Can you please tell me how to plot $\sqrt{r}$ in 3d sph polar plot?

In the spherical plot 3d there is no option I can put variable r. So how should I proceed.

  • $\begingroup$ So you want $\mathbb{R}^3\to \mathbb{R}$ function? In exactly this case maybe better just Plot[Sqrt[r],{r, 0, rmax}]. In more complicated example, mathematica.stackexchange.com/q/19575/5478 $\endgroup$ – Kuba Jan 24 '14 at 11:08
  • $\begingroup$ @Kuba yes. actually I have to draw $|\bar{x}|^\frac{1}{2}$ where $\bar{x}\in \mathbb{R}^3$ i.e $\bar{x}=(x,y,z)$. $\endgroup$ – ricci1729 Jan 24 '14 at 11:10
  • $\begingroup$ To plot $\sqrt{r}$ vs. $(x,y,z)$, you would need four coordinates. See for example these questions: 19575, 20023, 25277, 26636 - (Oh, Kuba beat to the first one while I was searching. :) $\endgroup$ – Michael E2 Jan 24 '14 at 11:19
  • $\begingroup$ @MichaelE2 About that link, shouldn't the title be corrected to "3D functions", I mean, th "n" in "n-D functions" usualy reffers to dimension of a domain, doesn't it? $\endgroup$ – Kuba Jan 24 '14 at 11:23
  • $\begingroup$ To MichaelE2 and Kuba Please correct me if I have mistakes. I need the plot in 3D. Please see this notes on sobolev space (pg 21, Fig 2.2)where I encountered this iecn.u-nancy.fr/~munnier/files/cours_edp.pdf I want to reproduce the plot and I got the problem. Thanking you both once again. $\endgroup$ – ricci1729 Jan 24 '14 at 11:27
ParametricPlot3D[{u, v, Sqrt@Sqrt[ u u + v v]}, {u, -2, 2}, {v, -2, 2},
                RegionFunction -> (Norm[{#4, #5}] < 1 &), AspectRatio -> 1, 
                MeshFunctions -> {(#3 &), (ArcTan[#2, #1] &)}]

Mathematica graphics

Compare with your book:

Mathematica graphics

  • $\begingroup$ if you want to avoid this real nuisance indeterminacy message you may replace ArcTan with (Arg[Complex[#2, #1]] &) $\endgroup$ – Stefan Jan 24 '14 at 14:05
  • $\begingroup$ @Stefan Or just enclose the whole thing in Quiet[] :) $\endgroup$ – Dr. belisarius Jan 24 '14 at 14:09

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