Problem 21 - 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 parabolas \(y=2 x^{2}-4 x\) and \(y=2 x-x^{2}\)

Step-by-Step Solution

Answer

The center of mass is at \( \left( \frac{9}{7}, \frac{24}{7} \right) \).

Step 1: Identify the Functions and Intersection Points

The region is bounded by the parabolas: \( y = 2x^2 - 4x \) and \( y = 2x - x^2 \). To find the area of intersection, we first equate the functions to find intersection points: \[ 2x^2 - 4x = 2x - x^2 \].

Step 2: Solve for Intersection Points

Rearrange and simplify the equation: \[ 2x^2 - 4x = 2x - x^2 \] becomes: \[ 3x^2 - 6x = 0 \].Factor out the common term: \[ 3x(x - 2) = 0 \].Thus, the intersection points are \( x = 0 \) and \( x = 2 \).

Step 3: Set Up the Integral for Mass Center

The mass center or the centroid \((\bar{x}, \bar{y})\) is calculated using the formula:\[ \bar{x} = \frac{1}{A} \int_{a}^{b} x[f(x) - g(x)] \, dx \]\[ \bar{y} = \frac{1}{A} \int_{a}^{b} \frac{[f(x) - g(x)]^2}{2} \, dx \].Here, \( f(x) = 2x - x^2 \) and \( g(x) = 2x^2 - 4x \), with \( a = 0 \) and \( b = 2 \).

Step 4: Calculate the Area of the Region (A)

Calculate the area \( A \) by integrating the difference of the functions from \( x = 0 \) to \( x = 2 \): \[ A = \int_{0}^{2} [(2x - x^2) - (2x^2 - 4x)] \, dx \]\[ = \int_{0}^{2} (-x^2 + 6x) \, dx \].This simplifies to:\[ A = \left[ -\frac{x^3}{3} + 3x^2 \right]_{0}^{2} \]\[ = \left(-\frac{8}{3} + 12 \right) - (0) = \frac{28}{3} \].

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

Using the formula for \( \bar{x} \):\[ \bar{x} = \frac{1}{A} \int_{0}^{2} x[-x^2 + 6x] \, dx \]\[ = \frac{3}{28} \int_{0}^{2} (-x^3 + 6x^2) \, dx \]\[ = \frac{3}{28} \left[ -\frac{x^4}{4} + 2x^3 \right]_{0}^{2} \]\[ = \frac{3}{28} \left( -\frac{16}{4} + 16 \right) \]\[ = \frac{3}{28} \times 12 = \frac{36}{28} = \frac{9}{7} \].

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

Using the formula for \( \bar{y} \):\[ \bar{y} = \frac{1}{A} \int_{0}^{2} \frac{(6x - x^2)^2}{2} \, dx \]Simplify:\[ = \frac{3}{28} \int_{0}^{2} (6x - x^2)^2 \, dx \]Expanding:\[ = \frac{3}{28} \int_{0}^{2} (36x^2 - 12x^3 + x^4) \, dx \]Calculate:\[ \bar{y} = \frac{3}{28} \left[ 12x^3 - 3x^4 + \frac{x^5}{5} \right]_{0}^{2} \]\[ = \frac{3}{28} \times \left( 96 - 48 + \frac{32}{5} \right) \]\[ = \frac{3}{28} \times \frac{336-16}{5} = \frac{3}{28} \times \frac{320}{5} = \frac{96}{28} = \frac{24}{7} \].

Step 7: Compile the Center of Mass Coordinates

The center of mass for the region is located at:\( \left( \bar{x}, \bar{y} \right) = \left( \frac{9}{7}, \frac{24}{7} \right) \).

Key Concepts

Centroid Calculation

To find the centroid of a region, it's crucial to first understand the concept of a centroid itself. The centroid can be thought of as the "average" position of all the points in a shape. Imagine placing the shape on a pinpoint - the location of the pinpoint where it perfectly balances is the centroid. In the context of a two-dimensional shape, like the region between curves in our exercise, the centroid is a point

• otated as \((\bar{x}, \bar{y})\), representing average horizontal and vertical positions.
• ot dependent on the shape's weight, but rather its geometric arrangement.

To calculate these coordinates, we need to separate the region into smaller parts where we can treat each part as having weight concentrated at its center. These parts are typically infinitesimally small vertical strips. By integrating over the total continuous region, we end up with the whole centroid coordinates. This involves two formulas,

• the first one for \(\bar{x}\), which finds the horizontal position,
• and the second for \(\bar{y}\), which determines the vertical position.

Intersection of Curves

To find the area between two curves, we start by determining where they intersect. In our exercise, we need the regions between the curves described by:

• \( y = 2x - x^2 \) and
• \( y = 2x^2 - 4x \).

The intersection points are found by setting these equations equal to each other and solving for \(x\). By rearranging and factoring the resulting equation, we find

• \(x = 0\) and
• \(x = 2\).

These points tell us where one curve crosses over or under the other in a graph, defining the boundaries of the area of interest. Understanding where these intersections happen is crucial as they define the limits of integration for calculating area and centroid.

Definite Integrals

Definite integrals play a vital role in calculating areas and centroids when dealing with functions and curves. They help in determining the accumulated value of a function between two specific bounds. In the exercise, the bounds were established as the intersection points, \(x = 0\) and \(x = 2\). The integral calculates the total "accumulated" area or other quantity between these two points.For the area under a curve, the definite integral sums up all the infinitesimally small areas contained between a curve and the x-axis. For the exercise, the definite integral extracts this area to help find the centroid:

• We first find the area \(A\) of the region, using the difference of the functions being integrated over \([0,2]\).
• Then we use definite integrals again to calculate \(\bar{x}\) and \(\bar{y}\) values, translating into the final centroid position for the bounded region.

Area Between Curves

Finding the area between curves may seem straightforward, but it requires careful attention to detail. We begin by identifying the correct mathematical expressions for each curve. In our exercise, these expressions are:

• \( f(x) = 2x - x^2 \), and
• \( g(x) = 2x^2 - 4x \).

The area between them is then calculated by integrating the difference between the two functions over their intersection limits. It's like layering one function over the other and calculating the space between:

• Integrate the top function (\(f(x)\)) minus the bottom function (\(g(x)\)).
• The integration bounds are the x-coordinates of intersection points.
• This sum of small strips from these bounds gives the exact area between the two curves.

The requirement to integrate is driven by the fact that the area isn't formed by simple geometric shapes. Thus, calculus provides us with a precise area calculation in seemingly complex setups.