Problem 4 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the mass \(m\) and center of mass \((\bar{x}, \bar{y})\) of the lamina bounded by the given curves and with the indicated density. \(y=1 / x, y=x, y=0, x=2 ; \delta(x, y)=x\)

Step-by-Step Solution

Answer

Mass: \(\frac{2}{3}\); Center of Mass: \((\bar{x}, \bar{y})\) is \(\left(\frac{-11}{2}, \frac{3(\ln 2 - \frac{7}{8})}{2}\right)\).

Step 1: Understand the Region

Determine the area bounded by the graphs of the functions. The given functions are \(y=1/x\), \(y=x\), \(y=0\) (the x-axis), and \(x=2\). The intersection of \(y=1/x\) and \(y=x\) happens at points where \(1/x = x\), which simplifies to \(x^2 = 1\), hence \(x = 1\). This gives the left boundary in terms of x, \(x=1\), and the right boundary is \(x=2\). Thus, the region under consideration lies between \(x=1\) and \(x=2\) bounded above by \(y=1/x\) and \(y=x\).

Step 2: Set Up the Expression for Mass

To find the mass \(m\) of the lamina, integrate the density function over the region. The density function given is \(\delta(x,y) = x\). The mass is given by the double integral over the region:\[m = \int_{x=1}^{x=2} \int_{y=x}^{y=1/x} x \; dy \; dx.\]

Step 3: Integrate with respect to y

First evaluate the integral with respect to \(y\):\[\int_{y=x}^{y=1/x} x \; dy = x \cdot \left[y\right]_x^{1/x} = x \left(\frac{1}{x} - x\right) = 1 - x^2.\]

Step 4: Integrate with respect to x

Now integrate the result with respect to \(x\):\[m = \int_{1}^{2} (1 - x^2) \; dx = \left[x - \frac{x^3}{3}\right]_1^2.\]Calculate:\[m = \left[ 2 - \frac{8}{3} \right] - \left[ 1 - \frac{1}{3} \right] = \frac{6}{3} - \frac{8}{3} - \left(\frac{3}{3} - \frac{1}{3}\right) = \frac{2}{3}.\]

Step 5: Set Up x-Coordinate Center of Mass Formula

The x-coordinate of the center of mass \(\bar{x}\) is given by:\[\bar{x} = \frac{1}{m} \int_{x=1}^{x=2} \int_{y=x}^{y=1/x} x^2 \; dy \; dx.\]

Step 6: Calculate x-Coordinate Center of Mass

Evaluate the inner integral:\[\int_{y=x}^{y=1/x} x^2 \; dy = x^2 \left[ \frac{1}{x} - x \right] = x^2 (1 - x^2).\]Evaluate the outer integral:\[\int_{1}^{2} (x^2 - x^4) \; dx = \left[ \frac{x^3}{3} - \frac{x^5}{5} \right]_{1}^{2} = \left(\frac{8}{3} - \frac{32}{5}\right) - \left(\frac{1}{3} - \frac{1}{5}\right).\]Calculate:\[\bar{x} = \frac{1}{m} \left(\frac{8}{3} - \frac{32}{5} + \frac{1}{15}\right).\]

Step 7: Simplify x-Coordinate Center of Mass Expression

Find common denominators and simplify:\[\frac{8}{3} - \frac{32}{5} = \frac{40}{15} - \frac{96}{15} = \frac{-56}{15},\]plus \(\frac{1}{15}\):\[\frac{-56}{15} + \frac{1}{15} = \frac{-55}{15} = \frac{-11}{3}.\]Thus,\[\bar{x} = \frac{1}{(2/3)} \cdot \left(\frac{-11}{3}\right) = \frac{-11}{2}.\]

Step 8: Set Up y-Coordinate Center of Mass Formula

The y-coordinate of the center of mass \(\bar{y}\) is given by:\[\bar{y} = \frac{1}{m} \int_{x=1}^{x=2} \int_{y=x}^{y=1/x} x \cdot y \; dy \; dx.\]

