Let $f(x,y,z)=0$ is a bounded closed surface defined in $D^3$ Where $D$ is a finite closed interval included in real number set. The planar projections of the surface are given:
- On $z=0$ plane the Projection is given.
- On $y=0$ the Projection is given.
- On $x=0$ the Projection is given.
From the above mentioned planar projections can we uniquely determine the original surface?
Let $D=[-1,1]$ and consider the surface $f(x,y,z)=0$ where $$ f(x,y,z)=(x^2+y^2+z^2-1)(x^2+y^2+z^2-r^2) $$ for some fixed $r\in (0,1)$.
Then the projections are independent of the choice of $r$.
To address your comment, to show an example where the surface is connected and bounds a connected region, we can modify the above example as follows . . .
Let $D=[-1,1]$ and consider the surface $f(x,y,z)=0$ where $$ f(x,y,z) = \Bigl(z-\sqrt{1-x^2-y^2}\Bigr) \Bigl(z-a\sqrt{1-x^2-y^2}\Bigr) $$ for some fixed $a\in (0,1)$.
Then the projections are independent of the choice of $a$, and the region bounded by the surface is connected.