I just have one simple question. How do i calculate the integral of this area?

r=@(z)2+sin(5*z+2.2*cos(5*z));
z=linspace(0,2*pi);
x=r(z).*cos(z);
y=r(z).*sin(z);
plot(x,y)

Answer 1

[Star Strider]

One approach:

r=@(z)2+sin(5*z+2.2*cos(5*z));
z=linspace(0,2*pi);
x=r(z).*cos(z);
y=r(z).*sin(z);

Ixy = cumtrapz(z, abs(r(z)));

Check = max(abs(r(z)))^2 * pi;                  % Area Must Be Less Than Maximum Circle Area

plot(x,y)
axis equal
text(0, 0, sprintf('Area = %.2f', Ixy(end)), 'HorizontalAlignment','center', 'VerticalAlignment','middle')

[Christian Nilsson]

Thank you very much. Is it possible to solve this by usinng the integral function in some way?

[Star Strider]

As always, my pleasure.

To use the integral function:

r=@(z)2+sin(5*z+2.2*cos(5*z));
integral_approach = integral(@(z) abs(r(z)), 0, 2*pi)

integral_approach =
    16.0596

[Christian Nilsson]

Thanks!

[Torsten]

I guess it's not that easy:

area=-integral_{z=0}^{z=2*pi} r(z)*sin(z)*(r'(z)*cos(z)-r(z)*sin(z)) dz

https://math.stackexchange.com/questions/889522/area-inside-curve-given-by-parametric-equation

Best wishes

Torsten.

[Walter Roberson]

Note that the original question was one of discrete calculation, a finite number of specific z values. The integral() approach is for continuous z.

[Torsten]

But the formula to calculate the area inside the parametric curve (x(z),y(z)) is not abs(r(z)), discrete or not ...

Best wishes

Torsten.