double integration of parametric function
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hello all,
I know how to plot a parametric surface, for example as in
syms u v
x = u * cos(v);
y = u * sin(v);
z = v;
fsurf(x, y, z, [0 5 0 4*pi])
but can someone point me to the appropriate function for the double integration that calculates the surface area, for example over the interval
0 <= u <= 5
0 <= v <= 4*pi
of the example above?
regards, Danny.
采纳的回答
David Goodmanson
2020-4-15
Hi Danny,
You can find the surface area by finding the vectors Du and Dv that are parallel to the surface when you vary u and v respectively. Taking their cross product gives the the normal unit vector n, times the area element dS of a parallelogram whose area is proportional to dudv. Integrating the area elements give the total area. Since the area element does not depend on v, you can multiply by 4*pi and just do the u integral.
This procedure is analogous to finding the Jacobian. It's almost easier to do this by hand than to use syms, but
syms u v a real
x = u * cos(v);
y = u * sin(v);
z = a*v; % in case you want to vary the pitch
zplot = v;
fsurf(x, y, zplot, [0 5 0 4*pi])
Du = diff([x,y,z],u) % Du is diff() times du
Dv = diff([x,y,z],v) % Dv is diff() times dv
dS = simplify(norm(cross(Du,Dv)))
Area = 4*pi*int(dS,0,5)
double(subs(Area,a,1))
% results
Du = [ cos(v), sin(v), 0]
Dv = [ -u*sin(v), u*cos(v), a]
dS = (a^2 + u^2)^(1/2)
Area = 4*pi*((a^2*log((a^2 + 25)^(1/2) + 5))/2 - (a^2*log(a^2))/4 + (5*(a^2 + 25)^(1/2))/2)
ans = 174.7199
2 个评论
Danny Van Elsen
2020-4-16
编辑：Danny Van Elsen
2020-4-16
David Goodmanson
2020-4-17
Hi Danny, I don't think so, although I would be happy to be proved wrong. For a volume integral you can at least get the Jacobian basically in one line with symbolic variables.
更多回答（1 个）
Rajat Tewari
2020-4-13
0 个投票
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