I know that the area of a shape generated as below
$R=r_0+a_1\cos(\theta)+a_2\cos(2\theta)+a_3\cos(3\theta)+...$
Where you can plot it and see the area value in matlab by:
th=0:0.01:2*pi;
R=r0+a1*cos(th)+a2*cos(2*th)+a3*cos(3*th);
x=R.*cos(th);
y=R.*sin(th);
plot(x,y);
Area=polyarea(x,y)
Could be computed by integral: $\frac{1}{2} \int R^2d\theta$
Which is equal to: $\pi{r_0}^2+\frac{1}{2}\pi({a_1}^2+{a_2}^2+{a_2}^2...)$
I am wondering how can I compute the area of a shape generated as below in a similar way analetically:
$R_x=r_0+a_1\cos(\theta)+a_2\cos(2\theta)+a_3\cos(3\theta)+...$ $R_y=r_0+{a_1}'\cos(\theta)+{a_2}'\cos(2\theta)+{a_3}'\cos(3\theta)+...$
Where you can plot it and see the area value in matlab by:
th=0:0.01:2*pi;
Rx=r0+a1*cos(th)+a2*cos(2*th)+a3*cos(3*th);
Ry=r0+a11*cos(th)+a22*cos(2*th)+a33*cos(3*th);
x=Rx.*cos(th);
y=Ry.*sin(th);
plot(x,y);
But what is the analytical expression for the area of the shape? Area=polyarea(x,y)
Use Green's Theorem: If $$\gamma:\quad t\mapsto \bigl(x(t),y(t)\bigr)\qquad(0\leq t\leq T)$$ is a simply closed curve bounding counterclockwise a region $B\subset{\mathbb R}^2$ then $${\rm area}(B)={1\over2}\int_0^T\bigl(x(t)\dot y(t)-y(t)\dot x(t)\bigr)\>dt\ .$$
By the way: If you have a parametric representation of the form $$x(\theta):=R_1(\theta)\cos\theta,\quad y(\theta)=R_2(\theta)\sin\theta\ ,$$ whereby the functions $R_1$, $R_2$ have been chosen independently, the variable $\theta$ no longer is the polar angle of the running point. It is just a parameter variable ("time") used to produce the curve in question.