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I am presented with the following problem: Sketch the region bounded by the surfaces $z=\sqrt{x^2+y^2}$ and $x^2+y^2=1$ for $1$ less than/equal to $z$ greater than/equal to $2$.

I am not sure how to go about plotting this. I have tried multiple different ways but nothing seems to be working.

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  • $\begingroup$ You realize this is just the unit circle in the x-y plane translated by (0,0,1)? In any case note that RegionPlot3D requires a thickness to plot. If you'd like I can provide an answer where I show the shape using RegionPlot3D. $\endgroup$
    – b3m2a1
    Mar 9, 2017 at 0:57
  • $\begingroup$ Its fine, because the dawggie wants the bound area between those planes, not the planes themselves. $\endgroup$
    – Tomi
    Mar 9, 2017 at 1:03

1 Answer 1

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Lets start with the first area (since we are considering the bound area, we can use equalities). Change the equality depending on a closed or open interval.

RegionPlot3D[Sqrt[x^2 + y^2] < z, {x, -4, 4}, {y, -4, 4}, {z, -4, 4}]

enter image description here

Bound it with the second

RegionPlot3D[
 Sqrt[x^2 + y^2] < z && x^2 + y^2 < 1, {x, -4, 4}, {y, -4, 4}, {z, -4,
   4}]

enter image description here

And then the final z condition

RegionPlot3D[
 Sqrt[x^2 + y^2] < z && x^2 + y^2 < 1 && 1 <= z <= 2, {x, -4, 
  4}, {y, -4, 4}, {z, -4, 4}]

enter image description here

And there you have it.

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  • $\begingroup$ The two commands return a blank plot for me? EDIT: All the commands return a blank plot :( $\endgroup$
    – naanman
    Mar 10, 2017 at 4:16
  • $\begingroup$ Can you post one command - also, what version of Mathematica :) $\endgroup$
    – Tomi
    Mar 10, 2017 at 10:47

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