Question

Find the center of mass of the lamina described by the region \(R\) in the plane and its density function \(\delta(x, y)\) Note: these are the same lamina as in Exercises \(11-18\). \(R\) is the triangle with corners \((-1,0),(1,0),\) and (0,1) \(\delta(x, y)=2 \mathrm{lb} / \mathrm{in}^{2}\)

Short answer

The center of mass is at \((0, \frac{2}{3})\).

Step-by-step solution

Understand the Region

The region \(R\) is a triangle. The vertices of the triangle are \((-1,0)\), \((1,0)\), and \((0,1)\). This is a right triangle with the hypotenuse connecting the point \((0,1)\) to points along the x-axis from \(-1\) to \(1\).

Calculate the Area of the Triangle

The base of the triangle along the x-axis is 2 units (from \(-1\) to \(1\)), and the height is 1 unit (from the x-axis to \((0,1)\)). The area \(A\) of the triangle is given by the formula \(A = \frac{1}{2} \cdot \, base \, \cdot \, height = \frac{1}{2} \cdot 2 \cdot 1 = 1\, \text{square units}.\)

Determine the Mass of the Lamina

The density function is a constant \(\delta(x, y) = 2\,\mathrm{lb/in}^{2}\). The mass \(M\) of the region can be calculated by multiplying the density by the area: \(M = \delta \cdot A = 2 \cdot 1 = 2\,\mathrm{lb}.\)

Integrate to Find \(\bar{x}\)

The centroid \(\bar{x}\) is given by \(\bar{x} = \frac{1}{M}\int_R x \, \delta(x, y) \, dA\). For the triangular region, the integral becomes \(\frac{1}{2} \int_{-1}^{1} \int_{0}^{1-x} x \cdot 2 \, dy \, dx\). Evaluating the inner integral with respect to \(y\), and then the outer integral with respect to \(x\) gives \(\bar{x} = 0\).

Integrate to Find \(\bar{y}\)

The centroid \(\bar{y}\) is given by \(\bar{y} = \frac{1}{M}\int_R y \, \delta(x, y) \, dA\). The integral becomes \(\frac{1}{2} \int_{-1}^{1} \int_{0}^{1-x} y \cdot 2 \, dy \, dx\). Evaluating the inner integral with respect to \(y\), and then the outer integral with respect to \(x\) gives \(\bar{y} = \frac{2}{3}\).

Key concepts

Lamina

A lamina is a thin, flat object with a consistent thickness, often analyzed in planar geometry problems. In mathematics and physics, we regard a lamina as having mass and density spread across its surface. It is useful to understand a lamina when studying center of mass, as each point contributes to the overall distribution of mass. This concept helps in calculating properties like centroids or moments of inertia. In our example, the lamina occupies a triangular region on the plane, and for simplicity, has a uniform density.

Triangular Region

In this problem, we describe the lamina's shape using a specific triangular region. This triangle has vertices at the points

• \((-1, 0)\)
• \((1, 0)\)
• \((0, 1)\)

. This triangle is right-angled and symmetric about the y-axis, with its base along the x-axis. Understanding this geometry is key to setting up and calculating relevant integrals. Moreover, the area, determined by the formula for a triangle, is used to evaluate the mass and further properties of the lamina.

Density Function

Density function, denoted as \(\delta(x, y)\), characterizes how mass is distributed over the area of a lamina. For simplicity in many exercises, it's often constant. Here, the density is given as \(2 \ \text{lb/in}^2\), indicating every unit area weighs 2 pounds. Calculating the mass of lamina involves multiplying this density by the area of the region. Understanding the density function concept is crucial for determining the center of mass or centroid.

Double Integrals

Double integrals extend the concept of integration to functions of two variables over a region in the plane. They accumulate values such as area under a surface or mass of an object with variable density. In this exercise, double integrals are crucial in calculating the centroid coordinates \(\bar{x}\) and \(\bar{y}\) of our triangular lamina. The centroid \(\bar{x}\) is found using \[ \bar{x} = \frac{1}{M} \int_R x \, \delta(x, y) \, dA \] The calculations separate into inner and outer integrals. The inner integral sums vertical strips of the triangle region, while the outer sums horizontally. Similarly, \(\bar{y}\) uses: \[ \bar{y} = \frac{1}{M} \int_R y \, \delta(x, y) \, dA \] These integrals help pin down how mass is centered over the triangular region, capturing both x and y components. Understanding and evaluating double integrals involves a layered approach of integrating within each variable's boundary constraints accurately.