Revisions to How to make a 3D plot of $(x^2+y^2-1)^2+(y^2+z^2-1)^2+(x^2+z^2-1)^2=0$ [closed]

added 419 characters in body

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edited Oct 25, 2015 at 12:16

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$(x^2+y^2-1)^2+(y^2+z^2-1)^2+(x^2+z^2-1)^2=0$

is satisfied by a set of points. This can be established:

f = (x^2 + y^2 - 1)^2 + (y^2 + z^2 - 1)^2 + (x^2 + z^2 - 1)^2;
FullSimplify[Reduce[f == 0, {x, y, z}, Reals]]
Reduce[x^2 + y^2 == 1 && z^2 + y^2 == 1 && x^2 + z^2 == 1, {x, y, z}]

i.e.

(x == -(1/Sqrt[2]) || x == 1/Sqrt[2]) && (y == -(1/Sqrt[2]) || 
   y == 1/Sqrt[2]) && (z == -(1/Sqrt[2]) || z == 1/Sqrt[2])

Note as expected the last 2 results are equivalent. f=0, is a set of 8 points (vertices of a cube). This is separate issue for surfaces of f=n that can be explored by ContourPlot3D.

This can be seen in many ways:

ir = ImplicitRegion[
  x^2 + y^2 == 1 && z^2 + y^2 == 1 && x^2 + z^2 == 1, {x, y, z}];
dr = DiscretizeRegion[ir]

enter image description here

This can be visualised by using ContourPLot3D (as alluded to by Nasser ) imagine the limiting process to contour value 0:

ContourPlot3D[f, {x, -3, 3}, {y, -3, 3}, {z, -3, 3}, 
 Contours -> {0.1, 0.2, 0.4}, 
 ContourStyle -> {Opacity[0.2], Opacity[0.2], Opacity[0.2]}, 
 Mesh -> False, PlotLegends -> Automatic]

enter image description here

Just another way to see this:

pts = Tuples[{-1/Sqrt[2], 1/Sqrt[2]}, 3];
Graphics3D[{Opacity[0.4], Cylinder[{{0, 0, -1}, {0, 0, 1}}, 1], 
  Cylinder[{{0, -1, 0}, {0, 1, 0}}, 1], 
  Cylinder[{{-1, 0, 0}, {1, 0, 0}}, 1],
  Sphere[{0, 0, 0}, Sqrt[3/2]],
  Opacity[1], PointSize[0.02], Red, Point[pts]}, Boxed -> False, 
 Background -> Black]

$(x^2+y^2-1)^2+(y^2+z^2-1)^2+(x^2+z^2-1)^2=0$

is satisfied by a set of points. This can be established:

f = (x^2 + y^2 - 1)^2 + (y^2 + z^2 - 1)^2 + (x^2 + z^2 - 1)^2;
FullSimplify[Reduce[f == 0, {x, y, z}, Reals]]
Reduce[x^2 + y^2 == 1 && z^2 + y^2 == 1 && x^2 + z^2 == 1, {x, y, z}]

i.e.

(x == -(1/Sqrt[2]) || x == 1/Sqrt[2]) && (y == -(1/Sqrt[2]) || 
   y == 1/Sqrt[2]) && (z == -(1/Sqrt[2]) || z == 1/Sqrt[2])

Note as expected the last 2 results are equivalent. f=0, is a set of 8 points (vertices of a cube). This is separate issue for surfaces of f=n that can be explored by ContourPlot3D.

This can be seen in many ways:

ir = ImplicitRegion[
  x^2 + y^2 == 1 && z^2 + y^2 == 1 && x^2 + z^2 == 1, {x, y, z}];
dr = DiscretizeRegion[ir]

enter image description here

This can be visualised by using ContourPLot3D (as alluded to by Nasser ) imagine the limiting process to contour value 0:

ContourPlot3D[f, {x, -3, 3}, {y, -3, 3}, {z, -3, 3}, 
 Contours -> {0.1, 0.2, 0.4}, 
 ContourStyle -> {Opacity[0.2], Opacity[0.2], Opacity[0.2]}, 
 Mesh -> False, PlotLegends -> Automatic]

enter image description here

$(x^2+y^2-1)^2+(y^2+z^2-1)^2+(x^2+z^2-1)^2=0$

is satisfied by a set of points. This can be established:

f = (x^2 + y^2 - 1)^2 + (y^2 + z^2 - 1)^2 + (x^2 + z^2 - 1)^2;
FullSimplify[Reduce[f == 0, {x, y, z}, Reals]]
Reduce[x^2 + y^2 == 1 && z^2 + y^2 == 1 && x^2 + z^2 == 1, {x, y, z}]

i.e.

