91Ó°ÊÓ

The density function, denoted here as \(\delta(x)\), is a mathematical expression that represents how the mass of an object is distributed along its length, area, or volume. In this exercise, we are dealing with a linear wire, and the density is given by \(\delta(x) = 1 + 3x^2\). This means at any point \(x\) along the wire, the density changes based on this function.
A density function can depend on various factors, like material composition or temperature, but in this case, it straightforwardly depends on the square of \(x\). As you can see, it is not a constant function; rather, it increases with \(x^2\), indicating that parts of the wire further from the center (\(x=0\)) are denser. Such an understanding is crucial for solving problems involving varying density as it will affect both the total mass and the center of mass calculations.
To calculate properties like mass and center of mass using this density function, integral calculus methods come into play, which we'll cover next.

Integral calculus is a fundamental part of mathematics that helps calculate properties like area, volume, and in this context, mass and the center of mass. When dealing with a density function, as we need to integrate the density over the given interval, it becomes indispensable.
Given a linear density function \(\delta(x) = 1 + 3x^2\), to determine the total mass \(M\) of the wire from \(x = -3\) to \(x = 3\), we use the integral:\[ M = \int_{-3}^{3} \delta(x) \, dx = \int_{-3}^{3} (1 + 3x^2) \, dx. \]This integral sums up all the infinitely small masses along the interval. By splitting it into simpler integrals, it was easy to solve:

• \( \int_{-3}^{3} 1 \, dx = 6 \)
• \( \int_{-3}^{3} 3x^2 \, dx = 54 \)

Adding these results gives the total mass \(M=60\). Using integral calculus in this way allows us to handle non-uniform density functions effectively.
For the center of mass, another integral helps balance the mass around the axis, which is critical if \(\delta(x)\) varies over the length of the object.

The mass calculation is perhaps the most straightforward yet crucial part of determining overall properties like the center of mass. In cases where density changes over the wire, it is done not by mere multiplication of density and length but through an integral that accounts for changes in the density along the interval.
In our problem, the density function \(\delta(x) = 1 + 3x^2\) varies with \(x^2\), requiring us to integrate it over the interval \([-3, 3]\) to find the total mass:\[ M = \int_{-3}^{3} (1 + 3x^2) \, dx = 60. \]To find the center of mass \(\bar{x}\), we must consider not just the amount of mass, but also how it is distributed:\[ \bar{x} = \frac{1}{60} \left( \int_{-3}^{3} x \, \delta(x) \, dx \right). \]The mass calculation itself, and how it is weighted over the integral balance point, helps keep the center of mass at 0 when the interval is symmetric and the density function is symmetric around it, as shown here.
Thus, we successfully determine \(\bar{x} = 0\) by integrating \(x\delta(x)\) over our interval, taking into consideration how the weight is spread along the wire.