What surface is represented by the following equation

Question

$$\sqrt[3]{x^{2}}+\sqrt[3]{y^{2}}+\sqrt[3]{z^{2}}=1$$ Taking cubes of both sides only leads to a more complicated formula. How should one interpret this one. And, also if you could point me to a tool that is useful for graphing these kinds of surfaces, I would be grateful.

Answer 1

The general form is $( |x|^p + |y|^p + |z|^p )^\frac{1}{p} = 1$, which is just the surface with $p$-norm equal to $1$. The $1$-norm gives an octahedron, the $2$-norm gives a sphere, the $\infty$-norm (defined as the limit of the $p$-norm as $p \to \infty$ gives a cube, and in-between looks like in-between shapes. For yours, it is the $\frac{2}{3}$-norm which is is like an octahedron with the faces pushed inward, so it would have $6$ sharp points and $12$ curved edges and $8$ concave faces. As for 3d plotting software, they abound online as others have suggested.

Answer 2

If $(u, v, w)$ lies on the unit sphere in $\mathbf{R}^{3}$, i.e., if $u$, $v$, and $w$ are real numbers such that $u^{2} + v^{2} + w^{2} = 1$, then $$ (x, y, z) = (u^{3}, v^{3}, w^{3}) = F(u, v, w) $$ lies on your surface. In fact, the mapping $F$ defines a "homeomorphism" between the unit sphere and your surface, i.e., a continuous bijection with continuous inverse. The easiest way to prove this is perhaps to write down the inverse mapping, i.e., to solve for $(u, v, w)$ in terms of $(x, y, z)$, an easy exercise left to you.

If you have a favorite parametrization of the unit sphere (perhaps spherical coordinates?), composing with $F$ gives a parametrization of your surface. However, you can get a good idea (and a fine sketch) without a computer: If $c < 1$ is a real number, the equation $$ x^{2/3} + y^{2/3} = 1 - c $$ defines an astroid curve in the plane. This has some bearing in the shape of the "slices" of your surface by planes $z = \text{constant}$. Similar remarks hold if you fix $x$ or $y$ to a constant value.

Answer 3

I always use mathematica for plotting things like these.

ContourPlot3D[ x^(2/3) + y^(2/3) + z^(2/3) == 1, {x, -1, 1}, {y, -1, 1}, {z, -1, 1}]