Problem 5 - 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 \(y\) -axis and the curve \(x=y-y^{3}\), \(0 \leq y \leq 1\)

Step-by-Step Solution

Answer

The center of mass is at \(\left( \frac{32}{105}, \frac{8}{15} \right)\).

Step 1: Understand the Problem and Set Up Coordinates

We need to find the center of mass of a region on the xy-plane bounded by the y-axis (thus x = 0) and the curve given by the equation \( x = y - y^3 \). The bounds for \( y \) are from \( 0 \) to \( 1 \). This region is a vertical strip along the y-axis.

Step 2: Formula for the Center of Mass

The center of mass for a thin plate with constant density \( \delta \) can be calculated using the formulas for the coordinates:\[\bar{x} = \frac{1}{M} \int_{a}^{b} x \cdot \delta \cdot dx \, dy,\]\[\bar{y} = \frac{1}{M} \int_{a}^{b} y \cdot \delta \cdot dx \, dy.\]Here, \( M \) is the total mass, defined as \( M = \delta \int_{a}^{b} dx \, dy \) since \( x(y) = y - y^3 \), the bounds of y are 0 to 1, and the density \( \delta \) is constant and cancels out in the ratio.

Step 3: Calculate Total Mass M

The mass \( M \) is obtained by integrating the area of the region:\[M = \delta \int_{0}^{1} (y - y^3) \, dy.\]Computing this integral:\[M = \delta \left( \frac{y^2}{2} - \frac{y^4}{4} \right) \Bigg|_0^1 = \delta \left( \frac{1}{2} - \frac{1}{4} \right) = \delta \frac{1}{4}.\]

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

Find \(\bar{x}\) using the formula:\[\bar{x} = \frac{1}{M} \int_{0}^{1} (y - y^3)^2 \, dy.\]Calculate the integral:\[\int_{0}^{1} (y - y^3)^2 \, dy = \int_{0}^{1} (y^2 - 2y^4 + y^6) \, dy = \left( \frac{y^3}{3} - \frac{2y^5}{5} + \frac{y^7}{7} \right) \Bigg|_0^1.\]This evaluates to:\[\left( \frac{1}{3} - \frac{2}{5} + \frac{1}{7} \right).\]

Step 5: Evaluate \(\bar{x}\) Using \(M\) and Integral Results

The value of the evaluated integral from Step 4 is \(\frac{1}{3} - \frac{2}{5} + \frac{1}{7} = \frac{35}{105} - \frac{42}{105} + \frac{15}{105} = \frac{8}{105}\).Now, calculate \(\bar{x}\):\[\bar{x} = \frac{8/105}{\delta/4} = \frac{32}{105}.\]

Step 6: Calculate \(\bar{y}\)

To find \(\bar{y}\), use the equation:\[\bar{y} = \frac{1}{M} \int_{0}^{1} y(y - y^3) \, dy = \frac{1}{M} \int_{0}^{1} (y^2 - y^4) \, dy.\]Integrate:\[\int_{0}^{1} (y^2 - y^4) \, dy = \left( \frac{y^3}{3} - \frac{y^5}{5} \right) \Bigg|_0^1 = \frac{1}{3} - \frac{1}{5} = \frac{5}{15} - \frac{3}{15} \]This simplifies to \(\frac{2}{15}\).

Step 7: Evaluate \(\bar{y}\) Using \(M\)

Substitute \(M\) and the evaluated integral into the formula for \(\bar{y}\):\[\bar{y} = \frac{2/15}{\delta/4} = \frac{8}{15}.\]

Step 8: Conclusion: Combine Results to Get the Center of Mass

The center of mass \((\bar{x}, \bar{y})\) of the plate is at the coordinates \(\left( \frac{32}{105}, \frac{8}{15} \right)\).

Key Concepts

Integral Calculus

Integral calculus is an essential mathematical tool for solving problems involving quantities that accumulate over a continuous range. In this exercise, integral calculus helps us determine the center of mass of a thin plate. The center of mass is like the balancing point of the plate.

The critical step to solve this exercise is understanding the integration of continuous functions over defined limits. The formulas for the x and y coordinates of the center of mass integrate the function representing the region. This is achieved through the definite integral across the bounds defined by the curve.

Here's a quick rundown of what integration does:

• It accumulates the total mass or area over the region.
• It helps find the weighted average position (center of mass) by integrating position times density (if present).
• It simplifies the calculation by removing the need to label individual points and sums small contributions into a continuous whole.

In simpler terms, integral calculus takes the summation concept to a continuous level, crucial in physics, engineering, and beyond. It allows us to solve problems involving areas, volumes, and other quantities derived from summing infinitely small parts.

Thin Plate

A thin plate, in mathematics and physics, refers to a two-dimensional object whose thickness is negligible compared to its other dimensions. For the purposes of calculating the center of mass, we treat the plate as a uniform surface such that its mass distribution is even across its area.

When dealing with a thin plate:

• We typically assume that the thickness doesn't affect calculations related to area or mass distribution.
• We only need to consider two-dimensional properties, like area, rather than volume.
• Mathematically, the plate is examined as a region on a plane, bounded by certain lines or curves, as in this exercise by the curve and the y-axis.

The focus remains on the shape defined in the plane—the relationship between x and y coordinates—rather than any third dimension. This greatly reduces the complexity of calculations and is precisely why integral calculus is applied to derive quantities like the center of mass in these scenarios.

Constant Density

Constant density is a situation where the mass distribution per unit area or volume remains uniform throughout the object. In the context of this problem, it simplifies calculations significantly because:

• We can factor out the density when calculating mass and subsequently solving for the center of mass.
• The density, denoted by \( \delta \), cancels out from the calculations of \( \bar{x} \) and \( \bar{y} \) when computing ratios.
• It allows us to focus solely on the shape and dimensions (area) of the object to understand its mass distribution behavior.

Imagine the thin plate as painted with a uniform coat of paint. The uniformity means calculations depend on geometry alone without needing to account for any variation in density. This elegant simplicity is why constant density plays a crucial role in finding parameters like the center of mass effortlessly.