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Visualize manifold specified by equalities
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Visualize manifold specified by equalities
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$begingroup$
Suppose I have two nonlinear equalities $x^3 = y^2, y = z^3$. How can I visualize the manifold in $mathbb{R}^3$ that is generated by simultaneously satisfying the two equalities? I think ContourPlot3D is the one to use but I couldn't get it to work show the set of points in $mathbb{R}^3$ that satisfy the two equalities. The best I can do is make it show the intersection of the surfaces:
How can I plot the curve defined by the intersection in 3D?
graphics3d
asked 32 mins ago
ITAITA
1575
$endgroup$
add a comment |
$begingroup$
Suppose I have two nonlinear equalities $x^3 = y^2, y = z^3$. How can I visualize the manifold in $mathbb{R}^3$ that is generated by simultaneously satisfying the two equalities? I think ContourPlot3D is the one to use but I couldn't get it to work show the set of points in $mathbb{R}^3$ that satisfy the two equalities. The best I can do is make it show the intersection of the surfaces:
How can I plot the curve defined by the intersection in 3D?
graphics3d
asked 32 mins ago
ITAITA
1575
$endgroup$
add a comment |
$begingroup$
Suppose I have two nonlinear equalities $x^3 = y^2, y = z^3$. How can I visualize the manifold in $mathbb{R}^3$ that is generated by simultaneously satisfying the two equalities? I think ContourPlot3D is the one to use but I couldn't get it to work show the set of points in $mathbb{R}^3$ that satisfy the two equalities. The best I can do is make it show the intersection of the surfaces:
How can I plot the curve defined by the intersection in 3D?
graphics3d
asked 32 mins ago
ITAITA
1575
$endgroup$
Suppose I have two nonlinear equalities $x^3 = y^2, y = z^3$. How can I visualize the manifold in $mathbb{R}^3$ that is generated by simultaneously satisfying the two equalities? I think ContourPlot3D is the one to use but I couldn't get it to work show the set of points in $mathbb{R}^3$ that satisfy the two equalities. The best I can do is make it show the intersection of the surfaces:
How can I plot the curve defined by the intersection in 3D?
graphics3d
graphics3d
asked 32 mins ago
ITAITA
1575
asked 32 mins ago
ITAITA
1575
asked 32 mins ago
ITAITA
1575
asked 32 mins ago
ITAITA
1575
asked 32 mins ago
ITAITA
1575
1575
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You can use the option BoundaryStyle to mark the intersection of the two contour surfaces as follows:
ContourPlot3D[{x^3 == y^2, y == z^3}, {x, -2, 2}, {y, -2, 2}, {z, -2, 2},
Mesh -> None, ContourStyle -> Opacity[.3],
BoundaryStyle -> {1 -> None, 2 -> None, {1, 2} -> Directive[Thick, Red]}]
Also
SliceContourPlot3D[y - z^3, x^3 == y^2, {x, -2, 2}, {y, -2, 2}, {z, -2, 2},
Contours -> {{0}}, BoundaryStyle -> None, ContourShading -> None,
ContourStyle -> Directive[Red, Thick]]
answered 23 mins ago
kglrkglr
186k10202421
$endgroup$
$begingroup$
That's exactly what I needed. I follow most of it: Since a list was passed as first argument the '{1 -> None, 2-> None ... }' but how Mathematica knew to handle {1,2} -> is just magic!
$endgroup$
– ITA
13 mins ago
add a comment |
$begingroup$
r = 1;
R = ImplicitRegion[{x^3 == y^2, y == z^3}, {{x, -r, r}, {y, -r, r}, {z, -r, r}}];
Region[R]
answered 22 mins ago
Henrik SchumacherHenrik Schumacher
55.1k475154
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can use the option BoundaryStyle to mark the intersection of the two contour surfaces as follows:
ContourPlot3D[{x^3 == y^2, y == z^3}, {x, -2, 2}, {y, -2, 2}, {z, -2, 2},
Mesh -> None, ContourStyle -> Opacity[.3],
BoundaryStyle -> {1 -> None, 2 -> None, {1, 2} -> Directive[Thick, Red]}]
Also
SliceContourPlot3D[y - z^3, x^3 == y^2, {x, -2, 2}, {y, -2, 2}, {z, -2, 2},
Contours -> {{0}}, BoundaryStyle -> None, ContourShading -> None,
ContourStyle -> Directive[Red, Thick]]
answered 23 mins ago
kglrkglr
186k10202421
$endgroup$
$begingroup$
That's exactly what I needed. I follow most of it: Since a list was passed as first argument the '{1 -> None, 2-> None ... }' but how Mathematica knew to handle {1,2} -> is just magic!
