91Ó°ÊÓ

In the world of mathematics, particularly in physics and engineering, the concept of a density function is crucial. The density function, often denoted as \[ \delta(x, y, z) \] in three-dimensional space, describes how a quantity (like mass) is distributed over a region. In our given problem, the density is specified as:\[ \delta(x, y, z) = \left(x^2 + y^2 + z^2\right)^{-1/2}. \]This density function indicates that as the distance from the origin increases, the density decreases. - Using spherical coordinates, which simplifies calculations involving spheres, this function transforms to:\[ \delta(\rho, \theta, \phi) = \rho^{-1}, \]where \(\rho\) is the radial distance. Thus, in spherical coordinates, the density directly depends on \(\rho\), which is useful when calculating mass over spherical volumes.

Finding the mass of a solid involves integrating the density over the volume the solid occupies. In the exercise, the solid is the space between two concentric spheres. The spheres are defined by:\[ x^2 + y^2 + z^2 = 1 \] \[x^2 + y^2 + z^2 = 4. \]In spherical coordinates, these are simply represented as \( \rho = 1 \) and \( \rho = 2 \). The mass calculation uses these limits to find the total mass of the space between these spheres.- The mass \( M \) is represented by the integral:\[ M = \int_0^{2\pi} \int_0^{\pi} \int_1^2 \delta(\rho, \theta, \phi) \cdot \rho^2 \sin \phi \ d\rho \ d\phi \ d\theta. \]This equation integrates the density function over the volume, which is determined by the spherical limits (\(\rho\), \(\theta\), and \(\phi\)). The term \(\rho^2 \sin \phi\) accounts for the change from Cartesian to spherical volume elements.

Integral calculus provides the tools necessary for finding quantities like mass, area, and volume by summing infinitesimal changes. In this problem, the integral\[ \int_0^{2\pi} \int_0^{\pi} \int_1^2 \rho^{-1} \cdot \rho^2 \sin \phi \ d\rho \ d\phi \ d\theta \]is used to calculate the mass of the solid. Here's how:- 1. **Inner Integral:** The innermost integral, \[ \int_1^2 \rho \ d\rho, \] simplifies the computation within the radial limits. Evaluating this gives: \[ \left[ \frac{\rho^2}{2} \right]_1^2 = \frac{3}{2}. \]2. **Middle Integral:** It calculates the contribution in the angle from the positive z-axis: \[ \int_0^{\pi} \sin \phi \, d\phi = 2. \]3. **Outer Integral:** Finally, it sums around the full revolution: \[ \int_0^{2\pi} \ d\theta = 2\pi. \]Putting these calculations together results in the total mass being:\[ M = 6\pi. \]These steps demonstrate how integral calculus plays a vital role in determining physical properties of a solid. Each integral takes into account a different part of the spherical coordinate space, ensuring comprehensive coverage of the volume.