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. x=0, x=4, y=0, y=3 ; \delta(x, y)=y+1

Short answer

Mass is 30, center of mass is \((2, \frac{7}{5})\).

Step-by-step solution

Understand the Problem

We need to find the mass and center of mass of a lamina bounded by the curves \(x=0\), \(x=4\), \(y=0\), and \(y=3\) with a density function \(\delta(x, y) = y+1\). The region is a rectangle in the xy-plane where \(0 \leq x \leq 4\) and \(0 \leq y \leq 3\).

Calculate the Mass

The mass \(m\) of the lamina is given by the double integral \(m = \iint_R \delta(x, y) \, dA\), where \(R\) is the region. \[ m = \int_0^4 \int_0^3 (y+1) \, dy \, dx \]First, solve the inner integral with respect to \(y\): \[ \int_0^3 (y+1) \, dy = \left[ \frac{y^2}{2} + y \right]_0^3 = \frac{9}{2} + 3 = \frac{15}{2} \] Then solve the outer integral with respect to \(x\):\[ m = \int_0^4 \frac{15}{2} \, dx = \left[ \frac{15}{2}x \right]_0^4 = 30 \]

Calculate \(\bar{x}\), the x-coordinate of the Center of Mass

The x-coordinate of the center of mass is given by the formula:\[ \bar{x} = \frac{1}{m} \iint_R x \delta(x, y) \, dA \]\[ \bar{x} = \frac{1}{30} \int_0^4 \int_0^3 x(y+1) \, dy \, dx \]First, solve the inner integral:\[ \int_0^3 x(y+1) \, dy = x \left[ \frac{y^2}{2} + y \right]_0^3 = x \left( \frac{9}{2} + 3 \right) = \frac{15}{2}x \]Then solve the outer integral:\[ \bar{x} = \frac{1}{30} \int_0^4 \frac{15}{2}x \, dx = \frac{1}{30} \left[ \frac{15}{4}x^2 \right]_0^4 = \frac{1}{30} \cdot \frac{240}{4} = 2 \]

Calculate \(\bar{y}\), the y-coordinate of the Center of Mass

The y-coordinate of the center of mass is given by the formula:\[ \bar{y} = \frac{1}{m} \iint_R y \delta(x, y) \, dA \]\[ \bar{y} = \frac{1}{30} \int_0^4 \int_0^3 y(y+1) \, dy \, dx \]Solve the inner integral first:\[ \int_0^3 y(y+1) \, dy = \int_0^3 (y^2 + y) \, dy = \left[ \frac{y^3}{3} + \frac{y^2}{2} \right]_0^3 = \frac{27}{3} + \frac{9}{2} = \frac{63}{6} = \frac{21}{2} \]Then solve the outer integral:\[ \bar{y} = \frac{1}{30} \int_0^4 \frac{21}{2} \, dx = \frac{1}{30} \cdot \frac{21}{2} \cdot 4 = \frac{42}{30} = \frac{7}{5} \]

Present Final Answers

The mass of the lamina is \(30\). The coordinates of the center of mass are \((\bar{x}, \bar{y}) = (2, \frac{7}{5})\).

Key concepts

Lamina

A lamina is a two-dimensional surface or layer with negligible thickness. Imagine it like a sheet of paper. In mathematics, we often deal with laminas to simplify complex three-dimensional objects. When dealing with laminas, especially in calculus, it's important to consider the concept of mass distribution across the surface. In this example, the lamina is bounded by the curves and straight lines:

• Vertical lines: from \(x = 0\) to \(x = 4\).
• Horizontal lines: from \(y = 0\) to \(y = 3\).

This defines a rectangular region in the xy-plane. Understanding this setup is crucial since it frames the limits for our calculations of mass and center of mass.

Density Function

The density function describes how mass is distributed over a given area or volume. In the context of our lamina, the density function is given by \(\delta(x, y) = y + 1\). This means the density varies with both the x and y coordinates, but dependent only on the y coordinate in this case, as the function is \(y + 1\).

This function indicates that the density increases with the y-coordinate and there's always an additional unit of mass at any point \((x, y)\). Knowing how the density function behaves is fundamental for finding the mass and the center of mass, as it tells us where the mass is more concentrated. This asymmetric distribution impacts calculations, making the center of mass not necessarily the geometric center.

Double Integral

To find the mass and the center of mass of a lamina, we employ the concept of a double integral. Double integrals allow us to integrate over a two-dimensional area, making them ideal for calculating properties of areas like laminas. In our exercise involving the lamina, the double integral is employed twice:

• First, to compute total mass, where we integrate the density function \(\delta(x, y)\) over the rectangle region.
• Next, to calculate the components of the center of mass, \(\bar{x}\) and \(\bar{y}\), by finding the weighted averages of coordinates.

To apply a double integral, we calculate the inner integral first, followed by the outer integral. The role of the double integral is crucial because it helps accumulate small areas (with density values) over the entire shape to find total mass and to locate the balancing point or center of mass.

Coordinate Geometry

Coordinate geometry, or analytic geometry, involves the study of geometric figures using a coordinate system. In our context, it means analyzing the lamina within the xy-plane using the given boundaries: \(x=0\), \(x=4\), \(y=0\), and \(y=3\). With these coordinates, we can define the region where our lamina exists.

Understanding these boundaries is critical when setting up integrals for mathematical calculations. The coordinate system allows us to describe the lamina as a rectangle and set up the limits of integration. This setup is necessary for calculating both the mass and center of mass of the lamina. By interpreting these coordinates correctly, we can precisely apply the principles of integration and utilize the density function effectively for accurate results.