# center of gravity

I know the center of gravity of a uniform cicular sector is d = (2r sin x)/(3x), where d is the distance from the center and x is half the central angle of the sector. Can someone prove it or anyone has a link for the proof?
I tried to use the 2 theorems of Pappus, but I couldn't reach an answer. Any idea?

### Re: center of gravity

Can you please do it with more details (how you derived the integral)? Why should we integrate P? Can you please show me how you carried out the integral? Is it double or single integral? And why did you consider the center of gravity as a vector?
I tried a different approach but didnt reach the correct answer?:
dm = s dA = s dx dy (s is the surface density)

y (of center of gravity) = (Integral y dm)/(Integral dm) = (Integral y s dA)/(Integral s dA) = (Double integral y dy dx)/A

y (of c.o.g.) = (1/A)* Integral (between 0 and r) Integral (between 0 and r sin(theta)) y dy dx

A = area of sector , r = radius , theta = angle of sector

What is wrong with this?

### Re: center of gravity

"And why did you consider the center of gravity as a vector?"

The center of gravity is defined as a point (in this case 2-d) where the masses balance.

Your expression for the y coordinate is probably correct, but you also need the x coordinate. Then to get the distance from the center of the circle, use Pythagorean theorem. It is much easier to carry out the integration in polar coordinates - dxdy=rdrda (where a is angle).

### Re: center of gravity

Ya you're right, I forgot the x coordinate. But anyways I found out that you cant do it using cartesian coordinates (taking dA = dxdy) because the boundaries aren't well defined (in fact you cant define them), so my boundaries were wrong. The central angle theta is arbitrary, so we cannot know whether the sector is in the first or second quadrant. If in first quadrant then x goes from 0 to R, if second quadrant, x goes from Rcos(theta) to R, so we cannot know. Integrating with polar coordinates is the only correct way this can be solved.

### Re: center of gravity

It can be done by a slightly messy, but straightforward, integration.

For notation purposes:
R=radius of sector, A=angle of sector.

To find center of gravity (a vector), integrate P=(x,y) over the sector and divide the result by the sector area. Your answer is the distance of the c.g. from the center of the circle.

If you carry out the integration in polar coordinates, it is straightforward.

The integral of P is the vector ((R^3)/3)(sinA,1-cosA)
The sector area is (R^2)A/2.