I want to plot a 3d-plane (x=0) in MATLAB, (-2<=y<=2) and limited by the line z=4-y^2. Here's my code :
y=linspace(-2,2,50);
z=4-y.^2;
[Y Z]=meshgrid(y,z);
X=0*Y+0*Z;
mesh(X,Y,Z);
axis tight;
Can you give me any hints???? thank you very much!!

2 件のコメント

Walter Roberson
Walter Roberson 2014 年 5 月 22 日
編集済み: Walter Roberson 2014 年 5 月 22 日
When you say that the plane is to be limited by that z, and that you want to draw plot it, then what difference is there compared to plotting a single curved line in 3 space?
Nghia
Nghia 2014 年 5 月 22 日
編集済み: Nghia 2014 年 5 月 22 日
I'm not clear what you're trying to ask but the code which I posted plot just a single curved line in 3 space, but what I want is to plot a plane x=0 and limited by both the line z=0 and z=4-y^2 I mean that I want to use to plot a plane surfc(x,y,z)

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Hugo
Hugo 2014 年 5 月 22 日

0 投票

This solution is based on patch
y=-2:.01:2;
z=4-y.^2;
numy=length(y);
% Constructing the vertices
V=zeros(numy,3);
for iy=1:numy,
V(iy,:)=[0,y(iy),z(iy)]; % Vertices in the parabola
V(iy+numy,:)=[0,y(iy),0]; % Vertices in the y axis
end
% Constructing faces
F=zeros(numy-1,4); for iy=1:numy-1, F(iy,:)=[iy,iy+1,iy+1+numy,iy+numy]; end
patch('Vertices',V,'Faces',F)
The plot may look like a line. You just need to rotate it and you will see the figure in 3D.
Hope this helps.

2 件のコメント

Nghia
Nghia 2014 年 5 月 22 日
thank you for your help!
Nghia
Nghia 2014 年 5 月 26 日
編集済み: Nghia 2014 年 5 月 26 日
I have tried your code then I rewrite like this
%code patch('Vertices',V,'Faces',F,'FaceColor',[1 0 0])
but it doesn't change the face's color and it's still black. How can I fix this?

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