how to find area enclosed by ezplot

I have bean curve equation and plotted it using ezplot. I wold like to find out the area enclosed by the curve. Tried to do get gca and extracting x and y only gives me a straight line and not a bean curve. Tried plotting CountourMatrix but gives error. syms x y; f=ezplot((a*x)^4+(a*x)^2*(b*y)^2+(b*y)^4-a*x*((a*x)^2+(b*y)^2));axis equal; title('bean');

code : get(f); Q=get(f,'ContourMatrix'); gives error message: Error using hg.axes/get The name 'ContourMatrix' is not an accessible property for an instance of class 'axes'

Please help

Answer 1

[Star Strider]

The ezplot function is not the best way to go with your problem. Use contour instead:

a = 0.3;
b = 0.5;
[X,Y] = meshgrid(linspace(-5, 5, 50));
beanfcn = @(x,y) (a*x).^4+(a*x).^2.*(b*y).^2+(b*y).^4-a*x.*((a*x).^2+(b*y).^2);
hc = contour(beanfcn(X,Y), [0; 0]);
BeanArea = polyarea(hc(1,2:end), hc(2,2:end))

produces (for these values of ‘a’ and ‘b’):

BeanArea =

     167.1371

[komal shah]

Thank you Star Strider. It worked fine. Thanks again.

I would like to know how to change the linspace so the curve will be centered at 0,0 axis. I changed it to (-1,1,5) and it changed the shape? Any comments are welcomed. Thanks in advance

[Star Strider]

I would keep the arguments to linspace symmetrical. The shape changed because ‘X’ and ‘Y’ changed. You would have to put in manual offsets for it to be centred at the origin, however changing the contour call to:

hc = contour(X, Y, beanfcn(X,Y), [0; 0]);

will put its origin at (0,0) and also result in a different value for the area (with that change, to 6.9611).

[komal shah]

Ok, Now I understand what is linespace. I created the curve with (-5,5,500) points and get a very smooth curve. Thanks Star Strider. My problem is solved and I learn something new with the contour plot. Thank you, much appreciated.

[Star Strider]

My pleasure!

I use contour frequently when I need to find zeros (or any other constant values) of functions of two variables.