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Magazine Chapter 7: Problem 47
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 \(x<0\).
Short Answer
Expert verified
The center of mass is at \(\left(-\frac{1}{3}, \frac{77}{39}\right)\).
Step by step solution
01
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 \(y = 3x\).
02
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 \(\int_{0}^{1} ((4 - x^2) - 3x)\, dx\).
03
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 \(A = A_1 + A_2 = \frac{16}{3} + \frac{7}{6} = \frac{39}{6}\).
04
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, \(\bar{x} = \frac{1}{\frac{39}{6}}(-4 + \frac{3}{4}) = -\frac{13}{39} = -\frac{1}{3}\).
05
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, \(\bar{y} = \frac{1}{\frac{39}{6}}\left(-\frac{112}{15} + \frac{7}{3}\right) = \frac{154}{78} = \frac{77}{39}\approx 1.974\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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.
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 \( y = 4 - x^2 \).
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鈥檛 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.
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鈥� results, necessary for finding the center of mass coordinates.
