# Question about finding the centre of mass.

I'm working through Paul's Online Notes, and I'm on this page. I don't understand where the equations of moments come from. Can someone explain to me in the simplest terms how to derive these?

He explains the equations as

the tendency of the region to rotate about the x and y-axis respectively.

So if I wanted to do this with some function $f$, how does the equation

$M_x=\frac{1}{2}\int_a^b f(x)^2 dx$

express the function $f$'s tendency to rotate around the x-axis?

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Suppose that you have a point mass $m$ located at the point $(x,y)$. By definition its moment about the $x$-axis is $my$, and its moment about the $y$-axis is $mx$. In other words, the moment about an axis is the signed product of the mass and its distance from the axis. Think of a seesaw (teeter-totter): the further you sit from the centre, and the more massive you are, the stronger the tendency for your end to go down.

The nice thing about moments is that they’re additive: the moment of a whole system of point masses about a given axis is the sum of the individual moments. Of course, when you have a whole region of point masses, you can’t simply: you have to integrate. Pick any $x$ between $a$ and $b$ and look at the vertical strip of mass there between $g(x)$ and $f(x)$. A point mass $m$ anywhere along that strip has moment $mx$ about the $y$-axis. The total mass of the strip is $$dm=\rho\big(f(x)-g(x)\big)dx\;,$$ the density times the ‘area’ of the strip, so its moment about the $y$-axis is $$dM_y=xdm=\rho x\big(f(x)-g(x)\big)dx\;.$$ To get the moment about the $y$-axis of the entire region, you have to ‘add up’ (integrate) these individual moments over the entire range of $x$ values, from $a$ to $b$: $$M_y=\rho\int_a^b\big(f(x)-g(x)\big)dx\;.$$

It’s a little harder to see where the formula for $M_x$ comes from, though the basic idea is actually the same. The problem is that each point mass along that vertical strip at $x$ has a different $y$-coordinate, so to get $dM_x$ you’re trying to ‘add up’ $y\;dm$’s with different values of $y$. It turns out that since the density is constant, you can pretend that the entire mass of the strip is concentrated at its midpoint, i.e., at the point $$\left(x,\frac{f(x)+g(x)}2\right)\;,$$ with $y$-coordinate $\bar y=\frac12\big(f(x)+g(x)\big)$. (You’ll be able to show this after you learn about multiple integrals.) The mass of the strip is the same as before, $dm=\rho\big(f(x)-g(x)\big)dx$, so its contribution to the moment about the $x$-axis is \begin{align*} dM_x&=\bar y dm\\ &=\frac12\big(f(x)+g(x)\big)\cdot\rho\big(f(x)-g(x)\big)dx\\ &=\frac12\rho\Big(\big(f(x)\big)^2-\big(g(x)\big)^2\Big)dx\;, \end{align*}

and the actual moment about the $x$-axis is $$M_x=\frac12\rho\int_a^b\Big(\big(f(x)\big)^2-\big(g(x)\big)^2\Big)dx\;,$$ the result of ‘adding up’ the moments for the vertical strips between $x=a$ and $x=b$.

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Your response made me understand enough to figure it out for myself. Thanks! –  Korgan Rivera Feb 4 '12 at 21:23