91Ó°ÊÓ

To understand the center of mass for a solid, we start with the concept of a volume integral. A volume integral is a method used to compute quantities over a three-dimensional region. In this case, we are interested in calculating the volume of a solid with respect to its projection onto the xy-plane. The volume integral is expressed as \( V = \iint_{D} \int_{a}^{b} dA \, dz \), where \( D \) represents the area of the base of the solid in the xy-plane. This involves integrating over the height of the solid from \( a \) to \( b \), effectively 'slicing' the solid into thin sections along the z-axis and summing up their contributions to find the entire volume.

Since the volume is calculated as \( (b-a) \times \text{Area}(D) \), where \( (b-a) \) is the height and \( \text{Area}(D) \) is the area of the base, we are essentially multiplying the height by the area to get the total volume of the solid.

The density distribution of a solid describes how mass is spread throughout the material. In this exercise, the critical observation is that the density is uniform along the z-axis. This uniformity implies that any variation in density across different slices of the solid on the z-axis can be ignored. As a result, when determining the center of mass, the density does not affect the integration over z, making calculations more straightforward.

If the density were to vary, it would require a more complex integration to account for changes in mass distribution along the z-axis. However, with a constant density distribution, those complications are removed, simplifying the process of finding the center of mass.

Integration over z plays a fundamental role in determining the center of mass. Here, the goal is to compute the average z-coordinate, which represents the z-component of the center of mass. The crucial step is to integrate the variable z over its bounds from \( a \) to \( b \). For a uniform density along z, this integration becomes \( \int_{a}^{b} z \, dz = \left[ \frac{z^2}{2} \right]_{a}^{b} = \frac{b^2-a^2}{2} \).

This result is substituted into the formula for the center of mass in the z-direction. Since the integral \( \frac{b^2-a^2}{2} \) is a constant, it simplifies the formula for the center of mass: \( CM_z = \frac{\frac{b^2-a^2}{2}}{V} = \frac{b+a}{2} \). Thus, the center of mass is located at the midpoint between \( a \) and \( b \), proving that it lies in the plane \( z = \frac{a+b}{2} \).

With a uniform density and this straightforward integration, we conclude the calculation for the center of mass in the z-direction is remarkably simple and elegant.