I want draw a graph correctly

I write code like this

 function z = F3(x,y)
 z = 4- x.^2- 2*y.^2;

 figure(1)
 [X,Y] = meshgrid(0:0.1:2);
 mesh(X,Y,F3(X,Y))
 axis([0 2 0 2 0 6])

and than

graph is came out like this

I don't want to see value of Z under '0'

How can I do that??

Thought I give the value of the Z axis range from 0 to 6, but it is steel came out like this ,,,

[C.J. Harris]

Which version of MATLAB are you using?

[pty0220]

R2014a for mac thanks ~

[Walter Roberson]

I confirm that with that release on Mac I see the same results. The clipping property of the axes is seemingly being ignored.

Answer 1

[Thorsten]

With your axis command you cut off the z values at 0 (with the third 0 in axis([0 2 0 2 0 6]). Instead, use:

 Z = F3(X,Y);
 mesh(X,Y,Z);
 axis([0 2 0 2 min(Z(:)) max(Z(:))])

[pty0220]

I want to cut off the Z values at 0

but it's not completely cut off at Z = 0

graph is came out out side of the graph that's my problem ,,, T.T

[Thorsten]

I see. You could set the negative Z values to 0 or NaN:

Z(Z<0) = 0;

or

Z(Z<0) = NaN;

In the first case you have the a mesh at the positions were Z < 0, with NaN the cut-off is ragged. So both solutions may be better than the original plot, but they are not perfect.

[Mike Garrity]

You're seeing something called

ClippingStyle = 'rectangle'

In that type of clipping, the geometry is clipped at the 2D rectangle that surrounds the axes. That was the only type of clipping that MATLAB Graphics supported until R2014b. Since that release, you have another choice:

ClippingStyle = '3dbox'

And I think that will actually be the default in this case. So if you upgrade to a newer version, you should get what you want.

In newer versions, if you don't like the clipping style that was chosen for you, you can change it like so:

set(gca,'ClippingStyle','3dbox')

or

set(gca,'ClippingStyle','rectangle')