Line integral over a 3D surface

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 Comments

What have you tried? Did you try interp2?
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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Answers (2)

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 Comments

The integral over the segment (the one in red in your plot, the one with x1 and x2 as ends in mine)
You mean length of the curve?

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No, the line integral from x1=(-19,1) to x2(-9;-9) of the function SF_Full.
Cattura.PNG

4 Comments

I could use this too!
Think darova's answer above is getting close.
If, instead of using 50 sample points for z, you worked out how many points you needed to replicatate the same sample (pixel) spacing as in the original x and y directions (form the angle of the line) and used that number instead, you'd be about there. Then just sum the values in z from these. Would that do it?
If you mean integrating over the straight line connecting x1 and x2, you can do
f = @(x,y) x.^2+y.^2;
x1 = [-19,1];
x2 = [-9,-9];
line_integral = norm(x2-x1)*integral(@(t)f(x1(1)+t*(x2(1)-x1(1)),x1(2)+t*(x2(2)-x1(2))),0,1)
line_integral = 3.2338e+03
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 on 16 Aug 2022
Edited: Torsten on 16 Aug 2022
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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Asked:

on 22 Jan 2020

Edited:

on 16 Aug 2022

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