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I've been trying to figure out how to find the centre of mass of some region under a curve. I figured out how to find the first pivot line parallel to the x-axis, but I can't figure out how to find the other one. But that's not my question. This is the function that gives me the first pivot:

$$\dfrac{\int_a^b f(x)^2dx}{2\int_a^b f(x)dx}$$

My question is, what does the numerator represent? The denominator is twice the total mass of the shape.

Also, if anyone can explain to me how to find that second pivot line, I'd be grateful. I thought it would be as easy but I guess not.

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up vote 2 down vote accepted

Think of the fraction as $$\frac{\frac12\int_a^b f(x)^2dx}{\int_a^b f(x)dx}\;;$$ then you can more easily recognize it as $\dfrac{M_x}{M}$, where $M$ is the mass and $M_x$ is the moment about the $x$-axis.

For the second pivot line you do exactly the same thing, but with respect to the $y$-axis. It helps to understand the idea behind it. What you’re really doing is solving for the vertical line on with the region would balance. Say that this line has the equation $y=\bar x$. The region will balance on that line if its moment around that line is $0$. Since that’s a vertical line, you calculate the moment around it very much as you’d calculate the moment about the $y$-axis, except that the distance from the strip at $x$ to the line $y=\bar x$ is $x-\bar x$ instead of $x$. Thus, the moment about the line $y=\bar x$ is $$\begin{align*}\int_a^b (x-\bar x)f(x)dx&=\int_a^b xf(x)dx-\bar x\int_a^bf(x)dx\\ &=M_y-\bar xM\;. \end{align*}$$

This moment represents the tendency of the region to rotate around the line $y=\bar x$; in order for the region to balance (i.e., for this line to go through the centre of mass), this moment must be $0$. That means that $M_y-\bar xM=0$, and when you solve for $\bar x$, you get $$\bar x=\frac{M_y}M\;.$$

Now just substitute into this your formulas for $M_y$ and $M$, and you’ll have your other coordinate of the centre of mass of the region.

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I'm having a hard time understanding moments, and this 'tendency to rotate around the x-axis". I understand that a moment is the mass x 'moment arm'. I don't understand why $\frac{1}{2}\int_a^b f(x)^2dx$ represents this. – Korgan Rivera Feb 8 '12 at 15:48
You know, I've just understood it. I noticed that it's the moment about the x-axis, but I've been thinking of it as the moment about the curve, which it is equally so. – Korgan Rivera Feb 8 '12 at 16:42
Ok ok, I totally get it now and it's very obvious: when I'm dividing moment by mass, I'm getting distance: distance of the moment from the pivot point. I can then either think of the pivot line as having moved the pivot to the moment and thereby making the moment zero, or the distance away from the mass where I would put another moment so as to balance. SO obvious I don't know why that took me so long. The ability to sum all moments and then treat the result as a single moment is fantastic. I can reduce a complicated balancing act into a simple seesaw. – Korgan Rivera Feb 8 '12 at 16:58
@Korgan: Now you’ve got it: those last two sentences are the key. – Brian M. Scott Feb 8 '12 at 20:48

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