(x == -(1/Sqrt[2]) || x == 1/Sqrt[2]) && (y == -(1/Sqrt[2]) || 
   y == 1/Sqrt[2]) && (z == -(1/Sqrt[2]) || z == 1/Sqrt[2])

Note as expected the last 2 results are equivalent. f=0, is a set of 8 points (vertices of a cube). This is separate issue for surfaces of f=n that can be explored by ContourPlot3D.

This can be seen in many ways:

ir = ImplicitRegion[
  x^2 + y^2 == 1 && z^2 + y^2 == 1 && x^2 + z^2 == 1, {x, y, z}];
dr = DiscretizeRegion[ir]

enter image description here

This can be visualised by using ContourPLot3D (as alluded to by Nasser ) imagine the limiting process to contour value 0:

ContourPlot3D[f, {x, -3, 3}, {y, -3, 3}, {z, -3, 3}, 
 Contours -> {0.1, 0.2, 0.4}, 
 ContourStyle -> {Opacity[0.2], Opacity[0.2], Opacity[0.2]}, 
 Mesh -> False, PlotLegends -> Automatic]

enter image description here

Just another way to see this:

pts = Tuples[{-1/Sqrt[2], 1/Sqrt[2]}, 3];
Graphics3D[{Opacity[0.4], Cylinder[{{0, 0, -1}, {0, 0, 1}}, 1], 
  Cylinder[{{0, -1, 0}, {0, 1, 0}}, 1], 
  Cylinder[{{-1, 0, 0}, {1, 0, 0}}, 1],
  Sphere[{0, 0, 0}, Sqrt[3/2]],
  Opacity[1], PointSize[0.02], Red, Point[pts]}, Boxed -> False, 
 Background -> Black]

edited the TeX formula

Source Link

edited Apr 11, 2015 at 17:44

Dr. belisarius

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$(x2+y2−1)^2+(y2+z2−1)^2+(x2+z2−1)^2=0$$(x^2+y^2-1)^2+(y^2+z^2-1)^2+(x^2+z^2-1)^2=0$

is satisfied by a set of points. This can be established:

f = (x^2 + y^2 - 1)^2 + (y^2 + z^2 - 1)^2 + (x^2 + z^2 - 1)^2;
FullSimplify[Reduce[f == 0, {x, y, z}, Reals]]
Reduce[x^2 + y^2 == 1 && z^2 + y^2 == 1 && x^2 + z^2 == 1, {x, y, z}]

i.e.

(x == -(1/Sqrt[2]) || x == 1/Sqrt[2]) && (y == -(1/Sqrt[2]) || 
   y == 1/Sqrt[2]) && (z == -(1/Sqrt[2]) || z == 1/Sqrt[2])

Note as expected the last 2 results are equivalent. f=0, is a set of 8 points (vertices of a cube). This is separate issue for surfaces of f=n that can be explored by ContourPlot3D.

This can be seen in many ways:

ir = ImplicitRegion[
  x^2 + y^2 == 1 && z^2 + y^2 == 1 && x^2 + z^2 == 1, {x, y, z}];
dr = DiscretizeRegion[ir]

enter image description here

This can be visualised by using ContourPLot3D (as alluded to by Nasser ) imagine the limiting process to contour value 0:

ContourPlot3D[f, {x, -3, 3}, {y, -3, 3}, {z, -3, 3}, 
 Contours -> {0.1, 0.2, 0.4}, 
 ContourStyle -> {Opacity[0.2], Opacity[0.2], Opacity[0.2]}, 
 Mesh -> False, PlotLegends -> Automatic]

enter image description here

$(x2+y2−1)^2+(y2+z2−1)^2+(x2+z2−1)^2=0$ is satisfied by a set of points. This can be established:

f = (x^2 + y^2 - 1)^2 + (y^2 + z^2 - 1)^2 + (x^2 + z^2 - 1)^2;
FullSimplify[Reduce[f == 0, {x, y, z}, Reals]]
Reduce[x^2 + y^2 == 1 && z^2 + y^2 == 1 && x^2 + z^2 == 1, {x, y, z}]

i.e.

(x == -(1/Sqrt[2]) || x == 1/Sqrt[2]) && (y == -(1/Sqrt[2]) || 
   y == 1/Sqrt[2]) && (z == -(1/Sqrt[2]) || z == 1/Sqrt[2])

Note as expected the last 2 results are equivalent. f=0, is a set of 8 points (vertices of a cube). This is separate issue for surfaces of f=n that can be explored by ContourPlot3D.

