Problem 17 - Thomas Calculus

Question

In Exercises \(18,\) find the center of mass of a thin plate of constant density \(\delta\) covering the given region. The region between the curve \(y=1 / \sqrt{x}\) and the \(x\) -axis from \(x=1\) to \(x=16\)

Step-by-Step Solution

Answer

The center of mass is at \((-7, -\frac{\ln 16}{12})\).

Step 1: Set up the Integral for Mass

To find the center of mass, we first need the mass of the plate. The mass of a region is given by the integral of the density over the area. Here, \( ext{mass} = \int_{1}^{16} \delta y \, dx = \int_{1}^{16} \delta \frac{1}{\sqrt{x}} \, dx \).

Step 2: Calculate the Integral for Mass

Integrate to find the mass: \( ext{mass} = \delta \left[-2\sqrt{x}\right]_{1}^{16} = \delta \left[-2\sqrt{16} + 2\sqrt{1}\right] = \delta \left[-8 + 2\right] = -6\delta \).

Step 3: Set up the Integral for Center of Mass (\(x\)-coordinate)

The \(x\)-coordinate of the center of mass \( \bar{x} \) is given by: \( \bar{x} = \frac{1}{M} \int_{1}^{16} x \delta y \, dx = \frac{1}{M} \int_{1}^{16} x \delta \frac{1}{\sqrt{x}} \, dx \).

Step 4: Calculate \(x\)-coordinate of Center of Mass

Integrate to find: \( \int_{1}^{16} x \delta \frac{1}{\sqrt{x}} \, dx = \delta \int_{1}^{16} \sqrt{x} \, dx = \delta \left[ \frac{2}{3} x^{3/2} \right]_{1}^{16} = \delta \left[\frac{2}{3}(64) - \frac{2}{3}(1)\right] = \delta \cdot 42 \). The \(x\)-coordinate is \( \bar{x} = \frac{42\delta}{-6\delta} = -7 \).

Step 5: Set up the Integral for Center of Mass (\(y\)-coordinate)

The \(y\)-coordinate of the center of mass \( \bar{y} \) is given by: \( \bar{y} = \frac{1}{2M} \int_{1}^{16} \delta y^2 \, dx = \frac{1}{2M} \int_{1}^{16} \delta \left( \frac{1}{\sqrt{x}} \right)^2 \, dx = \frac{1}{2M} \int_{1}^{16} \delta \frac{1}{x} \, dx \).

Step 6: Calculate \(y\)-coordinate of Center of Mass

Integrate to find: \( \int_{1}^{16} \frac{1}{x} \, dx = \left[ \ln|x| \right]_{1}^{16} = \ln 16 \). The \(y\)-coordinate is \( \bar{y} = \frac{\delta}{-12\delta} \ln 16 = -\frac{\ln 16}{12} \).

Step 7: Conclusion

With \( \bar{x} = -7 \) and \( \bar{y} = -\frac{\ln 16}{12}\), the center of mass of the plate is \( (-7, -\frac{\ln 16}{12}) \).

Key Concepts

Integral Calculus

Integral calculus is a key mathematical tool used to find quantities like areas under curves and the total accumulation of quantities across a region. In the context of finding the center of mass, integral calculus helps us calculate both the mass of an object and its distribution.

• To find the mass, we integrate the density of the object over the given region. This accounts for the fact that mass could be different in different areas depending on the density.
• For the center of mass, we use integrals to determine where this mass is concentrated. The process involves integrating the function describing the shape of the object and the density.

It's particularly useful when dealing with shapes that are not uniform, like the area under a curve, because it allows us to piece together many small parts to get the total picture. Learning integral calculus not only helps in physics, but also in any field that deals with physical quantities spread over areas.

Mass Calculation

Mass calculation is at the heart of finding the center of mass. It requires both the density function and the understanding of how mass is distributed across an area. In this exercise, the mass of the plate is determined by \[\text{mass} = \int_{1}^{16} \delta \frac{1}{\sqrt{x}} \, dx \]

• The variable \( \delta \) represents the density, which is assumed constant, making calculations simpler.
• The expression for \( y = \frac{1}{\sqrt{x}} \) shows how the height of the region changes with respect to \( x \).

Integrating this product gives the mass of the area between the curve and the \( x \)-axis from \( x = 1 \) to \( x = 16 \). The result, \(-6\delta\), reveals the total mass, which is used later to calculate each coordinate of the center of mass.

Coordinate System

A coordinate system allows us to assign a specific point to the center of mass in a two-dimensional space. Understanding how to work within a coordinate system is crucial for solving problems involving the center of mass.

• The \( x \)-coordinate of the center of mass, \( \bar{x} \), helps in determining how far left or right the balance point is. In this example, \( \bar{x} = -7 \), indicating that the center is located 7 units to one side of the origin within the system used here.
• The \( y \)-coordinate, \( \bar{y} \), indicates the vertical position. For the plate, it's found by \( \bar{y} = -\frac{\ln 16}{12} \), which is derived from integrating the function and adjusting by mass.

The center of mass coordinates \((-7, -\frac{\ln 16}{12})\) tells us exactly where the point is that would balance this plate if it were physically possible to support it at a single point.