2) = (1/4 - x^2)^2, \\ (x^2 + z^2) = (1/4 - y^2)^2.$$ Then we are looking for the volume common to their intersection. We are searching for the value of $x = y = c > 0$ when $z = 0$ corresponding to the intersection of the curves $y = 1/4 - x^2$, $x = 1/4 - y^2$, which is of course at $(x,y) = (c,c)$ where $c = \frac{\sqrt{2}-1}{2}$. Hence the volume corresponds to the integral $$\begin{align} V &= 16 \int_{x=0}^c \int_{y=x}^{1/4-x^2} \sqrt{(1/4 - x^2)^2 - y^2} \, dy \, dx \\ &= \frac{1}{4} \int_{x=0}^c \left(1-4 x^2\right)^2\left( \pi - 2 \tan ^{-1}\frac{4 x}{\sqrt{16 x^4-24 x^2+1}}\right) - 8 x \sqrt{16 x^4-24 x^2+1} \, dx \\ &= \frac{2\pi + 16 \log 2 - 13}{60} \\ &\approx 0.0728923. \end{align}$$