# Why is $I$ = $\int_0^L R^2 \, dm$ considered a differential equation?

Earlier today in my physics course, we were talking about rotational inertia and how to calculate it for various shapes using calculus. Having not taken diff eq. yet, can anyone explain why this given integral is a differential equation? Note that I have dealt with differential equations in the past with separation of variables, albeit not much.

Defining moment of inertia as:

$$I = \int_0^L R^2 \, dm$$

where $L$ is the length of the given object, $R$ is the perpendicular distance of a particle from the axis of rotation, and $dm$ represents the mass of any infinitesimal particle of the object.

Using the fact that $\lambda \, dR = dm$ such that $m = \lambda R$, I can rewrite the integral above as:

$$I = \int_0^L \lambda R^2 \, dr$$

You could pull the $\lambda$ out front since it is a constant and then simply integrate to get:

$$I = \dfrac{\lambda L^3}{3}$$

Then, you could substitute $\lambda$ back in as $\dfrac{m}{R}$ to get:

$$I = \dfrac {mL^2}{3}$$

Note that this is the rotational inertia for a long uniform rod where the axis is through the end only. My question is (after a bit of the derivation): why is this considered a differential equation? In a lecture earlier today, my professor stated something along the lines of we may be asked to set up, but do not solve, a differential equation, similar to the one above (except I solved it). Is it a differential equation simply because in $$I = \int_0^L R^2 \, dm$$

we are not integrating with respect to the same variable ($R$)? Having to rewrite the integrand in terms of $R$ and $dr$ - is that what makes this a diff. eq?

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A differential equation relates a function to its derivatives. A simple example is $$\frac{d y(x)}{d x} - y(x) = 0$$ with the boundary condition $y(0) = 1$. The solution is $y(x) = e^x$.
The equation $I = \int_0^L R^2 d m$ doesn't relate $I$ to it's derivatives. (For example, it doesn't look like $\frac{d^2 I}{d x^2} + 4 \frac{d I}{d x} + 3 I = 0$.) Thus, it is not a differential equation. The right-hand side is just an integral you must calculate to find the moment of inertia. Note that $I$ doesn't even vary, it is just some constant that tells you how hard it is to spin the object.
@JayElectronica: Glad to help. The saying about math professors is true about physics professors too. They say $A$, mean $B$, and write $C$. –  oen Apr 25 '12 at 23:35