91Ó°ÊÓ

Understanding density distribution is essential for calculating properties like mass and moment of inertia for objects like spheres. In this exercise, the density of the sphere varies with distance from the center, which is known as a radial density distribution. The given density function is \( \rho(r) = 3.00 \times 10^3 \, \text{kg/m}^3 - (9.00 \times 10^3 \, \text{kg/m}^4) \cdot r \). This function tells us how dense different parts of the sphere are as you move away from the center.

- **At the center**: The density is highest. At \( r = 0 \), the density is simply \( 3.00 \times 10^3 \, \text{kg/m}^3 \).
- **At the surface**: The density decreases, meaning the surface layer is less dense than the center. At \( r = R = 0.2 \) m, you substitute into the equation to find the density at the surface.

This type of linear decrease is common in problems involving density distribution, where density depends not just on the total mass but on how that mass is distributed within the volume.

To find the total mass of a sphere with a varying density, we need to integrate the mass elements over the entire volume of the sphere. Imagine slicing the sphere into thin spherical shells with thickness \( dr \). Each shell at a distance \( r \) from the center has a mass element \( dm \).

The volume of each shell is \( dV = 4\pi r^2 dr \) because a sphere's surface area scales as \( 4\pi r^2 \). To find the mass of each shell, multiply the volume by the density at each radius: \( dm = \rho(r) \cdot dV = (3.00 \times 10^3 - 9.00 \times 10^3 r) \cdot 4\pi r^2 dr \).

Finally, to find the sphere's total mass, integrate these mass elements from the center \( r = 0 \) to the sphere's surface \( r = 0.2 \) m:\[ M = 4\pi \int_{0}^{0.2} \left(3.00 \times 10^3 r^2 - 9.00 \times 10^3 r^3\right) dr \] This integration accounts for how the changing density contributes to the sphere's total mass as we consider each radius.

Integral calculus is the mathematical backbone to solving problems with continuously varying quantities, such as density in this sphere's problem. Here, the integral helps sum up all the tiny mass elements to find total quantities, like the sphere's mass or its moment of inertia.

The process involves two main steps:

• **Determine the integral expression**: For mass, it was integrating over \( r^2 \) and \( r^3 \), representing different powers of radius due to density's dependence on \( r \). The form \( 4\pi \left( \int 3.00 \times 10^3 r^2 \, dr - \int 9.00 \times 10^3 r^3 \, dr \right) \) results from substituting the density function.
• **Solving integrals**: You compute definite integrals by evaluating the antiderivatives at the limits. For example, \( \int r^2 dr = \frac{r^3}{3} \) and \( \int r^3 dr = \frac{r^4}{4} \), evaluated from 0 to 0.2 m. This makes it possible to compute the total mass or moment of inertia by solving simpler integrals.

Using these methods, integral calculus provides a complete and precise way to calculate complex physical properties from basic principles.