Line integral over a 3D surface
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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
2020 年 1 月 22 日
What have you tried? Did you try interp2?
Riccardo Giaffreda
2020 年 1 月 22 日
回答 (2 件)
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?
Riccardo Giaffreda
2020 年 1 月 22 日
0 投票
4 件のコメント
Nick Gibson
2022 年 8 月 15 日
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)
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?
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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