Problem 84 - Thomas Calculus

Question

a. Find the center of mass of a thin plate of constant density covering the region between the curve \(y=1 / \sqrt{x}\) and the \(x\) -axis from \(x=1\) to \(x=16\) . b. Find the center of mass if, instead of being constant, the density function is \(\delta(x)=4 / \sqrt{x} .\)

Step-by-Step Solution

Answer

(7, 1/3) for constant density and (7, 1/6) for varying density.

Step 1: Determine the area of the region

Firstly, calculate the area under the curve \(y = \frac{1}{\sqrt{x}}\) from \(x=1\) to \(x=16\). This is given by the integral \( \int_1^{16} \frac{1}{\sqrt{x}} \, dx \). Solving this integral.

Step 2: Evaluate the integral for area

The integral \( \int \frac{1}{\sqrt{x}} \, dx = 2\sqrt{x} + C \). Evaluating from 1 to 16 gives \(2 \sqrt{16} - 2 \sqrt{1} = 8 - 2 = 6 \). Thus, the area is 6.

Step 3: Find the moment about the y-axis (constant density)

The moment about the y-axis \(M_y\) is given by \( \int_1^{16} x \cdot \frac{1}{\sqrt{x}} \, dx \). Simplifying, we have \( \int_1^{16} \sqrt{x} \, dx \).

Step 4: Integrate to find the moment about the y-axis (constant density)

\( \int \sqrt{x} \, dx = \frac{2}{3}x^{3/2} + C \). Evaluating from 1 to 16 gives \( \frac{2}{3}(64) - \frac{2}{3}(1) = \frac{128}{3} - \frac{2}{3} = \frac{126}{3} = 42 \). Thus, \(M_y = 42\).

Step 5: Calculate x-coordinate of center of mass (constant density)

The x-coordinate of the center of mass \(\bar{x}\) is given by \( \frac{M_y}{ ext{area}} = \frac{42}{6} = 7 \).

Step 6: Find the moment about the x-axis (constant density)

The moment about the x-axis \( M_x \) is given by \( \int_1^{16} \frac{1}{2} \left( \frac{1}{\sqrt{x}} \right)^2 \, dx = \frac{1}{2} \int_1^{16} \frac{1}{x} \, dx \).

Step 7: Integrate moment about the x-axis (constant density)

\( \int \frac{1}{x} \, dx = \ln|x| + C \). Evaluating from 1 to 16 gives \( \frac{1}{2}(\ln(16) - \ln(1)) = \frac{1}{2}\ln(16) = 2\ln(4) = 2 \cdot 2 = 4 \). So, \(M_x = 2\).

Step 8: Calculate y-coordinate of center of mass (constant density)

The y-coordinate of the center of mass \(\bar{y}\) is given by \( \frac{M_x}{ ext{area}} = \frac{2}{6} = \frac{1}{3} \).

Step 9: Calculate moments for varying density

For \(\delta(x) = \frac{4}{\sqrt{x}}\), the moment about the y-axis \(M_y\) becomes \( \int_1^{16} x \cdot \frac{4}{\sqrt{x}} \, dx = 4 \int_1^{16} \sqrt{x} \, dx \).

Step 10: Evaluate moment about y-axis (varying density)

The result is \(4 \times 42 = 168\) since we computed \( \int_1^{16} \sqrt{x} \, dx = 42 \) earlier.

Step 11: Calculate x-coordinate of center of mass (varying density)

\(\bar{x} = \frac{168}{24} = 7\) where 24 is the adjusted area from \( \int_1^{16} \frac{4}{\sqrt{x}} \, dx = 24 \) since earlier area integral was 6 times \(4\).

Step 12: Calculate moment about x-axis (varying density)

For \(M_x\), multiply by the density function: \( \int_1^{16} \frac{2}{\sqrt{x}} \, dx \). Completing similar steps: \( 4 \), same as constant due to symmetry.

Step 13: Calculate y-coordinate of center of mass (varying density)

\(\bar{y} = \frac{4}{24} = \frac{1}{6}\).

Key Concepts

Calculus

Calculus is a branch of mathematics that deals with the study of change and motion. It is divided into two primary areas: differential and integral calculus. In this exercise, calculus helps find the center of mass by utilizing both differentiation and integration.

• Differential Calculus: Focuses on rates of change and slopes of curves. It is not directly used in this problem but helps in understanding how functions behave.
• Integral Calculus: This is the main tool used here to calculate the area under a curve and the moments about axes. It allows us to find total values like area and moment by summing infinitely small parts of the region or object.

Calculus provides the foundational understanding for finding the center of mass, which is the point where an object's weight is evenly distributed.

Integration

Integration is the process of calculating the integral, which is essentially the accumulation of quantities. In this exercise, we use integration to find areas and moments.
There are different types of integration:

• Definite Integration: In this problem, when we compute the area and moments, we're doing definite integration because we're finding total values over specific intervals, from 1 to 16.
• Indefinite Integration: It represents a general form of integration without limits and includes a constant, C, to encompass all potential antiderivatives.

Integration is key to understanding the physical characteristics of a region, such as its center of mass, by considering how values accumulate over a given range.

Density Function

The density function describes how mass is distributed over an object or region. It affects the calculation of the center of mass.
In this problem, we encounter two density scenarios:

• Constant Density: The simplest scenario where the mass is uniformly distributed, meaning the density value is constant across the region. This leads to straightforward calculations for the moments and center of mass.
• Varying Density: Mass distribution changes across the region. Specifically, the density function is \(\delta(x) = \frac{4}{\sqrt{x}}\). This requires re-evaluating integrations as shown when calculating moments about the x and y axes in the given exercise.

The density function plays a crucial role in defining how the shape and distribution of mass impact the center of mass position.

Coordinate System

The coordinate system is a reference framework crucial for defining the positions of points in space. In this exercise, a Cartesian coordinate system helps to plot and interpret points on a plane.
Using a coordinate system, we:

• Define Axes: The x-axis and y-axis in a 2D plane, where all calculations are centered around finding the position of mass relative to these axes.
• Calculate moments About Axes: These are the integrations performed in relation to the x-axis or y-axis to determine the center of mass.

Understanding how coordinates work allows us to resolve complex spatial problems, such as determining a region's center of mass, by providing a clear mathematical foundation for plotting and analysis.