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Line integral over a 3D surface

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Riccardo Giaffreda
Riccardo Giaffreda 2020 年 1 月 22 日
編集済み: Torsten 2022 年 8 月 16 日
I have a surface defined by its x,y,z coordinates and want to find, fixed two points (x1,y1) and (x2,y2) the line integral over the surface of this two points. How can I do it?
  2 件のコメント
darova
darova 2020 年 1 月 22 日
What have you tried? Did you try interp2?
Riccardo Giaffreda
Riccardo Giaffreda 2020 年 1 月 22 日
This is what I tried:
The first part of the code is what gives me the 3D plot and value of the function. The important part is the matrix SF_Full which gives tha value on the surface of the function. I am asking about the second part
clear all
clc
Sampling_Map = 1;
SF_Mean = 0;
SF_Sigma = 6;
R = 50;
x = [ -R : Sampling_Map : R];
y = [ -R : Sampling_Map : R];
Num_Sample_Map = length(x);
SF_Sample = [ 1: 15 : Num_Sample_Map];
Num_Sample_SF = length(SF_Sample);
[X,Y] = meshgrid(x,y);
SF = zeros(Num_Sample_Map,Num_Sample_Map);
SF(SF_Sample,SF_Sample)= 6*(randn(Num_Sample_SF,Num_Sample_SF));
figure (1)
SF_Full=griddata(x(SF_Sample)',y(SF_Sample)',SF(SF_Sample,SF_Sample),X,Y,'cubic');
mesh(x,y,SF_Full)
hold on
plot3(X(SF_Sample,(SF_Sample)),Y(SF_Sample,(SF_Sample)),SF(SF_Sample,SF_Sample),'o')
hold off
Given two points on the map, for example x1=(-19,1) and x2(-9;-9) this is what I did to get the integral. Do you think it can work for any couple of numbers?
hold on
t = [-19:1:-9];
v = [1:-1:-9];
plot(t,v)
SUM=zeros(1,length(v));
for i=1:length(v)
SUM(i)= SF_Full(v(i)+R+1, t(i)+R+1); % +R+1 is for mapping the value on the plot and the corresponding on the matrix
end
Q= trapz( SUM )

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回答 (2 件)

darova
darova 2020 年 1 月 22 日
Here is a short example
[X,Y,Z] = peaks(20);
x = linspace(-2,2,50);
y = linspace(-1,3,50);
z = interp2(X,Y,Z,x,y);
surf(X,Y,Z)
alpha(0.2)
hold on
plot3(x,y,z,'.-r')
hold off
I don't understand what do you mean by linear integral? What do you want to calculate?
  2 件のコメント
Riccardo Giaffreda
Riccardo Giaffreda 2020 年 1 月 22 日
The integral over the segment (the one in red in your plot, the one with x1 and x2 as ends in mine)
darova
darova 2020 年 1 月 22 日
You mean length of the curve?

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Riccardo Giaffreda
Riccardo Giaffreda 2020 年 1 月 22 日
No, the line integral from x1=(-19,1) to x2(-9;-9) of the function SF_Full.
Cattura.PNG
  4 件のコメント
Nick Gibson
Nick Gibson 2022 年 8 月 16 日
thanks, yeah, exactly, just integrate over a straight line.
though you've lost me in the last line of code there!
and the first line is I assume to generate an example function?
Think that would work if the data you want to integrate over is a nice function, but mine is just a sample 2-D grid of data, so not sure your suggestion will work for that - can it be adapted?
Torsten
Torsten 2022 年 8 月 16 日
編集済み: Torsten 2022 年 8 月 16 日
You will have to be able to evaluate your function over the line connecting x1 and x2 - be it that you have scattered data or an explicit function definition as above. So the code applies in any case - you will have to supply the f-values somehow.
A parametrization of the line connecting x1 and x2 is given by
p(t) = x1 + s*(x2-x1) for s in [0 1].
norm(p'(t)) = norm(x2-x1)
The above formula follows if you read Line integral of a scalar field under

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