Problem 11 - 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 bounded by the \(y\) -axis and the curve \(x=y-y^{3}\) , \(0 \leq y \leq 1\)

Step-by-Step Solution

Answer

\((\frac{8}{15}, \frac{44}{105})\)

Step 1: Understand the Problem

The problem requires us to find the center of mass of a region that is a thin plate with constant density. The region is bound by the y-axis, the curve given by \( x = y - y^3 \), and the limits \( 0 \leq y \leq 1 \).

Step 2: Identify the Formula for Center of Mass

The center of mass \((\bar{x}, \bar{y})\) of a thin plate with constant density is given by the formulas \( \bar{x} = \frac{1}{A} \int y \cdot x(y) \, dy \) and \( \bar{y} = \frac{1}{2A} \int x^2(y) \, dy \), where \( A \) is the area of the region.

Step 3: Calculate the Area A of the Region

The area \( A \) is computed as \( A = \int_{0}^{1} (y - y^3) \, dy \). First, integrate:\[ A = \int_{0}^{1} (y - y^3) \, dy = \left[\frac{y^2}{2} - \frac{y^4}{4} \right]_{0}^{1} = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}. \]

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

Use \( \bar{x} = \frac{1}{A} \int_{0}^{1} y \cdot (y-y^3) \, dy \):\[ \int_{0}^{1} y \cdot (y-y^3) \, dy = \int_{0}^{1} (y^2 - y^4) \, dy = \left[\frac{y^3}{3} - \frac{y^5}{5} \right]_{0}^{1} = \frac{1}{3} - \frac{1}{5} = \frac{2}{15}. \]\[ \bar{x} = 4 \times \frac{2}{15} = \frac{8}{15}. \]

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

Use \( \bar{y} = \frac{1}{2A} \int_{0}^{1} (y-y^3)^2 \, dy \):\[ (y-y^3)^2 = y^2 - 2y^4 + y^6. \]Integrate:\[ \int_{0}^{1} (y^2 - 2y^4 + y^6) \, dy = \left[\frac{y^3}{3} - \frac{2y^5}{5} + \frac{y^7}{7} \right]_{0}^{1} = \frac{1}{3} - \frac{2}{5} + \frac{1}{7} = \frac{22}{105}. \]\[ \bar{y} = 2 \times 4 \times \frac{22}{105} = \frac{44}{105}. \]

Step 6: Conclusion

The center of mass of the region is \( \left( \frac{8}{15}, \frac{44}{105} \right) \).

Key Concepts

Integration

Integration is a fundamental concept in calculus that involves finding the integral of a function. When calculating the center of mass, integration plays a crucial role, as it helps us determine specific values like area and weighted averages over a particular region.
It basically sums up infinitesimally small pieces to find a whole. In our exercise, we integrate to discover the area under the curve defined by the equation of our region and to find the moment about the axes. Integrals can be thought of as the opposite of derivatives. Where derivatives show the rate of change, integrals sum changes over intervals:

• Definite integrals calculate the accumulation from point a to b, giving a numerical value.
• They are used to calculate areas, volumes, and other metrics.

When dealing with curves like in our problem, definite integrals help us compute quantities like area and moments. They are key to determining where the center of mass lies along with the region's attributes. To evaluate the integrals effectively:

• Perform anti-differentiation: Find a function whose derivative matches the integrand.
• Apply the Fundamental Theorem of Calculus, which links antiderivatives with definite integrals.

Thus, integration is not only a tool but the foundation upon which we solve for physical concepts like mass and center of mass over continuous surfaces.

Area Calculation

Finding the area of a region bounded by curves is an application of integration. In our problem, we calculated the area under the curve described by the function: \( x = y - y^3 \) over the interval \( 0 \leq y \leq 1 \).The steps involved are straightforward:

• Integrate the function with respect to \( y \) across the specified limits. This gives us the total area \( A \) bounded by the curve and the y-axis.
• Ensure the function is integrable over the interval by checking for continuity and boundedness.

Calculating the area allows us to normalize other calculations like computing the center of mass. By determining the total area under the curve, we set a baseline for other physical properties.It is essential because:

• It provides a scalar measurement over a continuous region.
• It gives a scale for calculating centroids and centers of mass, particularly in uniform density cases.

In calculations like this, accuracy in integration is crucial to result in the correct physical interpretation and location of the center of mass for our thin plate.

Thin Plate with Constant Density

In the realm of physics and engineering, a thin plate is a solid with negligible thickness compared to its other dimensions. In this exercise, the thin plate exhibits a constant density \( \delta \), meaning the mass per unit area is uniform throughout the plate.
Understanding density allows us to simplify the calculation of physical properties.With constant density:

• The mass of each tiny sector of the plate depends solely on the area, making calculations easier.
• The physical property distribution, such as mass or charge, is uniform across the region.

In practice, solving for properties like the center of mass becomes more straightforward with constant density as it eliminates the need to account for variable distribution. This simplification leads to easy integration.Typically, the process involves:

• Calculating the total mass, which in our setup depends solely on the area and density.
• Using symmetry and constant density assumptions to reduce complex three-dimensional problems to two-dimensional calculations.

By maintaining a constant density assumption, we have ensured our calculations remain manageable and intuitive, aligning with the fundamental principles of calculus and physics.