integral2 failing to calculate area of a square inside a circle?
1 vue (au cours des 30 derniers jours)
Afficher commentaires plus anciens
Jesus Sanchez
le 10 Juin 2021
Commenté : Jesus Sanchez
le 15 Juin 2021
Hello all,
I am integrating the area contained within a square and a circle using integral2. I use a circle centered at (0,0) with radius 2.75 and two squares with edge 0.5 (blue on image) and 1 (red on the image), whose bottom left corner lies on (0.5, 0.5). As the circle is larger than the squares, I would expect the result of the integral to be the areas of each square, 0.25 for the blue and 1 for the red. However, the output for the red one is 0.75. Am I missinterpretating the result?
Thanks in advance!
The image:
The code:
Zfun1 = @(x,y) double((x.^2 + y.^2) <= 2.75); % Circle function
integral2(Zfun1,0.5,1.5,0.5,1.5,'Method','iterated',...
'AbsTol',1e-5,'RelTol',1e-5) % Red square
integral2(Zfun1,0.5,1,0.5,1,'Method','iterated',...
'AbsTol',1e-5,'RelTol',1e-5) % Blue square
rectangle1 = [0.5 , 0.5; % Coordinates red square (for plotting)
0.5 , 1.5;
1.5 , 1.5;
1.5 , 0.5;
0.5 , 0.5];
rectangle2 = [0.5 , 0.5; % Coordinates blue square (for plotting)
0.5 , 1;
1 , 1;
1 , 0.5;
0.5 , 0.5];
figure
circle(0,0,2.75); % circle
hold on; grid on
plot(rectangle1(:,1),rectangle1(:,2),'-.r'); % red square
plot(rectangle2(:,1),rectangle2(:,2),'-.b'); % blue square
axis equal
xlim([-0.1 2.8])
ylim([-0.1 2.8])
function h = circle(x,y,r)
hold on
th = 0:pi/50:2*pi;
xunit = r * cos(th) + x;
yunit = r * sin(th) + y;
h = plot(xunit, yunit);
hold off
end
0 commentaires
Réponse acceptée
Tarunbir Gambhir
le 15 Juin 2021
Modifié(e) : Tarunbir Gambhir
le 15 Juin 2021
Please correct the equation of the circular region in function 'Zfun1'. It should be of the form, x^2+y^2<=r^2.
Plus de réponses (0)
Voir également
Catégories
En savoir plus sur Modify Image Colors dans Help Center et File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!