Problem 1 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-10, 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. \(x=0, x=4, y=0, y=3 ; \delta(x, y)=y+1\)

Step-by-Step Solution

Answer

Mass is 30; Center of mass is \((2, \frac{9}{5})\)."

Step 1: Identify the Region

The lamina is bounded by the curves: \(x=0\), \(x=4\), \(y=0\), and \(y=3\). This defines a rectangle with vertices at \((0,0)\), \((4,0)\), \((4,3)\), and \((0,3)\).

Step 2: Express the Mass of the Lamina

The mass \(m\) of the lamina with given density \(\delta(x, y) = y + 1\) is calculated as an integral: \ \[m = \int_{0}^{4} \int_{0}^{3} (y+1) \, dy \, dx.\]

Step 3: Integrate with Respect to y

First, integrate the density function \(y + 1\) with respect to \(y\), from \(y=0\) to \(y=3\):\[\int_{0}^{3} (y+1) \, dy = \left[ \frac{y^2}{2} + y \right]_{0}^{3} = \left( \frac{9}{2} + 3 \right) = \frac{15}{2}.\]

Step 4: Integrate with Respect to x

Next, integrate the result from Step 3 with respect to \(x\), from \(x=0\) to \(x=4\):\[m = \int_{0}^{4} \frac{15}{2} \, dx = \left[ \frac{15}{2}x \right]_{0}^{4} = 30.\]

Step 5: Find the Moment about the y-axis

The moment about the y-axis, \(M_y\), is given by the integral: \[M_y = \int_{0}^{4} \int_{0}^{3} x (y+1) \, dy \, dx.\]

Step 6: Integrate for Moment about the y-axis

First, calculate the inner integral with respect to \(y\):\[\int_{0}^{3} x(y+1) \, dy = x \left[ \frac{y^2}{2} + y \right]_{0}^{3} = x \left( \frac{9}{2} + 3 \right) = \frac{15}{2}x.\]

Step 7: Integrate for Total Moment about the y-axis

Then, integrate with respect to \(x\):\[M_y = \int_{0}^{4} \frac{15}{2}x \, dx = \left[ \frac{15}{4}x^2 \right]_{0}^{4} = 60.\]

Step 8: Find the Moment about the x-axis

The moment about the x-axis, \(M_x\), is:\[M_x = \int_{0}^{4} \int_{0}^{3} y(y+1) \, dy \, dx.\]

Step 9: Integrate for Moment about the x-axis

First, calculate the inner integral with respect to \(y\):\[\int_{0}^{3} y(y+1) \, dy = \int_{0}^{3} (y^2 + y) \, dy = \left[ \frac{y^3}{3} + \frac{y^2}{2} \right]_{0}^{3} = \left(9 + \frac{9}{2}\right) = \frac{27}{2}.\]

Step 10: Integrate for Total Moment about the x-axis

Then, integrate with respect to \(x\):\[M_x = \int_{0}^{4} \frac{27}{2} \, dx = \left[ \frac{27}{2}x \right]_{0}^{4} = 54.\]

Step 11: Compute the Center of Mass Coordinates

Finally, use the formulas for the center of mass:\[\bar{x} = \frac{M_y}{m} = \frac{60}{30} = 2,\]\[\bar{y} = \frac{M_x}{m} = \frac{54}{30} = \frac{9}{5}.\]

Key Concepts

Lamina

A lamina is a flat, two-dimensional object with a certain shape and area. In the realm of calculus, a lamina becomes particularly interesting when density functions are introduced. For our problem, we are given a rectangular lamina, formed by boundaries defined by curves. These boundaries make the vertices of the rectangle lie at the points

• (0, 0)
• (4, 0)
• (4, 3)
• (0, 3)

The lamina is often used in calculus to explore topics like mass and center of mass, especially when it has a varying density, which brings us to our next concept.

Density Function

The density function, denoted by delta(x, y), represents how mass is distributed across an area. In this exercise, the lamina has a density function of delta(x, y) = y + 1. This means the density increases linearly with the y-coordinate.

• At y = 0, delta = 1
• At y = 3, delta = 4

Understanding the density function is crucial, as it helps determine the total mass of the lamina. Since the density isn't constant, this is where integration helps us effectively find areas under the curve, representing varying mass distributions.

Double Integration

Double integration is the process of computing the integral of a function over a two-dimensional region, usually in the plane. In our problem, we have a double integral set up to calculate the lamina's mass with respect to its density function. First, we integrate with respect to one variable, and then the resulting expression is integrated with respect to the second variable. To determine the mass of the lamina, we performed the integration:\[m = \int_{0}^{4} \int_{0}^{3} (y+1) \, dy \, dx.\]

• The inner integral integrates over y, finding the area under the density function vertically.
• The outer integral integrates this result horizontally across x.

By utilizing double integration, calculus allows us to calculate not only areas and volumes but also the mass of regions with variable density, which is foundational for finding centroid and moments.

Calculus Problem Solutions

Solving calculus problems often involves different kinds of integrals, as well as understanding geometry and algebra to set up those integrals. In this exercise, identifying the region, formulating the problem in integral form, and solving step-by-step were key to finding the mass and center of mass. The process began by defining the region of integration and calculating the mass. Then, two more integrals calculated the moments about the axes. Lastly, by plugging those into the center of mass formulas:\[\bar{x} = \frac{M_y}{m}\]\[\bar{y} = \frac{M_x}{m}\]The solution reached was:

• \(\bar{x} = 2\)
• \(\bar{y} = \frac{9}{5}\)

This systematic approach is crucial for tackling similar problems, reemphasizing our understanding of integration, density variations, and geometric shapes.