Step 9: Calculate y-Coordinate Center of Mass

Evaluate the inner integral:\[\int_{y=x}^{y=1/x} xy \; dy = x \cdot \frac{y^2}{2} \Bigg|_{x}^{1/x} = x \left(\frac{1}{2x^2} - \frac{x^2}{2}\right) = \frac{1}{2x} - \frac{x^3}{2}. \]

Step 10: Integrate y Formula and Calculate

Now evaluate the outer integral:\[\int_{1}^{2} \left( \frac{1}{2x} - \frac{x^3}{2} \right) \; dx = \left[ \frac{\ln|x|}{2} - \frac{x^4}{8} \right]_{1}^{2}.\]Calculate the values:\[\frac{\ln|2|}{2} - \frac{16}{8} - \left(0 - \frac{1}{8}\right).\]After simplification:\[\bar{y} = \frac{1}{(2/3)} \left( \frac{\ln 2}{2} - 1 + \frac{1}{8} \right).\]

Step 11: Sum Up Final Values for the Solution

The mass \(m = \frac{2}{3}\). The center of mass \((\bar{x}, \bar{y})\) is calculated as \(\bar{x} = \frac{-11}{2}\) and \(\bar{y} = \frac{1.5 \cdot (\ln 2 - \frac{7}{8})}{2} = \frac{3}{2}(\ln 2 - \frac{7}{8})\).

Key Concepts

Density Function

In mathematics, a density function, denoted as \(\delta(x, y)\), describes how mass is distributed across a region of space. It is an important concept when dealing with problems related to mass and center of mass. In our exercise, the density function is given by \(\delta(x, y) = x\). This means that for any point \((x, y)\) within the region we are considering, the density of the mass is directly proportional to \(x\).
This simplifies our calculations because we only need to consider how changes in \(x\) affect the density, allowing us to focus on integrating over the given region to find quantities like the total mass.

Double Integral

The double integral is a technique used to integrate over a region in a two-dimensional space. It gives us a way to compute the "volume" under a surface, which may be related to mass or another physical quantity.
For our problem, the double integral is used to find the mass of the lamina. The expression \[m = \int_{x=1}^{x=2} \int_{y=x}^{y=1/x} x \; dy \; dx\] represents a double integral over the area defined by the boundaries of the problem.
First, you integrate with respect to \(y\), holding \(x\) constant, and then with respect to \(x\). This stepwise approach sums up the contributions from tiny rectangular elements within the bounded region, each with a small area \(dy \cdot dx\). It's a powerful technique for finding mass when combined with a density function.

Bounded Region

Bounded regions are crucial when calculating integrals because they determine the limits of integration. For our problem, the region is determined by the curves \(y = 1/x\), \(y = x\), the x-axis \(y = 0\), and the vertical line \(x = 2\).

• The curve \(y = 1/x\) is a hyperbolic curve that starts high at \(x = 1\) and descends as \(x\) increases.
• \(y = x\) is a straight line which progresses at a 45-degree angle.
• The x-axis serves as the lower boundary for the region.
• The vertical line \(x = 2\) caps the region horizontally.

Understanding the shape of this region helps us correctly set up the integration limits. The region is between \(x=1\) and \(x=2\), and the values of \(y\) range from \(y=x\) to \(y=1/x\). Visualizing these intersections lets us determine the enclosure that bounds our region for integration.

Mass of Lamina

The concept of the mass of a lamina is an application of the double integral over a bounded region with a given density function. In this problem, after setting up the correct limits of integration, we use the density function \(\delta(x, y) = x\) to find the mass.
The expression for mass, \[m = \int_{x=1}^{x=2} \int_{y=x}^{y=1/x} x \; dy \; dx,\] allows us to sum tiny mass contributions across the lamina. After computing the double integral by first integrating with respect to \(y\), and then with respect to \(x\), we ultimately find that the lamina's mass \(m\) is \(\frac{2}{3}\).
The mass value is fundamental for further calculations such as the center of mass, which uses the same integration process adjusted for different functions of \(x\) and \(y\) over the same region.