Dgdrxrt

Use BFD on a Virtual-Template Interface

How to tell that you are a giant?

How to find out what spells would be useless to a blind NPC spellcaster?

Identify plant with long narrow paired leaves and reddish stems

How to deal with a team lead who never gives me credit?

What is the meaning of the new sigil in Game of Thrones Season 8 intro?

What's the purpose of writing one's academic biography in the third person?

What exactly is a "Meth" in Altered Carbon?

What causes the vertical darker bands in my photo?

Align equal signs while including text over equalities

Simplicity of the roots of a minimal polynomial

String `!23` is replaced with `docker` in command line

What does the "x" in "x86" represent?

Are two submodules (where one is contained in the other) isomorphic if their quotientmodules are isomorphic?

Can an alien society believe that their star system is the universe?

Why aren't air breathing engines used as small first stages

Fundamental Solution of the Pell Equation

Why didn't this character "real die" when they blew their stack out in Altered Carbon?

Why do people hide their license plates in the EU?

51k Euros annually for a family of 4 in Berlin: Is it enough?

What does this icon in iOS Stardew Valley mean?

Why was the term "discrete" used in discrete logarithm?

Resolving to minmaj7

What does an IRS interview request entail when called in to verify expenses for a sole proprietor small business?

Asymptote: 3d graph over a disc

5

Is there a straightforward way to draw a 3D graph over a disc domain? Say
z=x^2-y^2 for x^2+y^2<1.

[I just started to use asymptote; this page explained me how to do it for a rectangular domain. I hope it is an easy question.]

graphs asymptote

share|improve this question

asked Apr 1 at 3:36

Anton PetruninAnton Petrunin

542313

add a comment | 

5

Is there a straightforward way to draw a 3D graph over a disc domain? Say
z=x^2-y^2 for x^2+y^2<1.

[I just started to use asymptote; this page explained me how to do it for a rectangular domain. I hope it is an easy question.]

graphs asymptote

share|improve this question

asked Apr 1 at 3:36

Anton PetruninAnton Petrunin

542313

add a comment | 

Is there a straightforward way to draw a 3D graph over a disc domain? Say
z=x^2-y^2 for x^2+y^2<1.

[I just started to use asymptote; this page explained me how to do it for a rectangular domain. I hope it is an easy question.]

graphs asymptote

share|improve this question

asked Apr 1 at 3:36

Anton PetruninAnton Petrunin

542313

share|improve this question

asked Apr 1 at 3:36

Anton PetruninAnton Petrunin

542313

share|improve this question

asked Apr 1 at 3:36

Anton PetruninAnton Petrunin

542313

asked Apr 1 at 3:36

Anton PetruninAnton Petrunin

542313

asked Apr 1 at 3:36

Anton PetruninAnton Petrunin

542313

1 Answer
1

active

oldest

votes

5

One way to make sure that x^2+y^2<1 is to use polar coordinates. Then x=r cos(phi) and y=r sin(phi).

documentclass[variwidth,border=3.14mm]standalone
usepackageasypictureB
begindocument
beginasypicturename=discgraph
 usepackage("mathrsfs");
 import graph3;
 import solids;
 import interpolate;

 settings.outformat="pdf";


 size(500); 

 defaultpen(0.5mm);
 pen darkgreen=rgb(0,138/255,122/255);

 draw(Label("$x$",1),(0,0,0)--(1.2,0,0),darkgreen,Arrow3);
 draw(Label("$y$",1),(0,0,0)--(0,1.2,0),darkgreen,Arrow3);
 draw(Label("$f(x,y)$",1),(0,0,0)--(0,0,0.6),darkgreen,Arrow3);



 //function: call the radial coordinate r=t.x and the angle phi=t.y
 triple f(pair t) 
 return ((t.x)*cos(t.y), (t.x)*sin(t.y),
 ((t.x)*cos(t.y))^2-((t.x)*sin(t.y))^2);
 

 surface s=surface(f,(0,1),(0.49,2.5*pi),32,16,
 usplinetype=new splinetype[] notaknot,notaknot,monotonic,
 vsplinetype=Spline);
 pen p=rgb(0,0,.7); 
 draw(s,lightolive+white);
endasypicture 
enddocument

share|improve this answer

answered Apr 1 at 3:58

marmotmarmot

118k6153288

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thank you, but is there a direct way to make a condition x^2+y^2<1 for the arguments?
  
