Section 8-2 Summary

Chapter 8
Integral Calculus and Its Uses

8.2 Two-Dimensional Centers of Mass

Section Summary

In this section we extended the subdivide-and-conquer approach from the preceding section to the two-dimensional problem of finding the center of mass of a thin planar shape.

Assume the region is placed in an $xy$-coordinate system so that it may be described as the set of all $(x,y)$ such that

$a \leq x \leq b$ and $g (x) \leq y \leq f (x)$ .

Then the cross-sectional height $h(x)=f(x)-g(x)$. If $\delta$ is the density and $\theta$ the thickness, then the total mass is

$m$	$= δ θ \times$ area of the region
	$= θ δ \int_{a}^{b} [f (x) - g (x)] d x …$ .

The moment with respect to the $y$-axis is

$M_{y}$	$= \int_{a}^{b}$ (typical moment arm)(typical height)(width)
	$= θ δ \int_{a}^{b} x [f (x) - g (x)] d x, …$

and the $x$-coordinate of the center of mass is

$\bar{x} …$	$= \frac{M_{y}}{m} …$
	$= \frac{\int_{a}^{b} x [f (x) - g (x)] d x}{\int_{a}^{b} [f (x) - g (x)] d x} . …$

The corresponding moment with respect to the $x$-axis is

$M_{x}$	$= \int_{a}^{b}$ (typical moment arm)(typical height)(width)
	$= \frac{θ δ}{2} \int_{a}^{b} [{f (x)}^{2} - {g (x)}^{2}] d x,$

and the $y$-coordinate of the center of mass is

$\bar{y}$	$= \frac{M_{x}}{m}$
	$= \frac{\frac{1}{2} \int_{a}^{b} [{f (x)}^{2} - {g (x)}^{2}] d x}{\int_{a}^{b} [f (x) - g (x)] d x} . …$

Contents for Chapter 8

Chapter 8 Integral Calculus and Its Uses

8.2 Two-Dimensional Centers of Mass

Chapter 8
Integral Calculus and Its Uses