Implicit surface plot of a finite cylinder

Using fimplicit3, I can plot an infinite cylinder, like so:

args={'EdgeColor','none','FaceColor','c','FaceAlpha',0.3};

fimplicit3(@(x,y,z) x.^2+y.^2-1, args{:} ); axis([-3,+3,-3,+3,-3,+3])

However, what I would really like is for the cylinder to be of finite length, say 4. My first idea for how to do this was the following. This does produce a finite cylinder of the correct length, but it is closed at both ends, which I do not want.

fimplicit3(@(x,y,z) x.^2+y.^2-1 + (abs(z)>2), args{:} ); axis([-3,+3,-3,+3,-3,+3])

I believe I know why this is happening. There is a discontinuous jump from negative values to 1 at the end caps, and so the fimplicit3 algorithm assumes there are roots there. However, I wonder if there is a way to trick fimplicit3 somehow so that I get a cylinder with open ends. Any ideas?

Answer 1

[Matt J]

The solution, as it turns out, was fairly simple:

args={'EdgeColor','none','FaceColor','c','FaceAlpha',0.3};

fimplicit3(@fun, args{:} ); axis([-3,+3,-3,+3,-3,+3])

function val=fun(x,y,z)

  val=x.^2+y.^2-1;
  val(abs(z)>2)=nan;

end