Problem 32 - Thomas Calculus

Question

Average distance squared from a point in a disk to a point in its boundary Find the average value of the square of the distance from the point \(P(x, y)\) in the disk \(x^{2}+y^{2} \leq 1\) to the boundary point \(A(1,0) .\)

Step-by-Step Solution

Answer

The average squared distance is 2.

Step 1: Understand and Visualize the Problem

The problem asks us to find the average of the squared distance from a point \((x,y)\) within the unit disk \(x^2 + y^2 \leq 1\) to the fixed boundary point \(A(1,0)\). The disk is centered at the origin \((0,0)\) with radius 1, and \(A\) is on the positive x-axis.

Step 2: Express the Distance Formula

The distance between point \((x, y)\) and \((1, 0)\) is given by the distance formula: \[ d = \sqrt{(x - 1)^2 + (y - 0)^2} = \sqrt{(x-1)^2 + y^2} \]. We need the square of this distance, so \[ d^2 = (x-1)^2 + y^2 \].

Step 3: Setup the Average Calculation as an Integral

To find the average value of \(d^2\) over the unit disk, we need to integrate \((x-1)^2 + y^2\) over the area of the disk and divide by the area of the disk. The area of the disk is \(\pi\).

Step 4: Convert to Polar Coordinates

For easier integration, convert to polar coordinates: - \(x = r\cos(\theta)\), \(y = r\sin(\theta)\) - \(dx\, dy = r\, dr\, d\theta\).The limits for \(r\) are from 0 to 1 (the radius of the disk), and for \(\theta\) from 0 to \(2\pi\).

Step 5: Substitute and Integrate

Substitute into the integral: \[ \int_0^{2\pi} \int_0^1 \left((r\cos\theta - 1)^2 + (r\sin\theta)^2\right) r\, dr\, d\theta \].Simplify and solve this: \[ \int_0^{2\pi} \int_0^1 \left(r^2\cos^2\theta - 2r\cos\theta + 1 + r^2\sin^2\theta\right) r\, dr\, d\theta \]\[ = \int_0^{2\pi} \int_0^1 \left(r^3 - 2r^2\cos\theta + r)\, dr\, d\theta \].

Step 6: Separate and Solve the Integral

The expression inside the integral can be separated:\[ \int_0^{2\pi} \int_0^1 r^3\, dr\, d\theta - 2\int_0^{2\pi} \int_0^1 r^2\cos\theta\, dr\, d\theta + \int_0^{2\pi} \int_0^1 r\, dr\, d\theta \].Evaluate each part:- \( \int_0^1 r^3\, dr = \frac{1}{4}\)- \( -2\int_0^{2\pi} \cos\theta\, d\theta \int_0^1 r^2\, dr = 0\)- \( \int_0^1 r\, dr = \frac{1}{2}\).The result becomes: \[ \frac{1}{4} \times 2\pi + 0 + \frac{1}{2} \times 2\pi = \pi + \pi = 2\pi \].Divide by the area \(\pi\) for the average.

Step 7: Compute the Average Distance Squared

The final average squared distance is \(\frac{2\pi}{\pi} = 2\). This is the average value of the squared distance from a point in the disk to the point \((1, 0)\).

Key Concepts

Polar Coordinates

When solving problems in Integral Calculus that involve regions like circles or disks, polar coordinates become incredibly helpful. Polar coordinates are a way of representing points in a plane using two numbers:

• The radius, denoted as \(r\), is the distance from the point to the origin.
• The angle, denoted as \(\theta\), is the direction from the origin to the point, measured in radians from the positive x-axis.

This coordinate system simplifies the calculation in circular regions because it directly relates to them, unlike Cartesian coordinates (\(x, y\)).
In our exercise, converting from Cartesian to polar coordinates required using:

• \(x = r\cos(\theta)\)
• \(y = r\sin(\theta)\)
• The differential area element in polar coordinates: \(dx\, dy = r\, dr\, d\theta\)

Polar coordinates, therefore, allow us to integrate easily over circular regions, such as our unit disk.

Average Distance

The concept of finding an "average distance" involves some integral calculus. You find the average value of a function across a region by integrating and then dividing by the size (area or volume) of the region. Here we focus on the average squared distance:
1. First, set up the desired function, which in our case is the squared distance: \((x-1)^2 + y^2\).
2. Integrate this function over the unit disk.
3. Finally, divide the result by the area of the disk, \(\pi\), because its radius is 1.
This technique helps to find the average value of the squared distance over all possible points \((x, y)\) in a given disk.

Squared Distance

Understanding squared distance is vital, especially in this scenario where we need to calculate it over a disk. The squared distance between two points

• $(1,0)$
- the point on the boundary,
• \((x, y)\) - any point within the disk.

The standard distance formula between two points \((x_1, y_1) and (x_2, y_2)\) is:\[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\].
In this exercise, the distance squared is simply:
\[d^2 = (x-1)^2 + y^2\].
Squaring the distance simplifies the integral and removes the square root, making it easier to handle when setting up integrals over specific regions.

Unit Disk

A unit disk refers to a perfectly round two-dimensional area centered at the origin with a radius of 1. Its boundary is defined by the equation \(x^2 + y^2 = 1\). Points inside satisfy \(x^2 + y^2 \leq 1\).
When solving mathematical problems involving integration, a unit disk is crucial for simplifying computations, especially when using polar coordinates. Here's why:

• The transformation into polar coordinates aligns naturally with the circular shape, as \(r\) varies from 0 to 1.
• This transformation allows \(\theta\) to cover a full revolution from 0 to \(2\pi\), which corresponds to the full circle in the Cartesian plane.
• The separation of \(r\) and \(\theta\) aids in evaluating integrals.

The use of a unit disk also simplifies the calculation of the area as \(\pi\), allowing for straightforward normalization processes in integral calculations.