This can be seen in many ways:

ir = ImplicitRegion[
  x^2 + y^2 == 1 && z^2 + y^2 == 1 && x^2 + z^2 == 1, {x, y, z}];
dr = DiscretizeRegion[ir]

enter image description here

This can be visualised by using ContourPLot3D (as alluded to by Nasser ) imagine the limiting process to contour value 0:

ContourPlot3D[f, {x, -3, 3}, {y, -3, 3}, {z, -3, 3}, 
 Contours -> {0.1, 0.2, 0.4}, 
 ContourStyle -> {Opacity[0.2], Opacity[0.2], Opacity[0.2]}, 
 Mesh -> False, PlotLegends -> Automatic]

enter image description here

$(x^2+y^2-1)^2+(y^2+z^2-1)^2+(x^2+z^2-1)^2=0$

is satisfied by a set of points. This can be established:

f = (x^2 + y^2 - 1)^2 + (y^2 + z^2 - 1)^2 + (x^2 + z^2 - 1)^2;
FullSimplify[Reduce[f == 0, {x, y, z}, Reals]]
Reduce[x^2 + y^2 == 1 && z^2 + y^2 == 1 && x^2 + z^2 == 1, {x, y, z}]

i.e.

(x == -(1/Sqrt[2]) || x == 1/Sqrt[2]) && (y == -(1/Sqrt[2]) || 
   y == 1/Sqrt[2]) && (z == -(1/Sqrt[2]) || z == 1/Sqrt[2])

Note as expected the last 2 results are equivalent. f=0, is a set of 8 points (vertices of a cube). This is separate issue for surfaces of f=n that can be explored by ContourPlot3D.

This can be seen in many ways:

ir = ImplicitRegion[
  x^2 + y^2 == 1 && z^2 + y^2 == 1 && x^2 + z^2 == 1, {x, y, z}];
dr = DiscretizeRegion[ir]

enter image description here

This can be visualised by using ContourPLot3D (as alluded to by Nasser ) imagine the limiting process to contour value 0:

ContourPlot3D[f, {x, -3, 3}, {y, -3, 3}, {z, -3, 3}, 
 Contours -> {0.1, 0.2, 0.4}, 
 ContourStyle -> {Opacity[0.2], Opacity[0.2], Opacity[0.2]}, 
 Mesh -> False, PlotLegends -> Automatic]

enter image description here

deleted 412 characters in body

Source Link

edited Apr 11, 2015 at 15:20

ubpdqn

64.8k
3
65
154

$(x2+y2−1)2+(y2+z2−1)2+(x2+z2−1)2=0$$(x2+y2−1)^2+(y2+z2−1)^2+(x2+z2−1)^2=0$ is satisfied by a set of points. This can be established:

f = (x^2 + y^2 - 1)^2 + (y^2 + z^2 - 1)^2 + (x^2 + z^2 - 1)^2;
FullSimplify[Reduce[f == 0, {x, y, z}, Reals]]
Reduce[x^2 + y^2 == 1 && z^2 + y^2 == 1 && x^2 + z^2 == 1, {x, y, z}]

i.e.

(x == -(1/Sqrt[2]) || x == 1/Sqrt[2]) && (y == -(1/Sqrt[2]) || 
   y == 1/Sqrt[2]) && (z == -(1/Sqrt[2]) || z == 1/Sqrt[2])

Note as expected the last 2 results are equivalent. f=0, is a set of 8 points (vertices of a cube). This is separate issue for surfaces of f=n that can be explored by ContourPlot3D.

This can be seen in many ways:

ir = ImplicitRegion[
  x^2 + y^2 == 1 && z^2 + y^2 == 1 && x^2 + z^2 == 1, {x, y, z}];
dr = DiscretizeRegion[ir]

enter image description here

or more instructively,

pts = Tuples[{-(1/Sqrt[2]), 1/Sqrt[2]}, 3];
Show[ParametricPlot3D[{{u, v, u^2 + v^2}, {u^2 + v^2, u, v}, {u, 
    u^2 + v^2, v}, {u, v, -u^2 - v^2}, {-u^2 - v^2, u, 
    v}, {u, -u^2 - v^2, v}}, {u, -2, 2}, {v, -2, 2}, Mesh -> False, 
  PlotStyle -> 
   Join[Table[Opacity[0.2], {6}], Table[Opacity[0.4], {6}]]], 
 Graphics3D[{Red, PointSize[0.02], Point[pts]}]]

enter image description here

orThis can be visualised by using ContourPLot3D (as alluded to by Nasser ) imagine the limiting process to contour value 0:

