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Hi Erin,
Here is a totally different method, based on the equation for the cone. It finds the surface area of a cone cut by a plane, then subtracts the area under plane 1 from the area under plane 2. In each case the quantity Sbase is the area of the unperturbed cone, from the lowest height where the plane intersects the cone down to the apex. Phi is the cone opening half-angle, 45 degrees in your case.
I have not downloaded SurfaceIntersection so I can't compare results with the previous answer.
phi = pi/4;
beta = 0.414;
h = [0.406 0.609];
T = tan(phi)/cos(phi);
S = zeros(1,2);
for k = 1:2
z0 = h(k);
z1 = z0/(1+beta*tan(phi));
z2 = z0/(1-beta*tan(phi));
fun = @(z) 2*z.*acos((z-z0)./(beta*tan(phi)*z));
Sint = T*integral(fun,z1,z2);
Sbase = pi*T*z1^2;
S(k) = Sint + Sbase;
end
Sfinal = diff(S)
Best Answer