  – Anton Petrunin
  Apr 1 at 4:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @marmot: The x-axis near origin should be hidden from the given point of view. Is there any way to improve this issue? E.g., by setting some samples-option?
  
  – Marian G.
  Apr 1 at 5:28
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  A line has a thickness, a surface not. It is why you see the x-axis near origin. You can observe the same behavior with a simple square surface and the x-axis. Perhaps it is possible to avoid its by creating two z translated surfaces, but you have to manage the boundary...
  
  – O.G.
  Apr 1 at 13:16
  

  
  
  
  

add a comment | 

Your Answer

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "85"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2ftex.stackexchange.com%2fquestions%2f482530%2fasymptote-3d-graph-over-a-disc%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

5

One way to make sure that x^2+y^2<1 is to use polar coordinates. Then x=r cos(phi) and y=r sin(phi).

documentclass[variwidth,border=3.14mm]standalone
usepackageasypictureB
begindocument
beginasypicturename=discgraph
 usepackage("mathrsfs");
 import graph3;
 import solids;
 import interpolate;

 settings.outformat="pdf";


 size(500); 

 defaultpen(0.5mm);
 pen darkgreen=rgb(0,138/255,122/255);

 draw(Label("$x$",1),(0,0,0)--(1.2,0,0),darkgreen,Arrow3);
 draw(Label("$y$",1),(0,0,0)--(0,1.2,0),darkgreen,Arrow3);
 draw(Label("$f(x,y)$",1),(0,0,0)--(0,0,0.6),darkgreen,Arrow3);



 //function: call the radial coordinate r=t.x and the angle phi=t.y
 triple f(pair t) 
 return ((t.x)*cos(t.y), (t.x)*sin(t.y),
 ((t.x)*cos(t.y))^2-((t.x)*sin(t.y))^2);
 

 surface s=surface(f,(0,1),(0.49,2.5*pi),32,16,
 usplinetype=new splinetype[] notaknot,notaknot,monotonic,
 vsplinetype=Spline);
 pen p=rgb(0,0,.7); 
 draw(s,lightolive+white);
endasypicture 
enddocument

share|improve this answer

answered Apr 1 at 3:58

marmotmarmot

118k6153288

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thank you, but is there a direct way to make a condition x^2+y^2<1 for the arguments?
  
  – Anton Petrunin
  Apr 1 at 4:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @marmot: The x-axis near origin should be hidden from the given point of view. Is there any way to improve this issue? E.g., by setting some samples-option?
  
  – Marian G.
  Apr 1 at 5:28
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  A line has a thickness, a surface not. It is why you see the x-axis near origin. You can observe the same behavior with a simple square surface and the x-axis. Perhaps it is possible to avoid its by creating two z translated surfaces, but you have to manage the boundary...
  
  – O.G.
  Apr 1 at 13:16
  

  
  
  
  

add a comment | 

5

One way to make sure that x^2+y^2<1 is to use polar coordinates. Then x=r cos(phi) and y=r sin(phi).

documentclass[variwidth,border=3.14mm]standalone
usepackageasypictureB
begindocument
beginasypicturename=discgraph
 usepackage("mathrsfs");
 import graph3;
 import solids;
 import interpolate;

 settings.outformat="pdf";


 size(500); 

 defaultpen(0.5mm);
 pen darkgreen=rgb(0,138/255,122/255);

 draw(Label("$x$",1),(0,0,0)--(1.2,0,0),darkgreen,Arrow3);
 draw(Label("$y$",1),(0,0,0)--(0,1.2,0),darkgreen,Arrow3);
 draw(Label("$f(x,y)$",1),(0,0,0)--(0,0,0.6),darkgreen,Arrow3);



 //function: call the radial coordinate r=t.x and the angle phi=t.y
 triple f(pair t) 
 return ((t.x)*cos(t.y), (t.x)*sin(t.y),
 ((t.x)*cos(t.y))^2-((t.x)*sin(t.y))^2);
 

 surface s=surface(f,(0,1),(0.49,2.5*pi),32,16,
 usplinetype=new splinetype[] notaknot,notaknot,monotonic,
 vsplinetype=Spline);
 pen p=rgb(0,0,.7); 
 draw(s,lightolive+white);
endasypicture 
enddocument

share|improve this answer

answered Apr 1 at 3:58

marmotmarmot

118k6153288

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thank you, but is there a direct way to make a condition x^2+y^2<1 for the arguments?
  
