Problem 9 - 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 parabola \(y=x-x^{2}\) and the line \(y=-x\)

Step-by-Step Solution

Answer

The center of mass is at \((1, \frac{11}{40})\).

Step 1: Determine the Intersection Points

To determine the region bounded by the curves, we need to find the points where the parabola intersects the line. We'll set their equations equal to each other:\[ x - x^2 = -x \]This simplifies to:\[ x - x^2 = -x \rightarrow x^2 - 2x = 0 \rightarrow x(x-2) = 0 \]This gives us the intersection points at \(x = 0\) and \(x = 2\).

Step 2: Calculate the Area of the Bounded Region

The area (A) of the region is given by the integral of the top function minus the bottom function from their intersection points:\[ A = \int_{0}^{2} ((x - x^2) - (-x)) \, dx = \int_{0}^{2} (2x - x^2) \, dx \]Evaluate this integral:\[ A = \left[ x^2 - \frac{x^3}{3} \right]_{0}^{2} = \left[ 4 - \frac{8}{3} \right] - (0 - 0) = \frac{4}{3} \]

Step 3: Set Up the Integral for Center of Mass Coordinates

Now we set up the integrals for the \(x\) and \(y\) coordinates of the center of mass. The \(x\)-coordinate is:\[ \bar{x} = \frac{1}{A} \int_{0}^{2} x((x-x^2) - (-x)) \, dx = \frac{3}{4} \int_{0}^{2} x(2x - x^2) \, dx \]And the \(y\)-coordinate is:\[ \bar{y} = \frac{1}{A} \int_{0}^{2} \frac{(x-x^2) + (-x)}{2} ((x-x^2) - (-x)) \, dx \]

Step 4: Evaluate the Integral for \( \bar{x} \)

Evaluate the integral for \(\bar{x}\):\[ \int_{0}^{2} (2x^2 - x^3) \, dx = \left[ \frac{2x^3}{3} - \frac{x^4}{4} \right]_{0}^{2} = \left[ \frac{16}{3} - 4 \right] - 0 = \frac{4}{3} \]So, \(\bar{x} = \frac{3}{4} \times \frac{4}{3} = 1\).

Step 5: Evaluate the Integral for \(\bar{y}\)

First, simplify and evaluate the expression for \(\bar{y}\):\[ \int_{0}^{2} \frac{x - x^2 - x}{2} (2x - x^2) \, dx = \int_{0}^{2} (x - x^2) (2x - x^2) \, dx = \int_{0}^{2} (2x^2 - 2x^3 - x^3 + x^4) \, dx = \int_{0}^{2} (2x^2 - 3x^3 + x^4) \, dx \]Integrate term by term and evaluate over the interval:\[ \left[ \frac{2x^3}{3} - \frac{3x^4}{4} + \frac{x^5}{5} \right]_{0}^{2} = \left[ \frac{16}{3} - 12 + \frac{32}{5} \right] = \frac{11}{30} \]Thus, \(\bar{y} = \frac{3}{4} \times \frac{11}{30} = \frac{11}{40}\).

Step 6: Determine the Center of Mass

The center of mass of the region is at the point \( (\bar{x}, \bar{y}) \). From our calculations, \( \bar{x} = 1 \) and \( \bar{y} = \frac{11}{40} \).

Key Concepts

Intersection Points

To find where two curves intersect, we solve for the points where their equations are equal. In this problem, we are dealing with a parabola described by the equation \( y = x - x^2 \) and a straight line \( y = -x \). To determine where these two graphs meet, we set the equations equal:

• \( x - x^2 = -x \) simplifies to \( x^2 - 2x = 0 \).
• This factors to \( x(x-2) = 0 \).
• The solutions to this equation are \( x = 0 \) and \( x = 2 \).

These \( x \) values are the points of intersection along the x-axis. Then, by substituting back into either original equation, we can find the corresponding \( y \)-values. This means our points of intersection, or where the curves meet, are at (0,0) and (2,-2). These points will form the boundaries for our calculations of area and eventually our center of mass.

Integral Calculus

Integral calculus helps us find the area between curves and, in this case, is used to determine two key things: the area of the enclosed region and coordinates for the center of mass.
To find the area of the bounded region between our intersecting curves, we use the definite integral of the difference between the top function and the bottom function:

• The expression for the area \( A \) becomes: \( A = \int_{0}^{2} ((x - x^2) - (-x)) \, dx = \int_{0}^{2} (2x - x^2) \, dx \).
• Solving this integral gives us \( A = \frac{4}{3} \).

Next, we need to find the center of mass (\( \bar{x}, \bar{y} \)) of the region. Both coordinates require calculating integrals:

• For \( \bar{x} \), we compute the integral \( \frac{1}{A} \int_{0}^{2} x(2x - x^2) \, dx \).
• As calculated, \( \bar{x} = 1 \).
• For \( \bar{y} \), we use a more complex expression \( \bar{y} = \frac{1}{A} \int_{0}^{2} \frac{(x-x^2) + (-x)}{2} (2x - x^2) \, dx \).
• From solving, we obtained \( \bar{y} = \frac{11}{40} \).

Integral calculus, therefore, not only delineates the size of our area but also the balance point for masses across that area.

Parabola

A parabola is a symmetrical, curved shape which can open upwards, downwards, or sideways depending on the equation. The specific parabola in this exercise is given as \( y = x - x^2 \), signifying it is a concave-down parabola (it opens downward).

• In its standard form, a parabola is represented as \( y = ax^2 + bx + c \). Here, \( a = -1, b = 1, \) and \( c = 0 \).
• The vertex of the parabola \( y = x - x^2 \) is found using \( -\frac{b}{2a} \), here \( x = \frac{1}{2} \).
• Substituting \( x = \frac{1}{2} \) into the original parabola equation gives the \( y \)-coordinate of the vertex, so \( y = \frac{1}{4} \).
• Thus, the vertex is at \( \left( \frac{1}{2}, \frac{1}{4} \right) \).

The vertex represents the highest point on this downward-opening parabola. Understanding the shape and characteristics of this parabola is crucial, as it helps define the boundaries for integration and identifies significant points like the vertex needed for solving such problems in calculus.