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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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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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    @Korgan: Now you’ve got it: those last two sentences are the key.2012-02-08