# double integration of parametric function

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Danny Van Elsen 2020년 4월 10일
댓글: David Goodmanson 2020년 4월 17일
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.

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### 채택된 답변

David Goodmanson 2020년 4월 15일
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
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Danny Van Elsen 2020년 4월 16일
편집: Danny Van Elsen 2020년 4월 16일
Indeed, I have the same solution as
(manual)
r = cos(v)i + u*sin(v)j + vk
dr/du = cos(v)i + sin(v)j
dr/dv = u*sin(v)i + u*cos(v)j + k
||dr/du X dr/dv|| = sqrt(sin(v)^2 + cos(v)^2 + u^2)
which then leads to the double integral
(matlab)
fun = @(u,v) sqrt(1 + u.^2)
q = integral2(fun, 0, 5, 0, 4*pi)
q = 174.7199
But somehow, and probably naively, I was expecting there to be a function taking the r-vector and the limits as input, and then "just" calculate the area. In one line as it were. So this is probably not the case?
David Goodmanson 2020년 4월 17일
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.

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### 추가 답변 (1개)

Rajat Tewari 2020년 4월 13일
You can refer to integral2 function of MATLAB to calculate the double integraltion and area.

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