Plotting a parametric surface where the domain is non-rectangular

Question

Hey I'm teaching a calculus course, and for an example in my lecture on surface integrals I would like to generate a surface plot in MATLAB of a portion of a circular cylinder. Here is my MATLAB code for the case where the parameter domain is rectangular:

clc; clear  
syms theta z  
x(theta,z) = cos(theta)
y(theta,z) = sin(theta)  
z(theta,z) = z  
fsurf(x, y, z, [0 2*pi 0 1])

However, the surfaced required for the problem instead has 0<z<1+sin(theta)

I'm not sure what is the best way to modify this code for the case where the domain is not rectangular. Thanks!

Ramashalanka · Accepted Answer · 2025-08-11 02:14:50Z

3

You could continue to use a rectangular domain, but use a dummy variable t and a mapping to your z:

syms theta t
x(theta,t) = cos(theta);
y(theta,t) = sin(theta);
z(theta,t) = t .* (1 + sin(theta));
fsurf(x, y, z, [0 2*pi 0 1])

You can do it numerically with the same result:

theta = linspace(0, 2*pi, 100);
t = linspace(0, 1, 50)';
X = repmat(cos(theta), length(t), 1);
Y = repmat(sin(theta), length(t), 1);
Z = t * (1 + sin(theta));
surf(X, Y, Z)

To avoid introducing t, you can do it by component with the same result:

theta = linspace(0, 2*pi, 100);
Nz = 50;
Z = nan(numel(theta),Nz);
for ntheta = 1:numel(theta)
    Z(ntheta,:) = linspace(0,1+sin(theta(ntheta)),Nz);
end
THETA = repmat(theta',1,Nz);
X = cos(THETA);
Y = sin(THETA);
surf(X, Y, Z)

The previous option is just a vectorized version of this.

edited Aug 11 at 2:14

answered Aug 10 at 9:45

Ramashalanka

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2 Comments

ಠ_ಠ Aug 10 at 19:56

Thank you! Is there a way to do this without introducing the dummy variable? I plan on sharing the code with my students, and I think they might get confused by this, since it will look different to how we will be writing the parametrization on paper. I'm not married to the symbolic approach (I just do it out of habit), and I'm happy to use the usual numerical functions instead if needed .

Ramashalanka Aug 11 at 2:15

See my updated answer.

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