Center of Gravity and Coordinate Systems

Question

In below figure

I need to find the coordinates of center of gravity which are $x$' and $y$' in order to find distances of $OG$ and $OS$. To explain the figure; $x^q+y^q<=1$, area of $ROP$ is not circular because degree of $q$ is bigger than $2$. I need the solution for any $x^q+y^q<=1$ , for any $q$ which is between $3$ and $\infty$.

Answer 1

Hint:

If you can compute the area and first order moments under the curve, you are (nearly) done.

The explicit equation of the curve can be written

y = (1 - x^q)^(1/q)

The area is given by the integral of

(1 - x^q)^(1/q) dx

and the x/y moments, respectively by

x (1 - x^q)^(1/q) dx

(1 - x^q)^(2/q) dx

By a change of variable t:= x^q, all these integrands are turned to a form

t^α (1 - t)^ß dt

known as the incomplete Beta function. So there is no closed-form solution for general exponents.

You have two choices:

• use the Beta function from a mathematical library,

• use numerical integration (Simson's rule).