Problem 10 - Thomas Calculus

Question

In Exercises \(18,\) find the center of mass of a thin plate of constant density \(\delta\) covering the given region. The region enclosed by the parabolas \(y=x^{2}-3\) and \(y=-2 x^{2}\)

Step-by-Step Solution

Answer

The center of mass of the region is at (0, 1.1).

Step 1: Identify the region of integration

To find the region enclosed by the two parabolas, we first need to determine the points of intersection for the parabolic equations: \(y = x^2 - 3\) and \(y = -2x^2\). Setting them equal, \(x^2 - 3 = -2x^2\). Rearranging gives \(3x^2 - 3 = 0\), so \(x^2 = 1\). Therefore, the points of intersection are \(x = 1\) and \(x = -1\).

Step 2: Set up integrals for mass calculation

The mass, \(M\), of the thin plate is given by the integral of its density \(\delta\) over the region bounded by the parabolas. Since \(\delta\) is constant, the mass is simply \(M = \delta \int_{-1}^{1} \left((-2x^2) - (x^2 - 3)\right) \, dx\). The expression inside the integral simplifies to \(-2x^2 - x^2 + 3 = -3x^2 + 3\).

Step 3: Evaluate the mass integral

The integral to find the mass is therefore \( M = \delta \int_{-1}^{1} (3 - 3x^2) \, dx \). Calculating this antiderivative, we have: \(\int (3 - 3x^2) \, dx = 3x - x^3 \). Evaluating this from \(-1\) to \(1\), we get: \([3(1) - (1)^3] - [3(-1) - (-1)^3] = (3 - 1) - (-3 + 1) = 2 + 4 = 6\). Hence, \(M = 6\delta\).

Step 4: Calculate the centroid coordinates using integrals

For \( \bar{x} \), the center of mass in the x-direction, you use \( \bar{x} = \frac{1}{M} \delta \int_{-1}^{1} x(3 - 3x^2) \, dx\). The integrand \( x(3 - 3x^2)\) is an odd function, so the integral over the symmetric interval \([-1, 1]\) is zero. Therefore, \(\bar{x} = 0\). Now, calculate \( \bar{y} = \frac{1}{M} \delta \int_{-1}^{1} \frac{1}{2}((x^2 - 3)^2 - (-2x^2)^2) \, dx \). Simplifying the terms gives \((x^2 - 3)^2 - (-2x^2)^2 = (x^4 - 6x^2 + 9) - (4x^4) = -3x^4 - 6x^2 + 9\). Therefore, \( \bar{y} = \frac{1}{6\delta} \delta \int_{-1}^{1} \frac{1}{2}(-3x^4 - 6x^2 + 9) \, dx\).

Step 5: Evaluate \(\bar{y}\) integral

To find \(\bar{y}\), evaluate \(\bar{y} = \frac{1}{12} \int_{-1}^{1} (-3x^4 - 6x^2 + 9) \, dx\). Calculate the antiderivative: \(\int (-3x^4 - 6x^2 + 9) \, dx = -\frac{3}{5}x^5 - 2x^3 + 9x\). Evaluating from \(-1\) to \(1\), gives \([-\frac{3}{5}(1)^5 - 2(1)^3 + 9(1)] - [-\frac{3}{5}(-1)^5 - 2(-1)^3 + 9(-1)] = (-\frac{3}{5} - 2 + 9) - (\frac{3}{5} + 2 - 9) = (6.6) - (-6.6) = 13.2\). Hence, \(\bar{y} = \frac{1}{12} (13.2) = 1.1\).

Key Concepts

Constant Density

Imagine we have a thin plate, like a flat cookie lying on a table. We can describe its physical properties in different ways. One way to think about this plate is that it has a **constant density**. When we say the plate has constant density, we mean that its mass per unit area is the same throughout the entire plate.
This means no part of the plate is heavier than another if we compare areas of the same size. Why is constant density important?

• It makes the math simpler. Uniform properties mean less complexity in calculations.
• When density is constant, finding the center of mass becomes a straightforward task of integrating over the area.

Understanding constant density is crucial when calculating properties like mass and center of mass. In this exercise, the constant density extquote{ extdelta} acts as a multiplying factor in all of our calculations.

Thin Plate

In this context, a **thin plate** refers to a flat, two-dimensional object. Think of it like a sheet of paper or a piece of foil. The idea is that the thickness of the plate is negligible, so we only focus on its length and width.
Why do we model it this way?

• The simplification allows us to use two-dimensional integrals, which are easier to handle in calculus.
• By treating the plate as thin, we eliminate the need to consider variations in thickness, focusing only on the surface.

In practical terms, this means we only have to figure out the geometry of the two-dimensional shape to find the center of mass. The simplicity of the thin plate model helps when dealing with regions like those described in the exercise.

Integration

**Integration** plays a central role in finding the center of mass. It's the mathematical tool we use to add up tiny pieces of mass across the entire region.
In our exercise, the region is bounded by two parabolas. We set up integrals to calculate two main things: the total mass and the center of mass.

• The integral for mass sums all the mass contributions across the region by considering the density function and the area under consideration.
• To find the center of mass coordinates, we integrate weighted positions over the plate. For instance, integrating x times the density gives us information about distribution along the x-axis, and similarly for y.

By integrating functions over the specific region, we effectively "add up" all those little mass segments to find where the whole mass effectively concentrates.

Parabolic Region

The region of interest in this exercise is defined by two curves, each in the shape of a parabola. This **parabolic region** is determined by the equations:- \(y = x^2 - 3\) - \(y = -2x^2\)
Finding where these parabolas intersect tells us where the boundary of the region is. In solving our exercise, this involves setting the equations equal to each other and solving the resulting algebraic equation.
Once we know where they meet,

• We can set the limits for our integrals, specifically from \(x = -1\) to \(x = 1\), since those are the points of intersection.
• These two curves enclose the area, with one curve serving as the "top" and the other as the "bottom" within those x-limits when visualized on a graph.

This enclosed shape then becomes the area over which we integrate to find the desired physical properties.