Density function, calculating the mass

[Robin04]

Homework Statement

The density of a rod in function of space is given as $\rho (x)=\frac{c}{x^2}$
1. What kind of density is this?
2. What is the dimension of $c$?
3. What is the mass of the rod in the intervals
- [1 m, 2 m],
- (1 m, 2 m),
- (0 m, 1 m),
- [0 m, 1 m]?
4. Can a plate with such density exist?

Homework Equations

The Attempt at a Solution

1. Assuming this is a thin rod with infinitesimal cross section area, this should be a linear density function.
2. Therefore $[c]=kg \cdot m$
3. For the interval [1m,2m] it is obvious that $m_1 = \int_{1}^{2}\rho (x) dx = \frac{c}{2}$
But how do I evaluate the integral on an open interval? Also, the intervals having 0 as their beginning (either open or closed) should imply an infinite mass, right?

 

[RPinPA]

But how do I evaluate the integral on an open interval?

Since the function is finite and the interval differs by an infinitesimal amount between the open and closed interval, it doesn't make any difference. The integral is the same for the open and the closed interval.

Also, the intervals having 0 as their beginning (either open or closed) should imply an infinite mass, right?

Possibly, but you should evaluate the improper integral by integrating from $a$ to 1 and then taking the limit as $a \rightarrow 0$. Sometimes those limits do exist.

 

[Robin04]

Since the function is finite and the interval differs by an infinitesimal amount between the open and closed interval, it doesn't make any difference. The integral is the same for the open and the closed interval.

What do you mean by the function being finite? The function at x=0 is infinite.
Is this also true if the point that is left out from the interval is singular?

 

[PeroK]

What do you mean by the function being finite? The function at x=0 is infinite.
Is this also true if the point that is left out from the interval is singular?

There's no notation for $\int_0^1$ to distinguish between the intervals $[0, 1]$ and $(0, 1)$.

Note that if you take one function and change a finite number of its function values, then you don't change the integral. You could think of the integral as an area - and an area doesn't depend on any point. Or, you could go to the definition of the definite integral as a limit and check that changing a finite number of function values makes no difference.

Hence the integral doesn't depend on whether you think of the interval as open or closed. Generally, I'd say you consider the interval to be closed.

If a function tends towards $\pm \infty$ at a given point, then you must take the integral for points close to the singular point and take the limit. These integrals are called improper integrals, as mentioned above.

 

[haruspex]

2. Therefore [c]=kg⋅m

Don't confuse dimension with units. No units are specified for x or ρ.

With regard to improper integrals, bear in mind that integration is a limit process. If you go back to its definition as such you will find that if the limit as ε→0 of ∫_εf(x).dx exists then it is valid as the value of ∫₀f(x).dx.

 

[RPinPA]

What do you mean by the function being finite? The function at x=0 is infinite.

With that statement, I was responding specifically to the evaluation of the integral over [1, 2] or (1, 2), where the function is finite, since you'd just evaluated it over the closed interval.

Is this also true if the point that is left out from the interval is singular?

x = 0 is not part of the domain since the function goes to infinity there. The function is not continuous at 0. The rigorous explanation of what's going on is a little more complex but you'd still evaluate either integral, [0, 1] or (0, 1), by integrating over [a, 1] where 0 < a < 1 and then taking the limit as $a \rightarrow 0$.

It's possible to cook up an integration where there's a difference between the open and closed interval of integration, but my main point was that things were well-behaved enough here that you didn't have to worry about it.

One place where you'd see that difference is in formal probability theory using Lebesgue integration, which handles both discrete and continuous distributions. In continuous distributions, there's no difference between the probability $P(X > 0)$ and $P(X \geq 0)$, both involve the same integral. But you could have a distribution where there's a probability of 0.1 that X is exactly equal to 0. In that case those two probabilities and their associated integrals would differ by 0.1.

 

[Robin04]

The integration is clear now, thank you! :)

Don't confuse dimension with units. No units are specified for x or ρ.

Thank you for pointing this out, I didn't know about this difference. So the dimension of $c$ is $m \cdot l$. The $[c]$ notation is for units, right? Is there a standard notation for dimension?

 

[PeroK]

The integration is clear now, thank you! :)Thank you for pointing this out, I didn't know about this difference. So the dimension of $c$ is $m \cdot l$. The $[c]$ notation is for units, right? Is there a standard notation for dimension?

Try here:

https://en.wikipedia.org/wiki/Dimensional_analysis#Definition

 

[Robin04]

Try here:

https://en.wikipedia.org/wiki/Dimensional_analysis#Definition

Thank you :)

 

[haruspex]

The [c] notation is for units, right? Is there a standard notation for dimension?

No, it's the other way around. The [] notation is for dimension. The terms inside should be in uppercase.