Problem 13 - Thomas Calculus in SI Units

Question

Find the center of mass of a thin plate covering the region between the \(x\) -axis and the curve \(y=2 / x^{2}, 1 \leq x \leq 2,\) if the plate's density at the point \((x, y)\) is \(\delta(x)=x^{2}\).

Step-by-Step Solution

Answer

The center of mass is \((1.5, 1)\).

Step 1: Define the mass and density function

The density function given is \( \delta(x) = x^2 \), which depends only on the \(x\)-coordinate. To find the center of mass, we first need to calculate the total mass by integrating this density over the given region.

Step 2: Determine the limits for integration

The region of the plate is given between \(x = 1\) and \(x = 2\) under the curve \(y = \frac{2}{x^2}\). This implies the limits of integration for \(x\) are from \(x = 1\) to \(x = 2\). The curve defines the upper bound of \(y\), thus, integrating over \(y\) would use limits from 0 to \(\frac{2}{x^2}\).

Step 3: Calculate the total mass

To find the total mass \(M\) of the plate, calculate the double integral \( M = \int_{1}^{2} \int_{0}^{\frac{2}{x^2}} x^2 \, dy \, dx \). First, integrate with respect to \(y\), then with respect to \(x\).

Step 4: Evaluate mass integral with respect to \(y\)

The integration with respect to \(y\) is straightforward as \(x^2\) is constant with respect to \(y\). Thus, \[ \int_{0}^{\frac{2}{x^2}} x^2 \, dy = x^2 \cdot y \Big|_{0}^{\frac{2}{x^2}} = x^2 \left(\frac{2}{x^2}\right) = 2. \]

Step 5: Integrate with respect to \(x\) for total mass

Now integrate the result \(M = \int_{1}^{2} 2 \, dx \). This results in \[ M = 2x \Big|_1^2 = 2(2) - 2(1) = 4 - 2 = 2. \] Thus, the total mass is 2 units.

Step 6: Calculate the x-coordinate of the center of mass

To find \( \bar{x} \), the x-coordinate of the center of mass, use \( \bar{x} = \frac{1}{M} \int_{1}^{2} \int_{0}^{\frac{2}{x^2}} x \cdot x^2 \, dy \, dx \). Simplifying, \( x \cdot x^2 = x^3 \).

Step 7: Evaluate \( \bar{x} \) integral with respect to \(y\)

Integrate \( \int_{0}^{\frac{2}{x^2}} x^3 \, dy = x^3 \cdot y \Big|_{0}^{\frac{2}{x^2}} = x^3 \left(\frac{2}{x^2}\right) = 2x. \)

Step 8: Integrate with respect to \(x\) for \( \bar{x} \)

Substitute the integration result and total mass into \( \bar{x} = \frac{1}{2} \int_{1}^{2} 2x \, dx \). Evaluate to get \[ \bar{x} = \int_{1}^{2} x \, dx = \frac{x^2}{2} \Big|_1^2 = \frac{4}{2} - \frac{1}{2} = 2 - 0.5 = 1.5. \] Thus, \( \bar{x} = 1.5. \)

Step 9: Calculate the y-coordinate of the center of mass

To find \( \bar{y} \), use \( \bar{y} = \frac{1}{M} \int_{1}^{2} \int_{0}^{\frac{2}{x^2}} \frac{2}{x^2} \cdot x^2 \, dy \, dx \), simplifying to \( \bar{y} = \frac{1}{2} \int_{1}^{2} 2 \, dy \, dx \).

Step 10: Evaluate \( \bar{y} \)

Since the integration with respect to \(y\) yields the same result, this simplifies to \( \bar{y} = \frac{1}{2} \cdot 2 \cdot (2-1) = 1 \). Therefore, \( \bar{y} = 1 \).

Step 11: Conclusion and final expression for center of mass

The center of mass \((\bar{x}, \bar{y})\) is calculated as \( (1.5, 1) \).

Key Concepts

Density Function

In this exercise, the density function plays a key role in understanding how mass is distributed across the plate. The density function, often denoted by \( \delta(x) \), specifies how dense or heavy the material is at any given point \((x, y)\). Here, \( \delta(x) = x^2 \), meaning density varies with the square of the \( x \)-coordinate. This function informs us that as \( x \) increases, the density increases rapidly. Understanding the density function allows us to calculate the total mass by integrating it over the region where the object is defined.

Double Integral

The double integral is a powerful tool used for calculating quantities over a two-dimensional area, such as mass, area, or volume. In this problem, it's used to compute the total mass of the plate. The double integral \( \int \int \) signifies integrating over two different variables.

In this case, the order of integration is first with respect to \( y \) and then with respect to \( x \). This step-by-step integration helps in simplifying complex calculations. It is essential to remember that the variable being integrated should be considered constant for the other integration, making calculations manageable and less error-prone.

Mass Calculation

The calculation of mass in this problem involves the integration of the density function over the specified region. To find the total mass \( M \), the formula used for this double integral is:

• Integrate \( x^2 \) over \( y \), the section bound by 0 and \( \frac{2}{x^2} \)
• Resultant expression from the first integral simplifies to 2
• Integrate this result over \( x \) from 1 to 2

Upon integration, it is crucial to correctly perform substitutions and solve each step to ensure accuracy. The calculation results in a mass of 2 units, illustrating the total weight of the plate given the density function provided.

Integration Limits

Integration limits specify the boundaries over which integration is performed. They are vital because they dictate the region on which the function is evaluated.

• In this problem, the limits for \(x\) are between 1 and 2, as the plate stretches between these values on the \(x\)-axis
• The limits for \(y\) range from 0 to \(\frac{2}{x^2}\), dictated by the curve under which the plate lies, \(y = \frac{2}{x^2}\)

Choosing the correct integration limits is crucial as they directly affect the results of integrations. They also align with the physical constraints or dimensions of the problem's context, helping in acquiring meaningful, realistic outcomes from theoretical calculations.