center of mass: functions with respect to y

**infraRed** · February 9th 2013, 03:15 PM

Hi. In Calc 2 I have some homework questions for calculating the moments and center of mass for planar laminae. Fortunately, the book gives the exact equations necessary for the calculations, when there are functions f(x) and g(x), within bounds a and b:

$M_{x} = \rho\int_a^b\frac{f(x)+g(x)}{2}[f(x)-g(x)]\,dx$

$M_{y} = \rho\int_a^b{x[f(x)-g(x)]\,dx}$

$m = \rho\int_a^b{[f(x)-g(x)]\,dx}$

$\overline{x} = \frac{M_{y}}{m}$

$\overline{y} = \frac{M_{x}}{m}$

However, how do I adjust these formulae when the functions must be functions in respect to y? For example, I have a problem where the area is bounded by

$x = y + 2$
$x = y^2$

Which involves a sideways parabola.

**infraRed** · February 9th 2013, 05:01 PM

Okay, after trying for an hour or so to understand how the moments of mass are calculated, I think I understood it, and now this seems pretty simple. It should be...

$M_{y} = \rho\int_{a}^{b}\frac{f(y)+g(y)}{2}[f(y)-g(y)]\,dy$

$M_{x} = \rho\int_{a}^{b}y[f(y)-g(y)]\,dy$

$m = \rho\int_{a}^{b}[f(y)-g(y)]\,dy$

$\overline{x} = \frac{M_{y}}{m}$

$\overline{y} = \frac{M_{x}}{m}$

...where a and b are on the y axis.

**hollywood** · February 9th 2013, 07:26 PM

That seems to be correct. You simply exchange x and y, right?

- Hollywood

**infraRed** · February 9th 2013, 09:03 PM

It's a little more complicated than that. The definitions of $\overline{x}$ and $\overline{y}$ remain the same, except that the formulae for moment around the y axis and moment around the x axis get flipped around, and the formulae are all dependent on y instead of x. It is difficult to explain why this makes sense without drawing and illustrating a graph, but it does make sense, and it has worked well for me on several problems so far.

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