Double integrals and center of mass

[Jameson]

Finding the center of mass of a two dimensional object of constant density is a question that frequently occurs on my Mu Alpha Theta tests. I know of a long way to find it which I'll show in a moment using single integrals. I'm wondering if using double integrals can shorten this considerably.

Let f(x) be greater than or equal to g(x) on the domain [a,b]. The moments about the x and y axes are:

(1)$$M_{x}=p\int_{a}^{b}\left(\frac{f(x)+g(x)}{2}\right)[f(x)-g(x)]dx$$

(2)$$M_{y}=p\int_{a}^{b}x[f(x)-g(x)]dx$$

, where p is the planar lamina of uniform density.

The center of mass (x,y) is given by:

(3)$$x=\frac{M_y}{m}$$

(4)$$y=\frac{M_x}{m}$$

where $$m=p\int_{a}^{b}f(x)-g(x)dx$$

Does anyone know how to do this same process with double integrals?! This is just way too much memorizing for me, and I say memorizing because I don't understand the derivation of it.

 

[Tide]

I'm not sure that

$$<\vec x> = \frac {\int \vec x dS}{\int dS}$$

is any quicker or easier.

 

[Jameson]

My book says that for planar lamina of variable density, the following is true.

(1)$$M_{x}=\int_{R}\int yp(x,y)dA$$ and

(2)$$M_{y}=\int_{R}\int xp(x,y)dA$$

So if the density is constant, than doesn't that make this more simple?

 

[Tide]

That's basically what I wrote.

As to whether it's easier you may be able to convince yourself one way or the other by actually trying a few example in a direct comparison. Try some circles, triangles and so on.

 

[Jameson]

Well I'm asking how to do this using double integrals. If p(x,y) is a constant, what constant do I treat it by and how do I get my bounds for the double integral?

 

[Tide]

Here's a simple example:

Consider a triangle with vertices (0, 0), (a, 0), (a, b). The x component of the center of mass is

$$\bar x = \frac {\int_0^a x dx \int_0^{bx/a} dy}{\int_0^a dx\int_0^{bx/a}dy} = \frac {2}{3} a$$

Similarly, you can evaluate

$$\frac {\int_0^a dx \int_0^{bx/a} ydy}{\int_0^a dx\int_0^{bx/a}dy}$$

to arrive at $$\bar y = b/3$$.

 

[Jameson]

Thank you for your post. I'm fairly new to double integrals and am having trouble seeing where the bounds came from, but I'll read of up on it and look at your example.

 

[Jameson]

Here's an example that my Calc book used.

Find the center of mass of the lamina of uniform density p bounded by the graph of $$f(x)=4-x^2$$ and the x-axis.

This is the method that the authors used.

(1)$$m=p\int_{-2}^{2}4-x^2dx$$

(2)$$m=\frac{32p}{3}$$

(3)$$M_{x}=p\int_{-2}^{2}\frac{(4-x^2)^2}{2}dx$$

(4)$$M_{x}=\frac{256p}{15}$$

(5)$$y=\frac{M_{x}}{m}$$

(6)$$y=\frac{8}{5}$$

Because the center of mass lies on the axis of symmetry, x=0. Thus the center of mass is $$(0,\frac{8}{5})$$

Now, could you please explain this using double integrals?

 

[Tide]

By symmetry, the x value is zero (you can set up the integrals to verify).

For y:

$$\bar y = \frac {\int_{-2}^{2} dx \int_{0}^{4-x^2} y dy}{\int_{-2}^{2} dx \int_{0}^{4-x^2} dy} = \frac {8}{5}$$