Problem 1 - Thomas Calculus in SI Units

Question

Find the center of mass of a thin plate of constant density \(\delta\) covering the given region. The region bounded by the parabola \(y=x^{2}\) and the line \(y=4\)

Step-by-Step Solution

Answer

The center of mass is \((0, \frac{16}{15})\).

Step 1: Identify the region for integration

The problem involves the region between the parabola \(y = x^2\) and the horizontal line \(y = 4\). To find the bounds in \(x\), we need to solve for where these two curves intersect. Set \(x^2 = 4\), giving \(x = 2\) and \(x = -2\). Thus, the region is bounded by \(-2 \leq x \leq 2\).

Step 2: Set up the integrals for the center of mass

The center of mass \((\bar{x}, \bar{y})\) for a constant density region can be found using \[\bar{x} = \frac{1}{A} \int_{-2}^{2} x \cdot \text{Height}(x) \, dx\] and \[\bar{y} = \frac{1}{2A} \int_{-2}^{2} (\text{Height}(x))^2 \, dx\]. Here, \(A\) is the area of the region. The height is the difference between \(y=4\) and \(y=x^2\), which is \(4 - x^2\).

Step 3: Calculate the area of the region

The area \(A\) is given by \(\int_{-2}^{2} (4 - x^2) \, dx\). Calculate this by finding the antiderivative: \(\int 4 \, dx - \int x^2 \, dx = 4x - \frac{x^3}{3}\). Evaluate at the bounds to find \[A = \left[4x - \frac{x^3}{3}\right]_{-2}^{2} = (8 - \frac{8}{3}) - (-8 + \frac{8}{3}) = \frac{32}{3}.\]

Step 4: Calculate \(\bar{x}\)

Substitute into the formula for \(\bar{x}\): \[\bar{x} = \frac{1}{\frac{32}{3}} \int_{-2}^{2} x(4 - x^2) \, dx = \frac{3}{32} \int_{-2}^{2} (4x - x^3) \, dx.\] Find the antiderivative: \(\int 4x \, dx - \int x^3 \, dx = 2x^2 - \frac{x^4}{4}\). Evaluate this between \(-2\) and \(2\) to get \(0\). Therefore, \(\bar{x} = 0\).

Step 5: Calculate \(\bar{y}\)

Substitute into the formula for \(\bar{y}\): \[\bar{y} = \frac{1}{2 \cdot \frac{32}{3}} \int_{-2}^{2} (4 - x^2)^2 \, dx= \frac{3}{64} \int_{-2}^{2} (16 - 8x^2 + x^4) \, dx.\] Evaluate each term separately: \(\int 16 \, dx = 16x\), \(\int -8x^2 \, dx = -\frac{8}{3}x^3\), and \(\int x^4 \, dx = \frac{x^5}{5}\). Evaluate these at the bounds to find \[\frac{3}{64} \left[ 32 - \frac{64}{3} + \frac{64}{5} - (-32 + \frac{64}{3} - \frac{64}{5}) \right] = \frac{3}{64} \times \frac{256}{15} = \frac{16}{15}.\] Thus, \(\bar{y} = \frac{16}{15}.\)

Key Concepts

Integration

In mathematics, integration is a fundamental concept used to find areas, volumes, and other quantities related to functions. When calculating the center of mass for a region, integration helps us determine key properties of that region.
The integrals \[\bar{x} = \frac{1}{A} \int_{-2}^{2} x \cdot \text{Height}(x) \, dx\] and \[\bar{y} = \frac{1}{2A} \int_{-2}^{2} (\text{Height}(x))^2 \, dx\] are used to calculate the center of mass of the region between the parabola and the line. These integrals take into account not only the shape but the position of the region in question.
Here, the integration bounds \([-2, 2]\) are determined by where the line and the parabola intersect, highlighting the importance of integration in setting limits for finding accurate measurements.

Parabola

A parabola is a U-shaped curve that represents the graph of a quadratic function. In our problem, the parabola is given by the equation \( y = x^2 \). Understanding parabolas is crucial, as they frequently appear in various physics and math problems relating to trajectories and forces.
Here’s what you need to know about a parabola in this context:

• The vertex of the parabola \( y = x^2 \) is at the origin \((0,0)\).
• The parabola opens upwards, meaning its minimum point is the vertex, but it continues infinitely upward on both sides.
• In this problem, we are interested in the part of the parabola ranging from \(-2\) to \(2\), where it intersects with the horizontal line \( y = 4 \).

Understanding these intersections helps us to correctly set the boundaries for our integration process and determine the enclosed area.

Constant Density

In physics and mathematics, constant density refers to a situation where the density of a material is uniform across the entire region of interest. For a thin plate with constant density \( \delta \), the calculations become relatively straightforward, allowing us to use direct proportionality in finding the center of mass.
When a region has constant density, the mass distribution is even, and the center of mass calculations depend only on the geometric shape of the region. So, for finding the center of mass, we are essentially averaging the positions weighted by their area since the density \( \delta \) cancels out in this scenario.
This simplification is beneficial because it allows us to focus purely on geometric integrals, making the process more efficient and less prone to errors related to variable density functions.

Area Calculation

Calculating the area of a region between curves is essential for finding many physical quantities, such as the center of mass. For our specific problem, the area \( A \) of the region between the parabola \( y = x^2 \) and the line \( y = 4 \) is crucial.
We calculate this area using:\[A = \int_{-2}^{2} (4 - x^2) \, dx\]Here are the steps involved:

• Compute the antiderivative of each term in the integral: \( \int 4 \, dx = 4x \) and \( \int x^2 \, dx = \frac{x^3}{3} \).
• Apply the bounds to the antiderivative to find the area. This yields \( A = \left[4x - \frac{x^3}{3}\right]_{-2}^{2} = \frac{32}{3} \).

This calculation is foundational for determining other quantities, as the area plays a major role in defining the center of mass through averaging processes in integration.