Problem 41 - Thomas Calculus

Question

Average distance to a given point inside a disk Let \(P_{0}\) be a point inside a circle of radius \(a\) and let \(h\) denote the distance from \(P_{0}\) to the center of the circle. Let \(d\) denote the distance from an arbitrary point \(P\) to \(P_{0} .\) Find the average value of \(d^{2}\) over the region enclosed by the circle. (Hint: Simplify your work by placing the center of the circle at the origin and \(P_{0}\) on the \(x\) -axis.)

Step-by-Step Solution

Answer

The average of \(d^2\) over the disk is \(\frac{a^2}{2} + h^2\)."

Step 1: Set Up the Coordinate System

Place the center of the circle at the origin of a coordinate system. Assume that point \(P_0\) is located at \((h,0)\) on the \(x\)-axis, which simplifies calculations due to symmetry.

Step 2: Define Points and Distances

Let an arbitrary point \(P\) in the disk be denoted by its coordinates \((x,y)\). The distance \(d\) from \(P\) to \(P_0\) is given by the Euclidean distance formula: \(d^2 = (x-h)^2 + y^2\).

Step 3: Integrate Over the Circle

We need to calculate the average value of \(d^2\) over the disk of radius \(a\). This requires integrating \(d^2\) over the circular region \(x^2 + y^2 \leq a^2\):\[\frac{1}{\pi a^2} \int_{x^2 + y^2 \leq a^2} ((x-h)^2 + y^2) \, dx \, dy\].

Step 4: Convert to Polar Coordinates

Transform the integral from Cartesian \(x,y\) to polar coordinates \(r,\theta\) where \((x,y) = (r \cos \theta, r \sin \theta)\). Limits for \(r\) will be from 0 to a and for \(\theta\) from 0 to \(2\pi\). The distance formula becomes: \[(r \cos \theta - h)^2 + (r \sin \theta)^2 = r^2 - 2hr \cos \theta + h^2\].

Step 5: Calculate the Polar Integrals

Compute the integral for each component separately: 1. \(\int_0^{2\pi}\int_0^a r^2 \cdot r \, dr \, d\theta\)2. \(\int_0^{2\pi}\int_0^a -2hr \cos \theta \cdot r \, dr \, d\theta\)3. \(\int_0^{2\pi}\int_0^a h^2 \cdot r \, dr \, d\theta\)Each integral will be solved and summed to find the average \(d^2\).

Step 6: Solve the Integrals

1. Solve \(\int_0^{2\pi}\int_0^a r^3 \, dr \, d\theta = 2\pi\frac{a^4}{4}\)2. Solve \(\int_0^{2\pi}\int_0^a -2hr^2 \cos \theta \, dr \, d\theta = 0\) because \(\int_0^{2\pi} \cos \theta \, d\theta = 0\)3. Solve \(\int_0^{2\pi}\int_0^a h^2r \, dr \, d\theta = h^2\pi a^2\)

Step 7: Combine Results and Average

Add the results: \[\frac{1}{\pi a^2} \left(\pi \frac{a^4}{2} + h^2 \pi a^2\right) = \frac{a^2}{2} + h^2\].This is the average value of \(d^2\).

Key Concepts

Polar Coordinates Integration

Polar coordinates allow us to describe the location of a point using radius and angle. In our case, polar coordinates simplify the integration process over a circular region, because circular shapes naturally align with polar symmetry. Rather than defining a point by its Cartesian coordinates

• \(x = r \cos \theta\)
• \(y = r \sin \theta\)

we use the radius \(r\) from the origin and the angle \(\theta\) from the positive \(x\)-axis.
When integrating in polar coordinates, it's important to account for the "differential area element" in polar coordinates, which is \(r \, dr \, d\theta\). This arises because the tiny sectors of our circular region become slightly larger as \(r\) increases. This is reflected in the formula for area in polar integrations.
Switching to polar coordinates is particularly advantageous since it aligns perfectly with the geometry of the disk, making the evaluation of such integrals straightforward.

Euclidean Distance Formula

The Euclidean distance formula measures the straight-line distance between two points in a plane. It's a fundamental concept in geometry and crucial for problems involving average distances.
For a point \(P\) located at coordinates \((x, y)\) and a point \(P_0\) at \((h, 0)\), the Euclidean distance \(d\) is defined as:

• \(d = \sqrt{(x - h)^2 + y^2}\)

In this exercise, we focus on \(d^2\), avoiding the square root making integration easier as we compute:

• \(d^2 = (x - h)^2 + y^2\)

This reformulation aims to simplify calculations, particularly when integrating over a circular domain. The squared distance also directly ties into mean square calculations often used in statistical analyses.

Circle Geometry

Understanding circle geometry is essential, particularly when the region of interest is circular in shape. A circle is defined by its radius and center point, with every point lying a fixed distance (the radius) from this center.

• The equation \(x^2 + y^2 \leq a^2\) defines the interior of a circle with radius \(a\) centered at the origin.
• Many geometrical properties, such as symmetry, can simplify integration problems over circular regions.

Thanks to the circle's inherent symmetry, issues like polar symmetry ease the transition to polar coordinates. For integration, knowing that the geometric properties remain consistent can save significant time and help intuitively grasp integration boundaries.

Integration in Polar Coordinates

Integration in polar coordinates is a method particularly useful for problems involving circles or sectors of circles. By shifting from Cartesian coordinates, we align with the natural symmetry of circular objects, easing calculation and facilitating understanding.
When converting a Cartesian integral to polar coordinates, certain steps should be followed:

• Define each \(x, y\) term in terms of \(r, \theta\).
• Replace \(dx \, dy\) with \(r \, dr \, d\theta\).
• Adjust limits to match \(r\) from 0 to the radius of the circle, and \(\theta\) from 0 to \(2\pi\).

Once set up, break the integral into simpler terms that can be independently solved. For instance, integrating \(r^2 - 2hr \cos \theta + h^2\) separately allows for easier management of complex integrals. This step-by-step breakdown greatly simplifies often-intimidating integral problems, by transforming them into solvable pieces.