Given (x^2 + y^2 + z^2)^2 = 8z, can z be written as a single function f(x, y)? Describe the surface.

Step-by-step solution

1. 1

   Note the immediate sign restriction. The left side is a square, hence non-negative, so $8z\ge 0$ and therefore $z\ge 0$. The entire surface lies in the upper half-space.

2. 2

   Test the origin to settle the question. Setting $x=y=0$ gives

   $\left(z^{2}\right)^{2}=8z\ \Longrightarrow\ z^{4}-8z=0\ \Longrightarrow\ z\left(z^{3}-8\right)=0\ \Longrightarrow\ z=0\ \text{or}\ z=2$

   Two values of $z$ correspond to the single input $(0,0)$, so no single-valued function $f$ can describe the whole surface — it fails the vertical line test.

3. 3

   Solve for the radial distance instead. Taking the non-negative square root of both sides (legitimate because the left side is a square and $z\ge 0$):

   $x^{2}+y^{2}+z^{2}=\sqrt{8z}\ \Longrightarrow\ x^{2}+y^{2}=\sqrt{8z}-z^{2}$

   This describes the surface cleanly, as a circle of a computable radius at each height $z$.

4. 4

   Find the range of heights. The left side $x^{2}+y^{2}$ cannot be negative, so we need

   $\sqrt{8z}\ge z^{2}\ \Longrightarrow\ 8z\ge z^{4}\ \Longrightarrow\ z\left(8-z^{3}\right)\ge 0\ \Longrightarrow\ 0\le z\le 2$

   so the surface is bounded, sitting between the planes $z=0$ and $z=2$.

5. 5

   Describe the shape. At each height $z\in[0,2]$ the cross-section is a circle of radius $\sqrt{\sqrt{8z}-z^{2}}$. That radius is $0$ at both $z=0$ and $z=2$ and positive in between (it peaks at $z=2^{-1/3}\approx 0.7937$, where the radius reaches $\approx 1.3747$), so the surface is a closed, apple-like body of revolution about the $z$-axis, pinched to points at top and bottom.

6. 6

   State the conclusion. There is no single function $z=f(x,y)$; the surface must be given implicitly, or split into an upper and a lower branch over the disk $x^{2}+y^{2}\le\max\left(\sqrt{8z}-z^{2}\right)$. Reporting a single closed-form $f$ here would silently discard half the surface.