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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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    If you have trouble formatting, you could see http://meta.math.stackexchange.com/questions/1773/do-we-have-an-equation-editing-howto2012-07-19

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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:

enter image description here

and rotating it about the $z$-axis to produce a surface:

enter image description here

If that is the case, then the equation for the above surface is $z=\frac{1}{\sqrt{x^2+y^2}}$

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    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*}2012-07-19
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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?