Problem 45 - Thomas Calculus

Question

Centroid of unbounded region Find the centroid of the infinite region in the \(x y\) -plane bounded by the curves \(y=1 / \sqrt{1-x^{2}}\) , \(y=-1 / \sqrt{1-x^{2}},\) and the lines \(x=0, x=1\)

Step-by-Step Solution

Answer

The centroid of the region is at \(\left(\frac{2}{\pi}, 0\right)\).

Step 1: Understanding the Problem

We need to find the centroid of the region in the xy-plane bounded by the curves \(y = \frac{1}{\sqrt{1-x^2}}\) , \(y = -\frac{1}{\sqrt{1-x^2}}\), and the lines \(x=0\), \(x=1\). This infinite region forms between these boundaries.

Step 2: Setting the Limits for Integration

The region is symmetrical about the x-axis due to the functions \(y = \frac{1}{\sqrt{1-x^2}}\) and \(y = -\frac{1}{\sqrt{1-x^2}}\). Thus, we only focus on the part where \(x\) ranges from 0 to 1. For each \(x\), the vertical boundaries are given by these functions. Therefore, the limits of integration for \(y\) will be from \(-\frac{1}{\sqrt{1-x^2}}\) to \(\frac{1}{\sqrt{1-x^2}}\).

Step 3: Formula for the Centroid

The centroid \((\bar{x}, \bar{y})\) for a plane region between x-boundary \([a, b]\) and y-boundary \([f(x), g(x)]\) is given by the formula: \[ \bar{x} = \frac{1}{A} \int_a^b x [g(x) - f(x)] \, dx \] \[ \bar{y} = \frac{1}{2A} \int_a^b [g(x)^2 - f(x)^2] \, dx \] where \(A\) is the area of the region.

Step 4: Finding the Area (A)

The area \(A\) is calculated by \[ A = \int_0^1 \left( \frac{1}{\sqrt{1-x^2}} - \left(-\frac{1}{\sqrt{1-x^2}}\right) \right) \, dx = 2 \int_0^1 \frac{1}{\sqrt{1-x^2}} \, dx \] The integral evaluates to\[ A = 2 [\arcsin x]_0^1 = 2\left(\frac{\pi}{2} - 0\right) = \pi \].

Step 5: Calculating \(\bar{x}\)

The x-coordinate of the centroid is given by \[ \bar{x} = \frac{1}{\pi} \int_0^1 x \cdot 2 \cdot \frac{1}{\sqrt{1-x^2}} \, dx \] This becomes\[ \bar{x} = \frac{2}{\pi} \int_0^1 \frac{x}{\sqrt{1-x^2}} \, dx \].Substituting \( u = 1-x^2, \, du = -2x \, dx \), we have\[ \bar{x} = \frac{1}{\pi} \int_1^0 -\frac{du}{\sqrt{u}} = \frac{1}{\pi} [2\sqrt{u}]_1^0 = \frac{2}{\pi} (1 - 0) = \frac{2}{\pi} \].

Step 6: Calculating \(\bar{y}\)

The y-coordinate of the centroid is calculated as \[ \bar{y} = \frac{1}{2\pi} \int_0^1 \left( \left( \frac{1}{\sqrt{1-x^2}} \right)^2 - \left( -\frac{1}{\sqrt{1-x^2}} \right)^2 \right) \, dx \] which simplifies to \[ \bar{y} = \frac{1}{2\pi} \int_0^1 0 \, dx = 0 \]. The expression simplifies because the squares cancel each other, confirming the symmetry observed.

Key Concepts

Infinite Region

When we talk about an infinite region, we mean an area in the Cartesian plane that extends indefinitely along one or more directions. In the context of the given problem, the infinite region is bounded horizontally by the lines \(x = 0\) and \(x = 1\), and vertically by the functions \(y = \frac{1}{\sqrt{1-x^2}}\) and \(y = -\frac{1}{\sqrt{1-x^2}}\). This forms a strip of infinite height but with a finite base extending horizontally from 0 to 1.

This infinite nature can often make calculating properties like area and centroids more complex. Thankfully, through calculus, we can tame these infinite regions using integration to determine finite answers for seemingly unbounded areas.

Integration Limits

Setting integration limits is key to solving problems involving areas and centroids. The limits define the boundaries of the region we're interested in. For our problem:

• The horizontal limits are from \(x = 0\) to \(x = 1\).
• The vertical limits are defined by the curves \(y=\frac{1}{\sqrt{1-x^2}}\) for the upper boundary, and \(y=-\frac{1}{\sqrt{1-x^2}}\) for the lower boundary.

These limits ensure that we integrate over the entire desired region. By integrating first with respect to \(y\) from \(-\frac{1}{\sqrt{1-x^2}}\) to \(\frac{1}{\sqrt{1-x^2}}\), and then with respect to \(x\) from 0 to 1, we account for the entire specified region in a structured manner.

Symmetrical Boundaries

Symmetrical boundaries refer to regions where the shape on one side of a central axis is a mirror image of the shape on the opposite side. In this exercise, the symmetry lies about the x-axis. The functions \(y = \frac{1}{\sqrt{1-x^2}}\) and \(y = -\frac{1}{\sqrt{1-x^2}}\) are symmetrical, forming a region that is perfectly mirrored above and below the x-axis.

This symmetry is a crucial advantage in calculus:

• It simplifies the calculation of the centroid, as symmetry often means that either \(\bar{x}\) or \(\bar{y}\) is zero, depending on the axis of symmetry.
• In our problem, because of symmetry about the x-axis, \(\bar{y}\) is calculated to be zero.

This property allows us to often simplify computations and focus on reduced regions.

Centroid Formula

The centroid of a region is often described as the "center of mass" or "balance point" of the shape. For a two-dimensional region bounded by functions, the centroid can be calculated using specific formulas for \(\bar{x}\) and \(\bar{y}\).

These formulas typically involve integrals over the x-boundaries and y-boundaries of the region:

• \(\bar{x} = \frac{1}{A} \int_a^b x [g(x) - f(x)] \, dx\)
• \(\bar{y} = \frac{1}{2A} \int_a^b [g(x)^2 - f(x)^2] \, dx\)

Here, \(f(x)\) and \(g(x)\) define the boundaries, and \(A\) is the total area of the region. These formulas integrate over the product of the "coordinate weight" (\(x\) or \(y\)) and the width of the region along that coordinate. By applying these, we obtain coordinates \((\bar{x}, \bar{y})\) for the centroid, effectively capturing the region's balance point.