how to fill color in 1/4 th circle

3 visualizaciones (últimos 30 días)
Xiao yang
Xiao yang el 15 de Jun. de 2024
Comentada: Star Strider el 16 de Jun. de 2024

clear
syms x
figure
fun = sqrt(2*x-x.^2);
fplot(fun,[0,2.2]);
hold on
y1 = [0 1 0 0];
x1 = [0 1 1 0];
fill(x1,y1,'y');
hold on
x2 = 1:0.01:2;
y2 = sqrt(2*x2-x2.^2);
fill([x2,fliplr(x2)],[y2,fliplr(y2)],'r')

Respuestas (1)

Star Strider
Star Strider el 15 de Jun. de 2024
Editada: Star Strider el 15 de Jun. de 2024
A new function in R2024a is the polarregion function.
Try this —
figure
thetas1 = [0.5 1]*pi;
radii1 = [1 2];
polarregion(thetas1, radii1, 'FaceColor','y')
hold on
thetas2 = [0 0.5]*pi;
radii2 = [1 2];
polarregion(thetas2, radii2, 'FaceColor','r')
hold off
Ax = gca;
Ax.ThetaLim = [0 180];
Ax.RLim = [1 2];
Also see PolarAxes Properties for those details.
Doing it without using any of the polar functions is also straightforward, although easier using trigonometric functions,
Try this —
r1 = 1;
th1 = linspace(0, 90);
xy1 = [1+r1*cosd(th1); 1+r1*sind(th1)];
r2 = 1;
th2 = linspace(90, 180);
xy2 = [1+r2*cosd(th2); 1+r2*sind(th2)];
figure
patch([1 xy1(1,:)], [1 xy1(2,:)], 'r', 'EdgeColor','none' )
hold on
patch([1 xy2(1,:)], [1 xy2(2,:)], 'y', 'EdgeColor','none')
hold off
axis('equal')
Ax = gca;
Ax.Visible = 0;
Make appropriate changes to get the result you want.
.
EDIT — (15 Jun 2024 at 14:38)
Added second plot using the patch function.
.
  10 comentarios
Xiao yang
Xiao yang el 16 de Jun. de 2024
thank you so much.
Star Strider
Star Strider el 16 de Jun. de 2024
My pleasure!
If my Answer helped you solve your problem, please Accept it!
.

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