Problem 47 - Calculus Single Variable

Question

Find the center of mass of the given region \(\mathcal{R},\) assuming that it has uniform unit mass density. \(\mathcal{R}\) is the region bounded above by \(y=4-x^{2}\) for \(-2 \leq x \leq 1,\) below by \(y=3 x\) for \(0 \leq x,\) and below by the \(x\) axis for \(x<0\).

Step-by-Step Solution

Answer

The center of mass is at \(\left(-\frac{1}{3}, \frac{77}{39}\right)\).

Step 1: Identify the boundaries

Identify the boundaries of region \(\mathcal{R}\). For \(x < 0\), \(\mathcal{R}\) is bounded above by \(y = 4 - x^2\) and below by the \(x\)-axis. For \(x \geq 0\), \(\mathcal{R}\) is bounded above by \(y = 4 - x^2\) and below by \(y = 3x\).

Step 2: Set up integrals for area calculation

First, find the area of \(\mathcal{R}\) by setting up integrals. The area for \(x < 0\) is given by \(\int_{-2}^{0} (4 - x^2)\, dx\). The area for \(x \geq 0\) is \(\int_{0}^{1} ((4 - x^2) - 3x)\, dx\).

Step 3: Evaluate the area integrals

Calculate each integral:\[A_1 = \int_{-2}^{0} (4 - x^2)\, dx = [4x - \frac{x^3}{3}]_{-2}^{0} = \left[(0) -\left( -8 + \frac{8}{3}\right)\right] = \frac{16}{3}\]\[A_2 = \int_{0}^{1} (4 - x^2 - 3x)\, dx = [4x - \frac{x^3}{3} - \frac{3x^2}{2}]_{0}^{1} = \left[4 - \frac{1}{3} - \frac{3}{2}\right] = \frac{7}{6}\]The total area \(A = A_1 + A_2 = \frac{16}{3} + \frac{7}{6} = \frac{39}{6}\).

Step 4: Find the x-coordinate of the center of mass

The \(x\)-coordinate \(\bar{x}\) is given by \(\bar{x} = \frac{1}{A} \left( \int_{-2}^{0} x(4 - x^2)\, dx + \int_{0}^{1} x(4 - x^2 - 3x)\, dx \right)\). Evaluate:\[\int_{-2}^{0} x(4 - x^2)\, dx = \int_{-2}^{0} (4x - x^3)\, dx = [2x^2 - \frac{x^4}{4}]_{-2}^{0} = [0 - (-8 + 4)] = -4\]\[\int_{0}^{1} x(4 - x^2 - 3x)\, dx = \int_{0}^{1} (4x - x^3 - 3x^2)\, dx = [2x^2 - \frac{x^4}{4} - x^3]_{0}^{1} = [2 - \frac{1}{4} - 1] = \frac{3}{4}\]Therefore, \(\bar{x} = \frac{1}{\frac{39}{6}}(-4 + \frac{3}{4}) = -\frac{13}{39} = -\frac{1}{3}\).

Step 5: Find the y-coordinate of the center of mass

The \(y\)-coordinate \(\bar{y}\) is given by \(\bar{y} = \frac{1}{A} \left( \int_{-2}^{0} \frac{1}{2}(4 - x^2)^2\, dx + \int_{0}^{1} \frac{1}{2}((4 - x^2)^2 - (3x)^2)\, dx \right)\).\[\int_{-2}^{0} \frac{1}{2}(4 - x^2)^2\, dx= \int_{-2}^{0} \frac{1}{2}(16 - 8x^2 + x^4)\, dx = [8x - \frac{8}{3}x^3 + \frac{x^5}{10}]_{-2}^{0} = -\frac{112}{15}\]\[\int_{0}^{1} \frac{1}{2}((4 - x^2)^2 - (3x)^2)\, dx = \int_{0}^{1} (8 - 5x^2 - 9x^2)\, dx = \int_{0}^{1} (8 - 14x^2)\, dx = [7x - \frac{14}{3}x^3]_{0}^{1} = \frac{7}{3}\]Therefore, \(\bar{y} = \frac{1}{\frac{39}{6}}\left(-\frac{112}{15} + \frac{7}{3}\right) = \frac{154}{78} = \frac{77}{39}\approx 1.974\).

Key Concepts

Integrals

Integrals are a fundamental concept in calculus. They help us find areas under curves, volumes, central points, and more. In this context, we use integrals to calculate the area of a region and its center of mass. To do this, we need to set up definite integrals, which allow us to compute the total area or mass over a given interval. These integrals are crucial as they let us express the accumulation of quantities over a continuous range.

• Setting up integrals: To find the area, we establish integrals based on the boundaries. For this region, we compute two separate integrals because it is bounded differently for values of x less than zero and those equal to or greater than zero.
• Evaluating integrals: Calculating each integral involves applying fundamental arithmetic operations and known formulas for antiderivatives. Integrals such as \( \int x^n \, dx \) and \( \int a \, dx \) (where a is a constant) are used extensively.

Integrals help locate the region's center of mass by computing the x and y components separately. Each has its own setup based on the region's equations, ensuring accurate results.

Bounded Region

A bounded region in mathematics refers to an area enclosed within specified boundaries. In our scenario, the bounded region is defined by three different curves or lines across specific intervals of x-values. Understanding which curve provides the top and bottom bounds is essential to solving such problems correctly and setting up your integrals.

• For this given region \( \mathcal{R} \), notice how it's divided: one part for negative x-values and another for positive ones. For \( x < 0 \), it's between the curve \( y = 4 - x^2 \) and the x-axis.
• For \( x \geq 0 \), the curve \( y = 3x \) acts as the lower bound while the upper remains \( y = 4 - x^2 \).

Visualizing this helps you understand how the region looks and how to integrate over these segments. Being careful with the limits of integration is crucial to avoid errors in calculations.

Mass Density

Mass density is a measure of how much mass is contained in a particular space. In problems involving center of mass, especially those assuming uniform mass density, it simplifies calculations. With a uniform mass density, the mass per unit area is constant across the entire region, which enables easy computations.

• Since this problem assumes uniform density, if a material filled this region entirely, density won’t change regardless of where you sample.
• Consider it as a simplification that lets you directly relate to areas when calculating properties like the center of mass.

This assumption is why our calculation for center of mass mainly involves understanding the region's shape and setting up correct integrals, rather than dealing with varying density values across different points.

Area Calculation

Calculating area is a critical step when finding the center of mass of a given region. In this particular instance, there are two areas to compute because the boundaries change depending on x-values. Each part is tackled by establishing a unique integral, then evaluating it over the respective limits given by the bounding curves.

• For \( x < 0 \), the integral \( \int_{-2}^{0} (4 - x^2) \, dx \) calculates the area above the x-axis up to \( y = 4 - x^2 \).
• For \( x \geq 0 \), where the lower bound becomes \( y = 3x \), the integral is \( \int_{0}^{1} ((4 - x^2) - 3x) \, dx \). This accounts for the area between the two curves.
• Total area is the sum of these integrals’ results, necessary for finding the center of mass coordinates.

Once areas are calculated, they are used not only to find positions like \( \bar{x} \) and \( \bar{y} \) but also to better understand the extent of the region.