# Function of the surface obtained by rotating the graph of $\frac{1}{|x|}$

What is the function of the 3 dimensional plane created when the graph of 1/abs(x) is rotated in the z-axis around the origin?

I'm sorry for bad formatting and if this is a duplicate.

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"The graph of 1/abs(x)" What does that mean? Do you mean the SURFACE $z = 1/|x|$ or the plane curve $y = 1/|x|$? And rotating about the z-axis and the origin are very different things – Hawk Jul 19 '12 at 4:47
If you have trouble formatting, you could see meta.math.stackexchange.com/questions/1773/… – Ross Millikan Jul 19 '12 at 4:49

I'm not sure I understand your phrasing, but I assume that you are looking for the equation of the surface obtained by taking the graph of $z=\dfrac{1}{|x|}$ in the $xz$-plane:
and rotating it about the $z$-axis to produce a surface:
If that is the case, then the equation for the above surface is $$z=\frac{1}{\sqrt{x^2+y^2}}$$
As a note: parametric equations make the construction of a surface of revolution much easier. If you have some function $z=f(x)$ (i.e. your function is embedded in the $x$-$z$ coordinate plane), and you wanted to rotate about the $z$-axis, then the parametric equations are \begin{align*}x&=u\cos\,\theta\\y&=u\sin\,\theta\\z&=f(u)\end{align*} – J. M. Jul 19 '12 at 5:17
Hint: First, what is the other side of the equation you are graphing? I would guess $y=\frac 1{|x|}$. Second, can you convince yourself that the absolute value doesn't matter? If you have a point $(x,y,z)$, isn't $(-x,y,z)$ a rotated version?