What is a name of this 3D shape obtained from two circles?

Question

I drew two intersecting circles and then folded the 3D shape.

Question.

Does this 3D shape have a mathematical name? is there a formula that describes this 3D shape?

Edit.

I ploted an astroid and circle.

Answer 1

It can be done without cheating; and, as colt_browning has remarked, such boxes can even be bought.

The box has a star shaped top and bottom, and four lens shaped sidewalls. The following figure shows the bottom star before the bending. The idea is to bend the four spikes cylindrically up, so that the grey lines become level lines, and the small central square stays flat. The bending is performed in such a way that the arc $\gamma$ in the figure becomes an arc $\gamma'$ rising according to the same law $s\mapsto z(s)$ as the lower rim of a lens put on edge.

The arc $\gamma$ begins at $\bigl(1-{\sqrt{2}\over2},1-{\sqrt{2}\over2}\bigr)$ and ends at $(1,0)$. It can be parametrized by arc length as $$\gamma:\quad s\mapsto\left(1-\sin\left({\textstyle{\pi\over4}}-s\right),1-\cos\left({\textstyle{\pi\over4}}-s\right),\ 0\right)\qquad\left(0\leq s\leq{\textstyle{\pi\over4}}\right)\ .$$ After the bending we have an arc $\gamma'$ given by $$\gamma':\quad s\mapsto\left(x(s),1-\cos\left({\textstyle{\pi\over4}}-s\right),\ 1-\cos s\right)\qquad\left(0\leq s\leq{\textstyle{\pi\over4}}\right)\ .$$ The function $s\mapsto z(s)=1-\cos s$ models the slope of the lower lens rim, and $s\mapsto x(s)$ has to be determined from the condition $\dot x^2+\dot y^2+\dot z^2=1$. One obtains $$\dot x^2(s)={1\over2}\bigl(\cos(2s)+\sin(2s)\bigr)\ ,$$ so that $s\mapsto x(s)$ will not be an elementary function.

In the end the lens will be glued to $\gamma'$. It is then standing on edge and will be cylindrically bent as well.