ContourPlot3D[f, {x, -3, 3}, {y, -3, 3}, {z, -3, 3}, 
 Contours -> {0.1, 0.2, 0.4}, 
 ContourStyle -> {Opacity[0.2], Opacity[0.2], Opacity[0.2]}, 
 Mesh -> False, PlotLegends -> Automatic]

enter image description here

$(x2+y2−1)2+(y2+z2−1)2+(x2+z2−1)2=0$ is satisfied by a set of points. This can be established:

f = (x^2 + y^2 - 1)^2 + (y^2 + z^2 - 1)^2 + (x^2 + z^2 - 1)^2;
FullSimplify[Reduce[f == 0, {x, y, z}, Reals]]
Reduce[x^2 + y^2 == 1 && z^2 + y^2 == 1 && x^2 + z^2 == 1, {x, y, z}]

i.e.

(x == -(1/Sqrt[2]) || x == 1/Sqrt[2]) && (y == -(1/Sqrt[2]) || 
   y == 1/Sqrt[2]) && (z == -(1/Sqrt[2]) || z == 1/Sqrt[2])

Note as expected the last 2 results are equivalent. f=0, is a set of 8 points (vertices of a cube). This is separate issue for surfaces of f=n that can be explored by ContourPlot3D.

This can be seen in many ways:

ir = ImplicitRegion[
  x^2 + y^2 == 1 && z^2 + y^2 == 1 && x^2 + z^2 == 1, {x, y, z}];
dr = DiscretizeRegion[ir]

enter image description here

or more instructively,

pts = Tuples[{-(1/Sqrt[2]), 1/Sqrt[2]}, 3];
Show[ParametricPlot3D[{{u, v, u^2 + v^2}, {u^2 + v^2, u, v}, {u, 
    u^2 + v^2, v}, {u, v, -u^2 - v^2}, {-u^2 - v^2, u, 
    v}, {u, -u^2 - v^2, v}}, {u, -2, 2}, {v, -2, 2}, Mesh -> False, 
  PlotStyle -> 
   Join[Table[Opacity[0.2], {6}], Table[Opacity[0.4], {6}]]], 
 Graphics3D[{Red, PointSize[0.02], Point[pts]}]]

enter image description here

or using ContourPLot3D imagine the limiting process to contour value 0:

ContourPlot3D[f, {x, -3, 3}, {y, -3, 3}, {z, -3, 3}, 
 Contours -> {0.1, 0.2, 0.4}, 
 ContourStyle -> {Opacity[0.2], Opacity[0.2], Opacity[0.2]}, 
 Mesh -> False, PlotLegends -> Automatic]

enter image description here

$(x2+y2−1)^2+(y2+z2−1)^2+(x2+z2−1)^2=0$ is satisfied by a set of points. This can be established:

f = (x^2 + y^2 - 1)^2 + (y^2 + z^2 - 1)^2 + (x^2 + z^2 - 1)^2;
FullSimplify[Reduce[f == 0, {x, y, z}, Reals]]
Reduce[x^2 + y^2 == 1 && z^2 + y^2 == 1 && x^2 + z^2 == 1, {x, y, z}]

i.e.

(x == -(1/Sqrt[2]) || x == 1/Sqrt[2]) && (y == -(1/Sqrt[2]) || 
   y == 1/Sqrt[2]) && (z == -(1/Sqrt[2]) || z == 1/Sqrt[2])

Note as expected the last 2 results are equivalent. f=0, is a set of 8 points (vertices of a cube). This is separate issue for surfaces of f=n that can be explored by ContourPlot3D.

This can be seen in many ways:

ir = ImplicitRegion[
  x^2 + y^2 == 1 && z^2 + y^2 == 1 && x^2 + z^2 == 1, {x, y, z}];
dr = DiscretizeRegion[ir]

enter image description here

This can be visualised by using ContourPLot3D (as alluded to by Nasser ) imagine the limiting process to contour value 0:

ContourPlot3D[f, {x, -3, 3}, {y, -3, 3}, {z, -3, 3}, 
 Contours -> {0.1, 0.2, 0.4}, 
 ContourStyle -> {Opacity[0.2], Opacity[0.2], Opacity[0.2]}, 
 Mesh -> False, PlotLegends -> Automatic]

enter image description here

added 369 characters in body

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edited Apr 11, 2015 at 11:39

ubpdqn

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154

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added 369 characters in body

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edited Apr 11, 2015 at 11:32

ubpdqn

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added 70 characters in body

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edited Apr 11, 2015 at 10:53

ubpdqn

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created Apr 11, 2015 at 10:47

ubpdqn

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