# How can I prove the location of centre of gravity of centroid?

In a book, There is chapter of "Centre of Gravity" before "Moment of Inertia" and in the chapter Centre of Gravity, the list of some centroid and it's location of C.G (i.e x̅ and y̅) is given. You can refer similar table from wikipedia. And with the help of it, the examples of finding the location of C.G is explained.

But I want to prove the location of C.G for that centroids. For example, I want to prove that y̅=4r/3π for the semicircular. How can I prove it?

• I am not sure about on-topicness of this question here. If suitable for another site, tell me Jan 1 '16 at 13:15
• It's all in the integrals. See any first-year Mechanics (Physics) text. Jan 1 '16 at 17:21

For a planar surface, the centroid is defined by the equation $$\overline{y} =\dfrac{\int y\ \text{d}A}{\int\ \text{d}A} = \dfrac{\int y\ \text{d}A}{A}$$ where $y$ is the distance of each point from the origin and $A$ is the surface's area.
Meanwhile, the center of gravity/mass is defined by $$\text{C.G.}=\dfrac{\int y\rho(y)\ \text{d}A}{\int\ \rho(y)\text{d}A} = \dfrac{\int y\rho(y)\ \text{d}A}{M}$$ where $\rho(y)$ is the density profile of the surface and $M$ is the total mass of the surface.
Now, in the case of a uniformly-dense surface, $\rho(y) = \rho$ and $M = A\rho$, which means that $$\text{C.G.} = \dfrac{\int y\rho\ \text{d}A}{A\rho} = \dfrac{\int y\ \text{d}A}{A} = \overline{y}$$