Plot of Integration Volume

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Athanasios Paraskevopoulos
Athanasios Paraskevopoulos 2024 年 4 月 1 日
Hi everyone,
I'm working on visualizing an integration volume in MATLAB for the first time, and I'd appreciate some feedback on the correctness of my plot. Below is the MATLAB code I've used to generate the plot:
syms x y z;
F = [x*y, y^2 + log(x*z^2), sin(x*y)]; % Vector field
divF = divergence(F, [x, y, z]); % Divergence of F
% Limits of integration
z_lim = [0, 1-x^2];
y_lim = [0, 2-z];
x_lim = [-1, 1];
% Calculate the volume integral
volume_integral = int(int(int(3*y, y, y_lim(1), y_lim(2)), z, z_lim(1), z_lim(2)), x, x_lim(1), x_lim(2));
double(volume_integral)
ans = 5.2571
% Define the grid with a sparser density
[x, y, z] = meshgrid(linspace(-1, 1, 10), linspace(0, 2, 10), linspace(0, 1, 5));
% Define the vector field F
F_x = x .* y;
F_y = y.^2 + log(abs(x) .* z.^2 + 1); % Use abs(x) to avoid negative values inside log
F_z = sin(x .* y);
% Plot the vector field with arrows
figure;
quiver3(x, y, z, F_x, F_y, F_z, 'AutoScaleFactor', 0.8, 'MaxHeadSize', 0.2);
title('Vector Field F');
xlabel('x');
ylabel('y');
zlabel('z');
axis tight;
grid on;
view(3);
% Define the grid
[x, y, z] = meshgrid(linspace(-1, 1, 50), linspace(0, 2, 50), linspace(0, 1, 50));
% Define the conditions for the region
regionCondition = (z >= 0) & (z <= 1 - x.^2) & (y >= 0) & (y <= 2 - z);
% Visualize the region
figure;
% Use the isosurface function to plot the region
isosurface(x, y, z, regionCondition, 0.5); % The 0.5 is a level to define the boundary
title('Integration Volume');
xlabel('x');
ylabel('y');
zlabel('z');
axis tight;
grid on;
view(3);

採用された回答

SAI SRUJAN
SAI SRUJAN 2024 年 4 月 10 日
Hi Athanasios,
I understand that you are seeking feedback on visualizing the volume of integration plot.
This approach seems correct for visualizing the integration volume and the vector field. The use of 'isosurface' is appropriate for showing the region of integration based on the defined conditions. The modification to include 'abs(x)' within the logarithmic term of ('F_y') is a good practice to avoid complex values which can arise from taking the logarithm of negative numbers.
There are other ways to visualize integration volumes in MATLAB. One alternative method is to use the 'patch' function to create a 3D surface plot of the integration volume. Here's an example code snippet for the region condition attached :
% Define the grid
[x, y, z] = meshgrid(linspace(-1, 1, 50), linspace(0, 2, 50), linspace(0, 1, 50));
% Define the conditions for the region
regionCondition = (z >= 0) & (z <= 1 - x.^2) & (y >= 0) & (y <= 2 - z);
% Extract the coordinates of the region
x_region = x(regionCondition);
y_region = y(regionCondition);
z_region = z(regionCondition);
% Create a patch object to represent the integration volume
patch('Faces', convhulln([x_region(:), y_region(:), z_region(:)]), 'Vertices', [x_region(:), y_region(:), z_region(:)], 'FaceColor', 'blue', 'FaceAlpha', 0.5);
axis tight;
grid on;
view(3);
For a comprehensive understanding of the 'patch' MATLAB function, please go through the following documentation.
I hope this helps!
  1 件のコメント
Athanasios Paraskevopoulos
Athanasios Paraskevopoulos 2024 年 4 月 10 日
Oh!! Cool thank you very much! Yes I was seeking a feedback!

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