How to calculate integral along the boundary of closed curve?

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Romuald Kuras 2018-9-3

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评论： Romuald Kuras 2018-9-3

采纳的回答： Dimitris Kalogiros

fig1.png

Hi!

I would like to calculate the area under the curve (in direction to Z axis - see the attachment file), which is formed of discrete values on the edge of a closed curve. I have an idea, but it is a quite arduous method (using 'trapz' function), does anyone have any other suggestions?

Thanks in advance!

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Torsten 2018-9-3

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Take the sum of

sqrt((x(i+1)-x(i))^2+(y(i+1)-y(i))^2)*(z(i)+z(i+1))/2

Best wishes

Torsten.

Romuald Kuras 2018-9-3

Thanks a lot! I never dreamed that I would get the answer so quickly :)

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Dimitris Kalogiros 2018-9-3

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编辑：Dimitris Kalogiros 2018-9-3

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Provided than (x,y) points are very dense you can approximate the area you ask, good enough:

clear; clc;
% generation of input data
t=0:0.01:2*pi-0.01;
x=4*cos(t); 
y=2*sin(t);
z=sqrt(abs(x))+y.^2+exp(0.6*x);
stem3(x,y,z,'-b'); hold on;
plot3(x,y,0*z,'-k.'); hold on;
plot3(x,y,z,'r*');
xlabel('x'); ylabel('y'); zlabel('z');
% calculation of area (approximately)
A=0;
for n=1:length(x)-1
    A= A+ sqrt( (x(n+1)-x(n))^2 + (y(n+1)-y(n))^2 ) * (z(n)+z(n+1))/2 ;
end
fprintf('Area is %f \n', A)

If you run this script, you will receive something like this:

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Romuald Kuras 2018-9-3

Thank you so much! I did not expect that I would get the answer so quickly and that someone will create a similar function for the clarity of the problem. Thanks a lot! :)

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