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.
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.
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 $z=\frac{1}{\sqrt{x^2+y^2}}$
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?