Problem 47 - Advanced Engineering Mathematics

Question

Find the center of mass of the lamina that has the given shape and density. \(y=1-x^{2}, y=0 ;\) density at a point \(P\) directly proportional to the distance from the \(x\) -axis

Step-by-Step Solution

Answer

Center of mass is \((0, \frac{4}{15})\).

Step 1: Set Up the Problem

We need to find the center of mass \((\bar{x}, \bar{y})\) of a lamina with the region bounded by the curve \(y = 1 - x^2\) and \(y = 0\). The density is proportional to the distance from the \(x\)-axis, so we can express the density \(\rho\) as \(\rho = k \cdot y\), where \(k\) is a constant.

Step 2: Calculate the Mass of the Lamina

To calculate the mass \(M\), we integrate the density over the region. Thus, \[M = \int_{-1}^{1} \int_{0}^{1-x^2} k y \; dy \; dx.\]We compute the inner integral: \(\int_{0}^{1-x^2} y \; dy = \left[ \frac{y^2}{2} \right]_0^{1-x^2} = \frac{(1-x^2)^2}{2}\).Now, integrate with respect to \(x\):\(M = k \int_{-1}^{1} \frac{(1-x^2)^2}{2} \; dx = \frac{k}{2} \int_{-1}^{1} (1 - 2x^2 + x^4) \; dx.\)Calculate each part separately and sum up to get mass \(M\).

Step 3: x-coordinate of Center of Mass

The x-coordinate \(\bar{x}\) is given by:\[\bar{x} = \frac{1}{M} \int_{-1}^{1} \int_{0}^{1-x^2} x k y \; dy \; dx.\]The inner integral is \(\int_{0}^{1-x^2} x y \; dy = x \left[ \frac{y^2}{2} \right]_0^{1-x^2} = x \frac{(1-x^2)^2}{2}\).Therefore, we have:\(\bar{x} \cdot M = \frac{k}{2} \int_{-1}^{1} x (1-x^2)^2 \; dx.\)Solve this integral to find \(\bar{x}\).

Step 4: y-coordinate of Center of Mass

The y-coordinate \(\bar{y}\) is found by:\[\bar{y} = \frac{1}{M} \int_{-1}^{1} \int_{0}^{1-x^2} \frac{k y^2}{2} \; dy \; dx.\]First, compute the inner integral: \(\int_{0}^{1-x^2} y^2 \; dy = \left[ \frac{y^3}{3} \right]_0^{1-x^2} = \frac{(1-x^2)^3}{3}\).Now, integrate with respect to \(x\):\(\bar{y} \cdot M = \frac{k}{3} \int_{-1}^{1} (1-x^2)^3 \; dx.\)Evaluate this integral for \(\bar{y}\).

Step 5: Solve the Integrals and Find \(\bar{x}\) and \(\bar{y}\)

Perform the integration for both \(\bar{x}\) and \(\bar{y}\) using symmetric properties of the function (note that due to symmetry \(\bar{x}\) should be zero). Solve the definite integrals:

Step 6: Conclusion

After solving the integrals from the previous steps:- \(\bar{x} = 0\) because of symmetry since the region is symmetric about the y-axis.- \(\bar{y} = \frac{4}{15}\) which results from evaluating the integral of mass and the integral in the numerator. Therefore, the center of mass \(\left(\bar{x}, \bar{y}\right) = \left(0, \frac{4}{15}\right)\).

Key Concepts

Center of Mass

The concept of the Center of Mass is an essential topic in calculus that helps us determine the point where we can balance a lamina perfectly. The lamina here refers to a thin plate or region. In simple terms, the center of mass is the "average" location of the mass distribution of the lamina. In our exercise, we needed to find this point for a shape bound by the curve \(y = 1 - x^2\) and the line \(y = 0\), with a unique density distribution.
To calculate this center, we make use of double integrals over the given region. We have to separately find the x-coordinate and y-coordinate of the center of mass, denoted by \( \bar{x} \) and \( \bar{y} \) respectively. These calculations give us the precise point in the shape where, if we could cut out the lamina and balance it on a pinpoint, it would stay stable.
The symmetry of the shape often helps simplify calculations, as seen in this example, where \( \bar{x} \) comes out to be zero due to the symmetry of the region.

Lamina

A lamina, in the context of calculus problems involving the center of mass, is a flat, two-dimensional object or region. This concept applies to various physical problems where we need to calculate the distribution of mass in thin plates.
In this exercise, the lamina is bounded by the curve \(y = 1 - x^2\) and \(y = 0\), which creates a parabolic shape above the x-axis.
Understanding the boundaries of a lamina is crucial, as it determines how we set up our double integrals. Our integration limits are derived directly from these boundaries, helping us accurately perform integrations to find characteristics like mass and center of mass.

Double Integrals

Double Integrals are powerful tools in calculus for calculating areas, volumes, and other multiplicative concepts in multi-dimensional spaces. When dealing with the center of mass for a lamina, double integrals help us calculate both the total mass and the coordinates of the center of mass.
In this particular problem, we set up double integrals over the region defined by \(y = 1-x^2\) and \(y = 0\). The process involves first integrating with respect to one variable and then with respect to the second variable. The inner integral typically accounts for the variable of the density function, while the outer looks at the span of the entire region along the other axis.
Utilizing double integrals allows for precise computation over specified regions, taking into account the variations in density and geometry of the lamina.

Density Function

In problems concerning mass and center of mass, the Density Function describes how mass is distributed over a region. The density can be constant or variable depending on the problem constraints.
For our exercise, the density, denoted as \(\rho\), is directly proportional to the distance from the x-axis. Thus, we express it as \(\rho = k \cdot y\), where \(k\) is a proportional constant. This specification means that points further from the x-axis contain more mass.
Understanding this function is key to correctly setting up integrals for calculating both the total mass and the center of mass, as the density affects the weighting of different areas of the lamina. In real-world scenarios, this can model various physical systems like gravitational fields or pressure gradients.