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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo
Capitolo 7, Problema 7.AAE.19

19. Center of mass Find the center of mass of a thin plate of constant density covering the region in the first and fourth quadrants enclosed by the curves y=1/(1+x²) and y=-1/(1+x²) and by the lines x=0 and x=1.

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Identify the region described: it lies between the curves \(y=\frac{1}{1+x^{2}}\) and \(y=-\frac{1}{1+x^{2}}\), bounded by the vertical lines \(x=0\) and \(x=1\). This region is symmetric about the x-axis and lies in the first and fourth quadrants.
Since the plate has constant density, denote the density by \(\rho\). The center of mass coordinates \((\bar{x}, \bar{y})\) are given by the formulas: \(\displaystyle \bar{x} = \frac{1}{M} \int x \, dm\) and \(\displaystyle \bar{y} = \frac{1}{M} \int y \, dm\), where \(M\) is the total mass of the plate.
Express the mass element \(dm\) in terms of \(dx\) and \(dy\). Because the plate is thin and has constant density, \(dm = \rho \, dA\), where \(dA\) is the differential area element. Here, \(dA\) can be expressed as the vertical slice between the two curves: \(\displaystyle dA = \left( \frac{1}{1+x^{2}} - \left(-\frac{1}{1+x^{2}}\right) \right) dx = \frac{2}{1+x^{2}} \, dx\).
Calculate the total mass \(M\) by integrating the area times density over \(x\) from 0 to 1: \(\displaystyle M = \rho \int_{0}^{1} \frac{2}{1+x^{2}} \, dx\).
Find the moments needed for the center of mass: - For \(\bar{x}\), compute \(\displaystyle \int x \, dm = \rho \int_{0}^{1} x \cdot \frac{2}{1+x^{2}} \, dx\). - For \(\bar{y}\), note the symmetry of the region about the x-axis: the positive and negative parts cancel out, so \(\bar{y} = 0\). Finally, use the formulas \(\bar{x} = \frac{1}{M} \int x \, dm\) and \(\bar{y} = 0\) to express the center of mass.

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Center of Mass for a Lamina

The center of mass of a thin plate (lamina) with constant density is the point where the plate would balance perfectly. It is found by calculating the coordinates (x̄, ȳ) using the moments about the y-axis and x-axis divided by the total mass (area times density). For constant density, the density factor cancels out, simplifying calculations to area and moments.
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To find the center of mass, you compute the area and the moments using integrals over the given region. Since the region is bounded by curves and lines, you set up integrals with appropriate limits for x and y. The area integral gives the total mass, while the moment integrals involve multiplying coordinates by the density function (constant here) and integrating over the region.
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Symmetry and Its Effect on the Center of Mass

The given region is symmetric about the x-axis because the upper and lower boundaries are y = 1/(1+x²) and y = -1/(1+x²). This symmetry implies that the vertical coordinate of the center of mass (ȳ) is zero, simplifying the problem to finding only the horizontal coordinate (x̄). Recognizing symmetry reduces computational effort.
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