  – Anton Petrunin
  Apr 1 at 4:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @marmot: The x-axis near origin should be hidden from the given point of view. Is there any way to improve this issue? E.g., by setting some samples-option?
  
  – Marian G.
  Apr 1 at 5:28
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  A line has a thickness, a surface not. It is why you see the x-axis near origin. You can observe the same behavior with a simple square surface and the x-axis. Perhaps it is possible to avoid its by creating two z translated surfaces, but you have to manage the boundary...
  
  – O.G.
  Apr 1 at 13:16
  

  
  
  
  

add a comment | 

One way to make sure that x^2+y^2<1 is to use polar coordinates. Then x=r cos(phi) and y=r sin(phi).

documentclass[variwidth,border=3.14mm]standalone
usepackageasypictureB
begindocument
beginasypicturename=discgraph
 usepackage("mathrsfs");
 import graph3;
 import solids;
 import interpolate;

 settings.outformat="pdf";


 size(500); 

 defaultpen(0.5mm);
 pen darkgreen=rgb(0,138/255,122/255);

 draw(Label("$x$",1),(0,0,0)--(1.2,0,0),darkgreen,Arrow3);
 draw(Label("$y$",1),(0,0,0)--(0,1.2,0),darkgreen,Arrow3);
 draw(Label("$f(x,y)$",1),(0,0,0)--(0,0,0.6),darkgreen,Arrow3);



 //function: call the radial coordinate r=t.x and the angle phi=t.y
 triple f(pair t) 
 return ((t.x)*cos(t.y), (t.x)*sin(t.y),
 ((t.x)*cos(t.y))^2-((t.x)*sin(t.y))^2);
 

 surface s=surface(f,(0,1),(0.49,2.5*pi),32,16,
 usplinetype=new splinetype[] notaknot,notaknot,monotonic,
 vsplinetype=Spline);
 pen p=rgb(0,0,.7); 
 draw(s,lightolive+white);
endasypicture 
enddocument

share|improve this answer

answered Apr 1 at 3:58

marmotmarmot

118k6153288

documentclass[variwidth,border=3.14mm]standalone
usepackageasypictureB
begindocument
beginasypicturename=discgraph
 usepackage("mathrsfs");
 import graph3;
 import solids;
 import interpolate;

 settings.outformat="pdf";


 size(500); 

 defaultpen(0.5mm);
 pen darkgreen=rgb(0,138/255,122/255);

 draw(Label("$x$",1),(0,0,0)--(1.2,0,0),darkgreen,Arrow3);
 draw(Label("$y$",1),(0,0,0)--(0,1.2,0),darkgreen,Arrow3);
 draw(Label("$f(x,y)$",1),(0,0,0)--(0,0,0.6),darkgreen,Arrow3);



 //function: call the radial coordinate r=t.x and the angle phi=t.y
 triple f(pair t) 
 return ((t.x)*cos(t.y), (t.x)*sin(t.y),
 ((t.x)*cos(t.y))^2-((t.x)*sin(t.y))^2);
 

 surface s=surface(f,(0,1),(0.49,2.5*pi),32,16,
 usplinetype=new splinetype[] notaknot,notaknot,monotonic,
 vsplinetype=Spline);
 pen p=rgb(0,0,.7); 
 draw(s,lightolive+white);
endasypicture 
enddocument

share|improve this answer

answered Apr 1 at 3:58

marmotmarmot

118k6153288

answered Apr 1 at 3:58

marmotmarmot

118k6153288

answered Apr 1 at 3:58

marmotmarmot

118k6153288