I'm trying to shade the region over which the integral extends with vertical lines. This is my integral:
F = @(x,y) x.*y;
ymin = @(x) x.^2;
ymax = @(x) x;
q = integral2(F,0,1,ymin,ymax)
The code below produces a graph with vertical line shading over its region. This is an example of what i'm trying to do.
f = @(x) sin(x + 1); g = @(x) x.^3 - 3*x + 1;
fplot(f, [-3, 3]), hold on
fplot(g, [-3, 3], 'LineWidth', 2)
>> x1 = fzero(@(x) f(x) - g(x), -2);
x2 = fzero(@(x) f(x) - g(x), 0);
x3 = fzero(@(x) f(x) - g(x), 2);
>> xcoord = linspace(x1, x3, 10);
ycoord = [f(xcoord); g(xcoord)];
plot([xcoord;xcoord], ycoord), hold off

5 comentarios

Your code already works. What exactly is your issue?
(note that this code could be optimized to use a single line object, instead of having a separate object of every vertical line, but that doesn't affect the fundamental operation)
f = @(x) sin(x + 1); g = @(x) x.^3 - 3*x + 1;
fplot(f, [-3, 3]), hold on
fplot(g, [-3, 3], 'LineWidth', 2)
x1 = fzero(@(x) f(x) - g(x), -2);
x2 = fzero(@(x) f(x) - g(x), 0);
x3 = fzero(@(x) f(x) - g(x), 2);
xcoord = linspace(x1, x3, 100);
ycoord = [f(xcoord); g(xcoord)];
plot([xcoord;xcoord], ycoord,'k'), hold off
DGM
DGM el 25 de Abr. de 2022
I'm pretty sure the second code block is an example of what OP wants to apply to the first code block.
Rik
Rik el 25 de Abr. de 2022
The problem with the first block is that F is not a function of x, but of x and y, so the shading of the second block doesn't make sense, unless the function is converted to a function of x alone.
@Ikenna Iwudike Do you mean you want some sort of surface plot? The shading might be difficult to see.
Ikenna Iwudike
Ikenna Iwudike el 25 de Abr. de 2022
Yea, let me see how a surface plot would look. Also yes I was talking about the first block.
Rik
Rik el 25 de Abr. de 2022
See if my edited answer works for you. The loop can be optimized to plot all stems as a single object, but I first wanted to create it like this.

Iniciar sesión para comentar.

 Respuesta aceptada

Rik
Rik el 25 de Abr. de 2022
Editada: Rik el 25 de Abr. de 2022
I have put some inefficient code at the start to integrate over y so you end up with a function of x alone. I then used the same g function as you used in your second block. Is this what you mean?
F = @(x,y) x.*y;
ymin = @(x) x.^2;
ymax = @(x) x;
f_base =@(x) integral(@(y) F(x,y),ymin(x),ymax(x));
f=@(x) arrayfun(f_base,x);
g = @(x) x.^3 - 3*x + 1;
fplot(f, [-2, 2]), hold on
fplot(g, [-2, 2], 'LineWidth', 2)
x1 = fzero(@(x) f(x) - g(x), -2);
x2 = fzero(@(x) f(x) - g(x), 0);
x3 = fzero(@(x) f(x) - g(x), 2);
xcoord = linspace(x1, x3, 100);
ycoord = [f(xcoord); g(xcoord)];
plot([xcoord;xcoord], ycoord,'k'), hold off
If you want to plot a surface, you can use surf. plot3 will allow you to plot a line in a 3D axes.
F = @(x,y) x.*y;
ymin = @(x) x.^2;
ymax = @(x) x;
[X,Y]=ndgrid(linspace(0,1,100));
Y(Y<ymin(X))=NaN;
Y(Y>ymax(X))=NaN;
Z=F(X,Y);
surf(X,Y,Z)
Az=-5;El=65;view(Az,El) % set azimuth and elevation
hold on
for n=1:numel(Z)
if isnan(Z(n)),continue,end
plot3(X(n)*[1 1],Y(n)*[1 1],[0 Z(n)],'k')
end

Más respuestas (0)

Categorías

Más información sobre Vehicle Scenarios en Centro de ayuda y File Exchange.

Productos

Versión

R2021a

Preguntada:

el 24 de Abr. de 2022

Comentada:

el 25 de Abr. de 2022

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by