Question

Let D be a region bounded by a simple closed path C in the xy-plane. The coordinates of the centroid x, y of D are x = 1 2A x2 dy C y = − 1 2A y2 dx C where A is the area of D. Find the centroid of a quarter-circular region of radius a.

Answer 1

Answer:

$Given\\x=\frac{1}{2A}\int\limits_c {x^2} \, dy\\y=\frac{-1}{2A}\int\limits_c {y^2} \, dx$

The centroid is the same as the center of mass when the density ρ is constant.the density cancels in the center of mass formula, and it becomes this formula for the centroid.

$x=\frac{1}{A}\int\limits\, \int\limits_D {x} \, dx dy \\y=\frac{1}{A} \int\limits\int\limits_D {y} \, dxdy$

This can be interpreted as saying the coordinates of the centroid are the mean, or average values of x  and y on D. So we now have a double integral formula for (¯x, y¯) and a suggestion in the problem that the centroid can also be computed from line integrals on the boundary using Green’s theorem. So we just plug in Green’s theorem into these line integrals to convert them to double integrals and see what we get.

for the integral  

$\int\limits_C {P} \, dx +Qdy=\int\limits_C {x^2} \, dy$  

we have $P(x,y)=0;Q(x,y)=x^2;dQ/dx-dP/dy=2x$

then by Green's theorem

$\frac{1}{2A}\int\limits_C {x^2} \, dy=\frac{1}{2A}\int\limits\int\limits_D{2x} \, dx dy=\frac{1}{A}\int\limits\int\limits_D {x} \, dx dy=x$

in the same way we find

$-\frac{1}{2A}\int\limits_C{y^2} \, dx=\frac{1}{2A}\int\limits\int\limits_D{2y} \, dx dy=\frac{1}{A}\int\limits\int\limits_D {y} \, dxdy=y$

And that is all that the problem asks. Note that the area A is also a double integral that can be  converted by Green’s theorem to a line integral over C.  it is just assumed that one way or another the area is computed and given the name A.