Plot of Integration Volume

Question

Athanasios Paraskevopoulos 2024-4-1

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2101516-plot-of-integration-volume

评论： Athanasios Paraskevopoulos 2024-4-10

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);

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Answer 1

SAI SRUJAN 2024-4-10

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2101516-plot-of-integration-volume#answer_1439131

在 MATLAB Online 中打开

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.

https://www.mathworks.com/help/matlab/ref/patch.html

I hope this helps!