Problem 31 - Thomas Calculus

Question

Average distance from interior of disk to center Find the average distance from a point \(P(x, y)\) in the disk \(x^{2}+y^{2} \leq a^{2}\) to the origin.

Step-by-Step Solution

Answer

The average distance is \(\frac{2a}{3}\).

Step 1: Understand the Problem

We need to find the average distance from any point \( P(x, y) \) inside the disk defined by \( x^2 + y^2 \leq a^2 \) to the origin \((0,0)\). This involves integrating the distance formula over the entire disk and then normalizing by the area of the disk.

Step 2: Set Up the Integral

First, represent the distance from any point \((x, y)\) in the disk to the origin as \( d = \sqrt{x^2 + y^2} \). We convert this into polar coordinates, where \( x = r \cos\theta \) and \( y = r \sin\theta \), leading to the integral over \( r \) from \( 0 \) to \( a \) and over \( \theta \) from \( 0 \) to \( 2\pi \). This gives us the double integral \( \int_{0}^{2\pi} \int_{0}^{a} r \cdot r \, dr \, d\theta \).

Step 3: Calculate the Integral

The expression inside the integrals becomes \( \int_{0}^{2\pi} \int_{0}^{a} r^2 \, dr \, d\theta \). First, compute the inner integral with respect to \( r \): \( \int_{0}^{a} r^2 \, dr = \frac{a^3}{3} \). Then compute the outer integral with respect to \( \theta \), which leads to \( \int_{0}^{2\pi} \frac{a^3}{3} \, d\theta = \frac{2\pi a^3}{3} \).

Step 4: Determine the Average Distance

To find the average distance, divide the total integrated distance by the area of the disk. The area of the disk is \( \pi a^2 \). Therefore, the average distance is \( \frac{\frac{2\pi a^3}{3}}{\pi a^2} = \frac{2a}{3} \).

Key Concepts

Average Distance Calculation

Calculating the average distance involves determining how far, on average, points within a disk are from the center. Here, the disk is defined by the inequality \(x^2 + y^2 \leq a^2\). To calculate this average distance, follow these steps:

• Identify the formula for distance: for a point \((x,y)\), the distance to the origin is \(d = \sqrt{x^2 + y^2}\).
• Integrate this distance formula over the entire disk area to accumulate all the distances.
• Finally, divide by the total area of the disk to find the average, simplifying the expression to obtain the result.

Taking an average helps to determine a central tendency of distances within a set region, providing insights into spatial properties of the disk.

Polar Coordinates

Polar coordinates offer a different way to describe a location in the plane using a radius and angle rather than \(x\) and \(y\) coordinates. In polar coordinates:

• The position of a point is given by \((r, \theta)\), where \(r\) is the radial distance from the origin and \(\theta\) is the angle with respect to the positive x-axis.
• These coordinates are particularly useful for circular and disk regions because they naturally match the geometry of circles.
• Converting from Cartesian coordinates \((x, y)\) to polar involves \(x = r \cos\theta\) and \(y = r \sin\theta\).

In the disk problem, using polar coordinates simplifies the integration over circular areas because the limits of integration align with the shape's natural borders.

Double Integration

Double integration is used when calculating quantities across two-dimensional regions, like area or volume, and specifically in this case, to find total distance metrics over a disk.

• Start by setting up an integral that accounts for both radial and angular changes, reflecting the full area of the region.
• In polar coordinates, the integral is set as \(\int_{0}^{2\pi} \int_{0}^{a} r^2 \, dr \, d\theta\), where \(r\) and \(\theta\) variables are integrated sequentially.
• The inner integral accumulates values over the radius, while the outer integrates over the angle around the disk.

Double integration is powerful for complex shapes, breaking down the area into manageable parts that can be summed to find overall properties, like total mass or distance.

Disk Area Calculation

Calculating the area of a disk enables proper normalization when finding averages since the total area forms the denominator in such calculations. For a disk defined as \(x^2 + y^2 \leq a^2\):

• The formula for the area is \(\pi a^2\), derived from the disk's radius \(a\) and the fact that the area of a circle is \(\pi r^2\).
• This calculation is crucial as it sets the normalization baseline for any average taken over the disk.

With this area in hand, you can easily divide the total integrated distance by it, simplifying complex disk problems into intuitive averages usable in various applications, such as physics and engineering.