Problem 30 - Calculus Single Variable

Question

Find the center of \(\operatorname{mass}(\bar{x}, \bar{y})\) of the given region \(\mathcal{R},\) assuming that it has uniform unit mass density. \(\mathcal{R}\) is the region bounded above by \(y=2 x\) and below by \(y=x^{2}\)

Step-by-Step Solution

Answer

The center of mass is at \((1, \frac{3}{4})\)."

Step 1: Identify and sketch the region

To solve this, first understand the region \( \mathcal{R} \) is bounded by the curves \( y = 2x \) (a straight line) and \( y = x^2 \) (a parabola). These curves intersect at the points where \( 2x = x^2 \). Solving the equation \( x^2 - 2x = 0 \) gives \( x(x-2) = 0 \), leading to intersection points \( x = 0 \) and \( x = 2 \). The region \( \mathcal{R} \) is between these intersection points.

Step 2: Set up integrals for the mass (M)

Since the region \( \mathcal{R} \) has uniform unit mass density, the mass \( M \) is simply the area of the region. The area is found by integrating the difference between the upper and lower curves:\[ M = \int_{0}^{2} [(2x) - (x^2)] \, dx \].

Step 3: Calculate the mass (M)

Integrate the function \( (2x - x^2) \) over \([0, 2]\):\[ M = \int_{0}^{2} (2x - x^2) \, dx = \left[ x^2 - \frac{x^3}{3} \right]_{0}^{2} = \left[ 4 - \frac{8}{3} \right] - (0) = \frac{4}{3}.\]

Step 4: Calculate x-coordinate of the center of mass (\(\bar{x}\))

The formula for \( \bar{x} \) is:\[ \bar{x} = \frac{1}{M} \int_{0}^{2} x \, [(2x - x^2)] \, dx. \]Calculate the integral:\[ \int_{0}^{2} [x(2x - x^2)] \, dx = \int_{0}^{2} (2x^2 - x^3) \, dx. \]

Step 5: Integrate to find \(\int_{0}^{2} (2x^2 - x^3) \, dx \)

Execute the integral:\[ \int_{0}^{2} (2x^2 - x^3) \, dx = \left[ \frac{2}{3}x^3 - \frac{x^4}{4} \right]_{0}^{2} = \left[ \frac{16}{3} - 4 \right] - (0) = \frac{4}{3}.\]

Step 6: Solve for \(\bar{x}\)

Substitute back to find \( \bar{x} \):\[ \bar{x} = \frac{1}{\frac{4}{3}} \cdot \frac{4}{3} = 1. \]

Step 7: Calculate y-coordinate of the center of mass (\(\bar{y}\))

The formula for \( \bar{y} \) is:\[ \bar{y} = \frac{1}{M} \int_{0}^{2} \frac{(2x + x^2)}{2} [(2x - x^2)] \, dx. \]Simplify to:\[ \int_{0}^{2} (2x - x^2)^2 \, dx. \]

Step 8: Find and compute \(\int_{0}^{2} (2x - x^2)^2 \, dx \)

Expand and integrate:\[ (2x - x^2)^2 = 4x^2 - 4x^3 + x^4. \]\[ \int_{0}^{2} (4x^2 - 4x^3 + x^4) \, dx = \left[ \frac{4}{3}x^3 - x^4 + \frac{x^5}{5} \right]_{0}^{2} = \left[ \frac{32}{3} - 16 + \frac{32}{5} \right]. \]

Step 9: Evaluate the integral result for \( \bar{y} \)

Calculate the expression:\[ \frac{32}{3} - 16 + \frac{32}{5} = \frac{160}{15} - \frac{240}{15} + \frac{96}{15} = \frac{16}{15}. \]

Step 10: Solve for \(\bar{y}\)

Substitute back to find \( \bar{y} \):\[ \bar{y} = \frac{1}{M} \cdot \frac{16}{15} = \frac{3}{4}. \]

Step 11: Conclusion

The center of mass \((\bar{x}, \bar{y})\) of the region \( \mathcal{R} \) is at \((1, \frac{3}{4})\).

Key Concepts

Mass Density and Center of Mass

Mass density is a measure of how much mass is spread over a particular area or volume. In cases where the mass density is uniform and equals one unit, like in this problem, the center of mass can be directly found by considering the geometric center based on the mass distribution.

For any area or region, especially in two dimensions, the center of mass is like the 'balance point' of the shape. Here, the center of mass for region \( \mathcal{R} \) defined by the curves \( y = 2x \) and \( y = x^2 \) is explored. By leveraging the concept of a uniform mass density, calculations for the center are directly related to the shape's geometry and position.

To find the center of mass, one typically follows two steps:

• Calculate the total mass, which is equivalent to the area for a uniform density.
• Find the positions \( \bar{x} \) and \( \bar{y} \), the weighted averages of the region's x and y coordinates.

In this exercise, these calculations are facilitated by integration, making it crucial to understand how integration helps in determining mass distributions.

Integration and its Role

Integration plays a crucial role in calculating areas under curves and, ultimately, the center of mass for a region. The process of integration allows you to combine an infinite number of infinitesimal slices of a function to find the total area, which is particularly useful for finding the total mass when the density is uniform.

In this problem, you use definite integrals to find both the area of region \( \mathcal{R} \) and the weighted averages for the x and y coordinates:

• The mass \( M \) as \( \int_{0}^{2} (2x - x^2) \, dx \) gives the total area between the parabola \( y = x^2 \) and the line \( y = 2x \).
• \( \bar{x} \) is found through \( \int_{0}^{2} x (2x - x^2) \, dx \), which measures how mass spreads along the x-axis.
• \( \bar{y} \) involves \( \int_{0}^{2} \frac{(2x + x^2)}{2}(2x - x^2) \, dx \), focusing on the y-axis distribution.

Each integral contributes to understanding the region's geometry, exploring how mass is laid out within \( \mathcal{R} \). Calculating these integrals accurately determines the region's center of mass.

Area Under Curve

The area under a curve is fundamental when discussing mass density and center of mass. It helps in calculating how much space or 'mass' a given shape occupies, especially when bounded by functions like straight lines or curves.

By focusing on areas between curves, like in this exercise, you can find how two-dimensional spaces contribute to overall mass. Calculating the area under \( y = 2x \) and \( y = x^2 \) involves:

• Identifying endpoints; here, \( x=0 \) and \( x=2 \), where the parabola intersects the line.
• Subtracting between the curves to get the function \( 2x - x^2 \), representing the region \( \mathcal{R} \).
• Integrally processing this function to find the actual space occupied, leading to \( M \), the region's mass.

Understanding these spaces between curves is crucial for accurately calculating centers of mass, especially when dealing with more complex shapes. The exercise transforms these abstract concepts into practical applications, showing the importance of calculus in real-world geometric problems.