Problem 22 - APEX Calculus

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 \((0,0),(1,0),\) and (0,1)\(;\) \(\delta(x, y)=\left(x^{2}+y^{2}+1\right) \mathrm{Ib} / \mathrm{in}^{2}\)

Step-by-Step Solution

Answer

Center of mass is \(\left(\frac{2}{5}, \frac{2}{5}\right)\).

Step 1: Identify the Vertices and Density Function

The problem states that the triangle with vertices is \((0,0), (1,0),\) and \((0,1)\). The density function given is \(\delta(x,y) = x^2 + y^2 + 1\ lb/in^2\).

Step 2: Calculate the Area of the Triangle

The area of a triangle with vertices at \((0,0), (1,0), (0,1)\) can be calculated using the formula for the area of a triangle with base and height, \(\frac{1}{2} \times \text{base} \times \text{height}\). Here, both base and height are 1, so the area \(A = \frac{1}{2} \times 1 \times 1 = \frac{1}{2} \).

Step 3: Set Up the Integral for Mass

To find the mass \(M\) of the lamina, integrate the density over the region \(R\). The mass \(M\) can be calculated using: \[M = \int_{0}^{1} \int_{0}^{1-x} (x^2 + y^2 + 1) \, dy \, dx\]

Step 4: Evaluate the Inner Integral

Integrate \(x^2 + y^2 + 1\) with respect to \(y\) from \(0\) to \(1-x\): \[\int_{0}^{1-x} (x^2 + y^2 + 1) \, dy = x^2(1-x) + \frac{(1-x)^3}{3} + (1-x)\]

Step 5: Evaluate the Outer Integral for Mass

Substitute the result of the inner integral and integrate with respect to \(x\) from \(0\) to \(1\): \[M = \int_{0}^{1} \left[ x^2(1-x) + \frac{(1-x)^3}{3} + (1-x) \right] dx\] Calculate this integral to find the mass \(M = \frac{5}{6}\).

Step 6: Calculate the Moments

Find \(M_x\) and \(M_y\) to locate the center of mass:For \(M_x\): \[M_x = \int_{0}^{1} \int_{0}^{1-x} y(x^2 + y^2 + 1) \, dy \, dx\]For \(M_y\):\[M_y = \int_{0}^{1} \int_{0}^{1-x} x(x^2 + y^2 + 1) \, dy \, dx\]

Step 7: Evaluate \(M_x\) and \(M_y\)

Calculate the integrals for \(M_x\) and \(M_y\) using similar steps to those used for \(M\). The results are:\(M_x = \frac{1}{3}\) and \(M_y = \frac{1}{3}\).

Step 8: Determine Center of Mass

The coordinates of the center of mass \((\bar{x}, \bar{y})\) are found using:\[\bar{x} = \frac{M_y}{M}, \quad \bar{y} = \frac{M_x}{M}\]Substitute \(M = \frac{5}{6}\), \(M_x = \frac{1}{3}\), and \(M_y = \frac{1}{3}\) into the formulas to get:\[\bar{x} = \frac{2}{5}, \quad \bar{y} = \frac{2}{5}\]

Key Concepts

Center of Mass

The center of mass is a fundamental concept in physics and calculus. It is the point in a body or system where the entire mass can be considered to be concentrated. This is crucial for understanding balance and motion. In the context of a planar lamina, which is a flat object with a specified area and density, finding the center of mass involves integrating the density function over the region.
For example, in the exercise above, we are finding the center of mass for a triangular lamina with a given density function. The center of mass \((\bar{x}, \bar{y})\) is calculated using the moments, \(M_x\) and \(M_y\), and the total mass, \(M\). Calculating these components allows us to precisely determine where the center of mass lies, giving us complete understanding of how this body or system would behave under forces.

Density Function

A density function is a mathematical representation of how mass is distributed over a given area or volume. It's crucial for calculating mass properties such as the center of mass. In this problem, the density function is defined as \(\delta(x, y) = x^2 + y^2 +1\ lb/in^2\). This tells us that the density at any point \((x, y)\) on our lamina alters based on its coordinates.
Density functions can be simple or complex, depending on how the mass varies over the region considered. By integrating this density function over the region defined, we compute the total mass of the lamina, which is a necessary step in determining other physical properties like the center of mass.

Integration

Integration is a core concept in calculus that allows us to find quantities like area, volume, and total mass. It is particularly useful for problems involving variable densities, like the one in this exercise. Through integration, we can sum an infinite number of infinitesimally small parts to find a whole quantity.

• In the exercise, we set up integrals to find the mass of the lamina: \(M = \int_{0}^{1} \int_{0}^{1-x} (x^2 + y^2 + 1) \, dy \, dx\).
• The strategy involves evaluating inner and outer integrals to compute the overall mass.

By breaking the integration into manageable parts and addressing each component step by step, we ensure accurate calculation of both mass and moments, which are fundamental to understanding how the entire system behaves.

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, pairs algebra with geometry, allowing us to solve geometrical problems with mathematical tools. In our problem, we consider a triangular region in a plane defined by three points: \((0,0), (1,0), (0,1)\).
The advantage of coordinate geometry is its ability to simplify complex calculus problems involving spatial figures. The vertices of the triangle define the region of interest, and the use of coordinates allows us to apply integration over this specific area.

• With the vertices known, we set boundaries for integration, crucial for calculating the mass and centre of mass efficiently.
• This precision helps with accurate computations, ensuring the integration limits truly reflect the shape and size of the region.

Through coordinate geometry, we can leverage formulas like the area of a triangle \(\frac{1}{2} \times \text{base} \times \text{height}\) to facilitate and validate our calculations, aligning algebraic techniques with geometric understanding.