Question

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.

Accepted Answer

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.

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.

Verified video answer for a similar problem:

Key Concepts

Center of Mass for a Lamina

Setting up Double Integrals for Area and Moments

Symmetry and Its Effect on the Center of Mass