Problem 14 - Thomas Calculus

Question

In Exercises \(13-24\) , find the center of mass of a thin plate of constant density \(\delta\) covering the given region. The region bounded by the parabola \(y=25-x^{2}\) and the \(x\) -axis

Step-by-Step Solution

Answer

The center of mass is at \((0, \frac{50}{3})\)."

Step 1: Identify the Bounded Region

The problem describes a region bounded by the parabola \(y = 25 - x^2\) and the \(x\)-axis. This means the region is below the parabola and above the \(x\)-axis.

Step 2: Find the Limits of Integration

Determine where the parabola intersects the \(x\)-axis by setting \(y = 0\). Solve \(25 - x^2 = 0\) to get \(x = \pm 5\). Thus, the region is bounded from \(x = -5\) to \(x = 5\).

Step 3: Set Up the Mass Integral

The mass \(M\) of the plate is calculated using \(M = \int_{-5}^{5} \delta \cdot (25 - x^2)\, dx\). Here, \(\delta\) is the constant density of the plate.

Step 4: Calculate the Mass

Evaluate the integral: \[ M = \delta \left( \int_{-5}^{5} 25\, dx - \int_{-5}^{5} x^2\, dx \right) \]. This simplifies to \[ M = \delta \left( 25[x]_{-5}^{5} - \frac{x^3}{3}|_{-5}^{5} \right) \].Computing these, we get \(M = \delta \times 500\).

Step 5: Find the Center of Mass on the x-axis (\(\bar{x}\))

Calculate \(\bar{x}\) using: \[ \bar{x} = \frac{1}{M} \int_{-5}^{5} x \cdot \delta \cdot (25-x^2)\, dx \]. Since the function is symmetric around the y-axis, \(\bar{x} = 0\).

Step 6: Find the Center of Mass on the y-axis (\(\bar{y}\))

Calculate \(\bar{y}\) using: \[ \bar{y} = \frac{1}{2M} \int_{-5}^{5} \delta \cdot (25-x^2)^2\, dx \]. Evaluate the integral: \[ \bar{y} = \frac{1}{2 \times 500 \times \delta} \left( \frac{25^3}{3}x - 50x^3 + \frac{x^5}{5} \right)_{-5}^{5} \].After integrating and computing, we get \(\bar{y} = \frac{50}{3}\).

Step 7: State the Center of Mass

The center of mass of the region is located at \((0, \frac{50}{3})\).

Key Concepts

Understanding the Parabola

A parabola is a U-shaped curve that is symmetric and is described mathematically by a quadratic equation. In this exercise, the parabola we are interested in is given by the equation \(y = 25 - x^2\). Here's what these elements mean:

• \(x^2\) is the basic parabola shape element, which determines how wide or narrow the parabola is.
• The \(-x^2\) indicates that the parabola opens downward, forming an upside-down U-shape.
• \(+25\) shifts the entire parabola upward by 25 units on the y-axis, positioning the vertex at (0, 25).

The region of interest in this problem is the area between the parabola and the x-axis. This area forms a sort of "bowl" shape lying between the points where the parabola intersects the x-axis (at \(x = -5\) and \(x = 5\)). This shape is symmetrical around the y-axis, which will simplify our calculations later on.

Calculating the Mass Integral

To find the center of mass, one foundational concept is the mass integral. This relates to how mass is distributed over the shape. We're asked to compute the total mass \(M\) of the plate covering the described region, assuming a constant density \(\delta\). The formula used is:\[M = \int_{-5}^{5} \delta \cdot (25 - x^2)\, dx\]

• This integral sums up tiny mass elements from \(x = -5\) to \(x = 5\).
• Since density \(\delta\) is constant, it can be factored out of the integral.

Breaking it down further, we separate the integral into two parts—one for the constant portion \(25\) and the other for \(x^2\). Evaluating each part separately and combining gives us a way to find the overall mass \(M\), which in this case is \(500 \cdot \delta\). This value reflects the total impact of the assumed density across the whole parabolic region.

Role of Density in Finding Center of Mass

Density is a measure of mass per unit area and plays a critical role in determining the center of mass. In problems like this, we often assume a uniform or constant density, which simplifies calculations significantly. Here’s why it's important:

• Uniform density \(\delta\) means every part of the region contributes equally to the total mass, simplifying measurements.
• It allows us to factor \(\delta\) out of our integrals when calculating mass or coordinates of the center of mass, reducing the complexity of these equations.

When finding the centers of mass on both the x-axis \((\bar{x})\) and y-axis \((\bar{y})\), the constant density helps assure symmetry, especially around the y-axis for this parabola. This symmetry gives us \(\bar{x} = 0\), simplifying our task to focus mainly on calculating \(\bar{y}\). Ultimately, the density ensures that the calculated mass and subsequent position calculations are consistent across the entire region.