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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.

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closed as unclear what you're asking by belisarius, bobthechemist, Artes, m_goldberg, Yves Klett Jan 24 at 16:07

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question.If this question can be reworded to fit the rules in the help center, please edit the question.

    
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 –  Kuba Jan 24 at 11:08
    
@Kuba yes. actually I have to draw $|\bar{x}|^\frac{1}{2}$ where $\bar{x}\in \mathbb{R}^3$ i.e $\bar{x}=(x,y,z)$. –  ricci1729 Jan 24 at 11:10
    
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. :) –  Michael E2 Jan 24 at 11:19
    
@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? –  Kuba Jan 24 at 11:23
    
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. –  ricci1729 Jan 24 at 11:27

1 Answer 1

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

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

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