how to fill color in 1/4 th circle

Question

Xiao yang 2024-6-15

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2128941-how-to-fill-color-in-1-4-th-circle

评论： Star Strider 2024-6-16

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')

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Answer 1

Star Strider 2024-6-15

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2128941-how-to-fill-color-in-1-4-th-circle#answer_1472291

编辑：Star Strider 2024-6-15

在 MATLAB Online 中打开

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.

.

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Star Strider 2024-6-16

在 MATLAB Online 中打开

My pleasure!

If my Answer helped you solve your problem, please Accept it!

The patch function is essentially ‘universal’. To use it, it is necessary to create a closed contour, although the beginning and ending points do not need to be repeated, since the function connects them itself.

More examples of it are —

t = linspace(0, 2*pi, 500);

figure

patch(2*cos(3*t), 3*sin(2*t), 'y')

axis('equal')

figure

patch([3*t flip(3*t)], [2+cos(3*t) flip(sin(3*t))], 'm', 'FaceAlpha',0.5)

figure

patch([t flip(-t)], [exp(-t) -flip(exp(-t))], 'c')

figure

y = cos(t);

Lv = y >= -0.5; % Logical Index Vector

patch([t(Lv) flip(t(Lv))], [y(Lv) ones(size(t(Lv)))*2], 'r')

figure

y = sin(t);

Lv = y >= 0; % Logical Index Vector

patch([t(Lv) flip(t(Lv))], [sin(t(Lv)) zeros(size(t(Lv)))], 'g', 'DisplayName','Positive')

hold on

patch([t(~Lv) flip(t(~Lv))], [sin(t(~Lv)) zeros(size(t(~Lv)))], 'r', 'DisplayName','Negative')

hold off

legend('Location','best')

.

There are all sorts of things you can do with it, providing that you always express its arguments as closed contours.

.

Star Strider 2024-6-16

编辑：Star Strider 2024-6-16

在 MATLAB Online 中打开

Just trun them as I wrote them. Do not use the Symbolic Matrh Toolbox with them, since it is not necessary here.

If you have symbolic problems you need to solve and that use the patch function, please provide more details.

EDIT — (16 Jun 2024 at 14:17)

Added ‘Example’.

Note my changes to your original code.

Example —

syms x

figure

fun = sqrt(2*x-x.^2);

hfp = fplot(fun,[0,2.2], 'ShowPoles','off');

axis('equal')

x1 = hfp.XData;

y1 = hfp.YData;

hold on

% y1 = [0 1 0 0];

% x1 = [0 1 1 0];

fill(x1,y1,'y');

% hold on

Lv = x1 >= 1; % Logical Index Vector

% x2 = 1:0.01:2;

x2 = x1(Lv);

y2 = sqrt(2*x2-x2.^2);

fill([x2,fliplr(x2)],[zeros(size(y2)),fliplr(y2)],'r')

hold off

Doing this outside the Symbolic Math Toolbox is still easier. If you are deriving functions from symbolic calculations and derivations, use the matlabFunction function to create them as anonymous functions that you can use in numeric (rather than symbolic) calculations, and then call the functions with the appropriate arguments in your numeric code.

Be sure first to clear the symbolic variables before using the anonymous functions after you create all of them, so there will be no confusion regarding them in your numeric code.

.