Q:

Prove expression for center of mass of a curve

While reading a book, I have come across the following expression for the abscissa of the center of mass of a curve $y=y(x)$, $a\leq x\leq b$:
  $$\frac{\int_a^bx\sqrt{1+y'^2}\ dx}{\int_a^b\sqrt{1+y'^2}\ dx}.$$

How can this expression be proven? I know that $\int_a^b\sqrt{1+y'^2}\ dx$ is the length of the curve, but I can't figure out how to go on from there. Any ideas would be appreciated.

A:

The general equation for the x-coordinate of the center of mass is 
$$x_{CM}=\frac{1}{M}\sum_{\Delta m} x\cdot\Delta m$$
where we have split the big mass $M$ in many little $\Delta m$. Assume the (linear) density of the line is constant. 
$$\rho=\frac{M}{L}=\frac{\Delta m}{\Delta l}$$
Here $\Delta l$ and $L$ are the lengths of respectively a small line piece and the entire line. Substituting $M$ and $\Delta m$ gives a sum dependent only on $\Delta l$ and $L$.
$$x_{CM}=\frac{1}{\rho L}\sum_{\Delta l} x\cdot\rho\Delta l=\frac{1}{L}\sum_{\Delta l} x\cdot\Delta l$$
If you know how to calculate the length of the line rewriting $\Delta l$ in terms of $\Delta x$ should be familiar:
$$\Delta l=\sqrt{\Delta x^2+\Delta y^2}=\Delta x\sqrt{1+\left(\frac{\Delta y}{\Delta x}\right)^2}$$
Finally let $\Delta x$ become infinitesimally small; the sum over all $\Delta x$ becomes an integral.
$$x_{CM}=\frac{1}{L}\sum_{\Delta x} x\cdot\sqrt{1+\left(\frac{\Delta y}{\Delta x}\right)^2}\Delta x\rightarrow\frac{1}{L}\int x\sqrt{1+y'(x)^2}dx$$

