Problem 79 - Thomas Calculus

Question

a. Find the centroid of the region between the curve \(y=1 / x\) and the \(x\) -axis from \(x=1\) to \(x=2 .\) Give the coordinates to two decimal places. b. Sketch the region and show the centroid in your sketch.

Step-by-Step Solution

Answer

The centroid coordinates are approximately (1.44, 0.50).

Step 1: Define the Area (A) of the Region

To find the centroid, we need to calculate the area under the curve from \(x=1\) to \(x=2\). We do this by integrating the function \( y = \frac{1}{x} \) from 1 to 2. \[ A = \int_{1}^{2} \frac{1}{x} \, dx \].

Step 2: Calculate the Integral for Area

Calculate the integral: \[ \int_{1}^{2} \frac{1}{x} \, dx = \left[ \ln|x| \right]_{1}^{2} = \ln(2) - \ln(1) = \ln(2) \]. Therefore, the area \( A = \ln(2) \approx 0.693.\)

Step 3: Find the x-coordinate of the Centroid (\(\bar{x}\))

The x-coordinate of the centroid is given by \( \bar{x} = \frac{1}{A} \int_{1}^{2} x \left( \frac{1}{x} \right) \, dx \). Thus, \[ \bar{x} = \frac{1}{\ln(2)} \int_{1}^{2} 1 \, dx = \frac{1}{\ln(2)} [x]_{1}^{2} = \frac{1}{\ln(2)} (2 - 1) = \frac{1}{\ln(2)} \].

Step 4: Approximate the Value of \(\bar{x}\)

Using \( \ln(2) \approx 0.693 \), we find that \( \bar{x} = \frac{1}{0.693} \approx 1.4427 \).

Step 5: Find the y-coordinate of the Centroid (\(\bar{y}\))

The y-coordinate of the centroid is \( \bar{y} = \frac{1}{A} \int_{1}^{2} \frac{1}{2} \left( \frac{1}{x} \right)^2 x \, dx \). This integral evaluates to \[ \bar{y} = \frac{1}{0.693} \cdot \frac{1}{2} \left( \int_{1}^{2} \frac{1}{x} \, dx \right) = \frac{1}{0.693} \cdot \frac{1}{2} \cdot \ln(2) = \frac{1}{2} \].

Step 6: Approximate the Value of \(\bar{y}\)

Since \( \bar{y} = \frac{1}{2} \), no approximation is needed, and \( \bar{y} = 0.5 \).

Step 7: Conclude the Coordinates of the Centroid

The coordinates of the centroid to two decimal places are \( (1.44, 0.50) \).

Step 8: Sketch and Indicate the Centroid

Draw the region bounded by \( y = \frac{1}{x} \), the x-axis, and the vertical lines at \( x=1 \) and \( x=2 \). Place a point at the coordinates \( (1.44, 0.50) \) to identify the centroid.

Key Concepts

Integration

Integration is a fundamental concept in calculus used to find the area under curves, among other applications. In this exercise, we integrate the function \( y = \frac{1}{x} \) from \( x = 1 \) to \( x = 2 \) to determine the area bounded by this curve and the x-axis.
This is done using a definite integral, which sums up infinitesimal elements over an interval. The integral for our region is expressed as \( A = \int_{1}^{2} \frac{1}{x} \, dx \).
By evaluating this integral:

• The antiderivative of \( \frac{1}{x} \) is \( \ln|x| \).
• Compute the definite integral: \( \left[ \ln|x| \right]_{1}^{2} = \ln(2) - \ln(1) \).
• Since \( \ln(1) = 0 \), the area is \( \ln(2) \approx 0.693 \).

Understanding how to evaluate such integrals is crucial for solving problems related to finding centroids and determining areas under curves.

Area under a curve

The area under a curve between two points on the x-axis represents the accumulation of values and in this context is crucial for calculating the centroid of the region.
For the given curve \( y = \frac{1}{x} \), the area from \( x = 1 \) to \( x = 2 \) is calculated using the previously mentioned definite integral.
It translates to finding the sum of vertical strips beneath the curve within these boundaries, which gives us:

• The area \( A = \ln(2) \).
• This area value helps in finding the x and y coordinates of the centroid by applying additional integration rules outlined in centroid formulas.

Thus, the area not only gives us a literal size of the region but is also a critical component used to find the centroid's coordinates.

Coordinate geometry

Coordinate geometry is the study of geometric figures through a coordinate system. It links algebraic concepts to geometry. In this exercise, it helps in finding the centroid of the region.
Finding the centroid involves calculating both x and y coordinates using integrations that represent weighted averages of the x and y coordinates of all the points in the region with respect to area.

The formulas used are:

• \( \bar{x} = \frac{1}{A} \int_{1}^{2} x \left( \frac{1}{x} \right) \, dx \) results in \( \bar{x} \approx 1.4427 \).
• \( \bar{y} = \frac{1}{A} \int_{1}^{2} \frac{1}{2} \left( \frac{1}{x} \right)^2 x \, dx \) gives \( \bar{y} = 0.5 \).

These calculations yield the centroid coordinates \( (1.44, 0.50) \), demonstrating the beauty and interconnectivity of coordinate geometry with calculus through the calculation of centroids.