$endgroup$
– ITA
13 mins ago
add a comment |
$begingroup$
You can use the option BoundaryStyle to mark the intersection of the two contour surfaces as follows:
ContourPlot3D[{x^3 == y^2, y == z^3}, {x, -2, 2}, {y, -2, 2}, {z, -2, 2},
Mesh -> None, ContourStyle -> Opacity[.3],
BoundaryStyle -> {1 -> None, 2 -> None, {1, 2} -> Directive[Thick, Red]}]
Also
SliceContourPlot3D[y - z^3, x^3 == y^2, {x, -2, 2}, {y, -2, 2}, {z, -2, 2},
Contours -> {{0}}, BoundaryStyle -> None, ContourShading -> None,
ContourStyle -> Directive[Red, Thick]]
answered 23 mins ago
kglrkglr
186k10202421
$endgroup$
$begingroup$
That's exactly what I needed. I follow most of it: Since a list was passed as first argument the '{1 -> None, 2-> None ... }' but how Mathematica knew to handle {1,2} -> is just magic!
$endgroup$
– ITA
13 mins ago
add a comment |
$begingroup$
You can use the option BoundaryStyle to mark the intersection of the two contour surfaces as follows:
ContourPlot3D[{x^3 == y^2, y == z^3}, {x, -2, 2}, {y, -2, 2}, {z, -2, 2},
Mesh -> None, ContourStyle -> Opacity[.3],
BoundaryStyle -> {1 -> None, 2 -> None, {1, 2} -> Directive[Thick, Red]}]
Also
SliceContourPlot3D[y - z^3, x^3 == y^2, {x, -2, 2}, {y, -2, 2}, {z, -2, 2},
Contours -> {{0}}, BoundaryStyle -> None, ContourShading -> None,
ContourStyle -> Directive[Red, Thick]]
answered 23 mins ago
kglrkglr
186k10202421
$endgroup$
You can use the option BoundaryStyle to mark the intersection of the two contour surfaces as follows:
ContourPlot3D[{x^3 == y^2, y == z^3}, {x, -2, 2}, {y, -2, 2}, {z, -2, 2},
Mesh -> None, ContourStyle -> Opacity[.3],
BoundaryStyle -> {1 -> None, 2 -> None, {1, 2} -> Directive[Thick, Red]}]
Also
SliceContourPlot3D[y - z^3, x^3 == y^2, {x, -2, 2}, {y, -2, 2}, {z, -2, 2},
Contours -> {{0}}, BoundaryStyle -> None, ContourShading -> None,
ContourStyle -> Directive[Red, Thick]]
answered 23 mins ago
kglrkglr
186k10202421
edited 11 mins ago
answered 23 mins ago
kglrkglr
186k10202421
answered 23 mins ago
kglrkglr
186k10202421
answered 23 mins ago
kglrkglr
186k10202421
186k10202421
$begingroup$
That's exactly what I needed. I follow most of it: Since a list was passed as first argument the '{1 -> None, 2-> None ... }' but how Mathematica knew to handle {1,2} -> is just magic!
$endgroup$
– ITA
13 mins ago
add a comment |
$begingroup$
That's exactly what I needed. I follow most of it: Since a list was passed as first argument the '{1 -> None, 2-> None ... }' but how Mathematica knew to handle {1,2} -> is just magic!
$endgroup$
– ITA
13 mins ago
$begingroup$
That's exactly what I needed. I follow most of it: Since a list was passed as first argument the '{1 -> None, 2-> None ... }' but how Mathematica knew to handle {1,2} -> is just magic!
$endgroup$
– ITA
13 mins ago
$begingroup$
That's exactly what I needed. I follow most of it: Since a list was passed as first argument the '{1 -> None, 2-> None ... }' but how Mathematica knew to handle {1,2} -> is just magic!
$endgroup$
– ITA
13 mins ago
add a comment |
$begingroup$
r = 1;
R = ImplicitRegion[{x^3 == y^2, y == z^3}, {{x, -r, r}, {y, -r, r}, {z, -r, r}}];
Region[R]
answered 22 mins ago
Henrik SchumacherHenrik Schumacher
55.1k475154
$endgroup$
add a comment |
$begingroup$
r = 1;
R = ImplicitRegion[{x^3 == y^2, y == z^3}, {{x, -r, r}, {y, -r, r}, {z, -r, r}}];
Region[R]
answered 22 mins ago
Henrik SchumacherHenrik Schumacher
55.1k475154
$endgroup$
add a comment |
$begingroup$
r = 1;
R = ImplicitRegion[{x^3 == y^2, y == z^3}, {{x, -r, r}, {y, -r, r}, {z, -r, r}}];
Region[R]
answered 22 mins ago
Henrik SchumacherHenrik Schumacher
55.1k475154
$endgroup$
r = 1;
R = ImplicitRegion[{x^3 == y^2, y == z^3}, {{x, -r, r}, {y, -r, r}, {z, -r, r}}];
Region[R]
answered 22 mins ago
Henrik SchumacherHenrik Schumacher
55.1k475154
answered 22 mins ago
Henrik SchumacherHenrik Schumacher
55.1k475154
answered 22 mins ago
Henrik SchumacherHenrik Schumacher
55.1k475154
answered 22 mins ago
Henrik SchumacherHenrik Schumacher
55.1k475154
55.1k475154
add a comment |
add a comment |
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