how to fix my code and how to find the area of the regon

Consider the region inside the outer loop but outside the inner loop of the limacon r = 3 - 6 sin theta . produce a complete graph of the limacon and find the area of the region.

syms  thet  s
y =  3-6*sin(thet )
ezpolar(3-6*sin(thet ) , [ 0 , 2*pi] )
figure
ths = 0 : 0.01 : 2*pi ;
polar(ths , 3-6*sin(ths))

[Geoff Hayes]

Youssef - what exactly is wrong with your code? Please clarify what you believe is incorrect (and so requires to be "fixed"). The code

 figure
 ths = 0 : 0.01 : 2*pi ;
 polar(ths , 3-6*sin(ths))

produces a limaçon. Are you unsure how to calculate the area?

[Image Analyst]

Is the inner part of the curlicue to be considered as being part of the area or excluded from the area?

[Carlos Guerrero García]

As the statement, the area to calculate is the green one in the following fig:

Answer 1

[Carlos Guerrero García]

The area is 9*(pi+3sqrt(3))=75.0397..., but I wrote the following script to estimate that value via the Montecarlo method. I know that the code is not the best one, but I was only checking the result obtained by integration (in R2 or using the Green theorem)

function A=limacon(n)
outside=0;
P=[6-12*rand(n,1) 1-10*rand(n,1)]; % Choosing n random points in [-6,6]x[-9,1]
polar=[sqrt(P(:,1).^2+P(:,2).^2) atan2(P(:,2),P(:,1))+2*pi*(atan2(P(:,2),P(:,1))<0)]; % Cartesian coordenates to polar coordinates with Arg in [0,2*pi)
plot(P(:,1),P(:,2),'.','MarkerSize',10); hold on; % Ploting the random points
t=0:pi/100:2*pi; plot((3-6*sin(t)).*cos(t),(3-6*sin(t)).*sin(t),'k','LineWidth',2); % Ploting the limacon
axis([-7 7 -10 2]); axis equal;
for k=1:n; % Cheking it the point aren't in the limacon: outside the outer loop or in the inner loop
    if polar(k,2)<=pi/6 && polar(k,1)>=3-6*sin(polar(k,2))
        outside=outside+1;        
    elseif polar(k,2)>pi/6 && polar(k,2)<=5*pi/6
        outside=outside+1;    
    elseif polar(k,2)>5*pi/6 && polar(k,2)<=7*pi/6 && polar(k,1)>=3-6*sin(polar(k,2))
        outside=outside+1;
    elseif polar(k,2)>7*pi/6 && polar(k,2)<=11*pi/6 && polar(k,1)>=3-6*sin(polar(k,2))
        outside=outside+1;
    elseif polar(k,2)>7*pi/6 && polar(k,2)<=11*pi/6 && polar(k,1)<=-6*sin(polar(k,2))-3
        outside=outside+1;
    elseif polar(k,2)>11*pi/6 && polar(k,1)>=3-6*sin(polar(k,2))
        outside=outside+1;
    end
end
A=120*(n-outside)/n; % Calculating the ratio and estimating the area

Answer 2

[Carlos Guerrero García]

The area can be found as follows (the explanation can be found in any Calculus text. I think that here is not the right place for an extended document about. Also I don't know if you want the value, the explanation, the code, ...or all of them. If you don't find the reason of that equality in your favorite Calculus text, I don't have any problem to send an extended document with the equality):

syms t;
A=int((3-6*sin(t))^2,t,5*pi/6,13*pi/6)/2-int((3-6*sin(t))^2,t,13*pi/6,17*pi/6)/2

A =