# Density and integrals

1. Feb 16, 2007

### -EquinoX-

1. Find the mass of a rod length 10 cm with density d(x) = e^-x gm/cm at a distance of x cm from the left.
2. Find the center of mass of a system containing three point masses of 5gm, 3gm, and 1 gm located respectively at x = -10, x = 1, and x =2.

for number two what I did is just this:

(5(-10) + 3(1) + 1(2)) / (-10+1+(-2)) and I got 45/7

Last edited: Feb 16, 2007
2. Feb 16, 2007

### neutrino

Hint For the first: Imagine the rod to be made of tiny mass elements $dm$. Each mass element is equal to $\rho dx$, where $\rho$ is the linear density within the differential $dx$. If you sum up all such $dm$'s you arrive at the mass of the rod.

2) How do you define, mathematically, the centre of mass of a system of particles?

3. Feb 16, 2007

### -EquinoX-

so here's what I did for number one.

I take the integral of 0 to 10 of e^-x dx, is that all?

what's wrong with my number 2??

4. Feb 16, 2007

### neutrino

Yes, that's right

Again, I ask you, what's the definition of the centre of mass?

5. Feb 16, 2007

### -EquinoX-

the center of mass as of my understanding is the point/position where there's a balance/equilibrium.

6. Feb 16, 2007

### neutrino

I was talking about the mathematical definition, which is: $$\vec{R}_{cm} = \frac{\sum_{i=1}^{n}m_i\vec{r}_i}{\sum_{i=1}^n m_i}$$

If you look at your answer in the first post, you may notice that it is not dimensionally correct.

Last edited: Feb 16, 2007
7. Feb 16, 2007

### -EquinoX-

Oh clumsy mistake, so it should be:

(5(-10) + 3(1) + 1(2)) / (5+3+1)) and results as -45/9 = -5 right??

I have one more question and this is kind of hard:

A cardboard figure has a region which is bounded on the left by the line x = a, on the right by the line x=b, above by f(x), and below by g(x). If the density d(x) gm/cm^2 varies only with x, find an expression for the total mass of the figure, in terms of f(x), g(x), and d(x)

8. Feb 16, 2007

### neutrino

The principle's the same as in post #2, except that, here you have an area instead of a line. Use double integrals.

9. Feb 17, 2007

### -EquinoX-

what do you mean here as double integral?? so I still use the same formula as what I did in number 2?