Problem 35 - Thomas Calculus

Question

In polar coordinates, the average value of a function over a region \(R\) (Section 15.3\()\) is given by $$\frac{1}{\text { Area(R) }} \iint_{R} f(r, \theta) r d r d \theta$$ 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 from a point in the disk to the origin is \(\frac{2a}{3}\).

Step 1: Interpret the Problem in Polar Coordinates

The given disk is centered at the origin with radius \( a \). The distance from any point \( P(x, y) \) in the disk to the origin \( (0, 0) \) is \( r = \sqrt{x^2 + y^2} \). In polar coordinates, this region is described by \( 0 \leq r \leq a \) and \( 0 \leq \theta \leq 2\pi \). Our function \( f(r, \theta) \) that describes the distance is simply \( r \).

Step 2: Set Up the Integral

The average distance is given by the formula \[\frac{1}{\text{Area}(R)} \iint_{R} f(r, \theta) \cdot r \; dr \, d\theta.\]Substitute \( f(r,\theta) = r \), setting up the integral:\[\frac{1}{\text{Area}(R)} \int_{0}^{2\pi} \int_{0}^{a} r^2 \; dr \, d\theta.\]

Step 3: Calculate the Area of the Disk

The area \( \text{Area}(R) \) of the disk \( x^2 + y^2 \leq a^2 \) is \( \pi a^2 \) as it's a circle with radius \( a \). This value will be used to find the average by dividing the integral by this area.

Step 4: Solve the Double Integral

Perform the integration of \( r^2 \) with respect to \( r \) first:\[\int_{0}^{a} r^2 \, dr = \left[ \frac{r^3}{3} \right]_{0}^{a} = \frac{a^3}{3}.\]Next, perform the integration of \( \frac{a^3}{3} \) with respect to \( \theta \):\[\int_{0}^{2\pi} \frac{a^3}{3} \, d\theta = \frac{a^3}{3} \cdot \theta \bigg|_{0}^{2\pi} = \frac{a^3}{3} \cdot 2\pi = \frac{2\pi a^3}{3}.\]

Step 5: Divide by the Area to Find the Average Distance

Divide the result of the integral by the area of the disk to find the average distance:\[\text{Average Distance} = \frac{1}{\pi a^2} \cdot \frac{2\pi a^3}{3} = \frac{2a}{3}.\]

Key Concepts

Average Value of a Function

The concept of the average value of a function is quite similar to finding an average in everyday math—like an average score, for instance. In calculus, though, it involves taking the sum of function values over a region and dividing it by the area of that region. This can be calculated using integration.
For a function defined in polar coordinates within a region \( R \), the average value is given by:

• **Average Value Formula:** \[ \frac{1}{\text{Area}(R)} \iint_{R} f(r, \theta) r \, dr \, d\theta \]

The formula uses a double integral to sum all values of the function \( f(r, \theta) \), weighted by the radius \( r \). Then, you divide by the total area of region \( R \) to obtain the average. This makes the concept especially handy when dealing with circular regions, like disks, where the region isn't a simple rectangle or square.

Double Integral

Double integrals extend the concept of integration to two-dimensional regions, allowing you to accumulate quantities over an area. In the context of polar coordinates, a double integral lets you calculate things like area, mass, or in this case—the average value across a circular disk.
When we set up a double integral in polar coordinates, it looks like this:

• **Integral Setup:**\[ \iint_{R} f(r, \theta) \, r \, dr \, d\theta \]

The parts \( f(r, \theta) \) and \( r \, dr \, d\theta \) tell you what to sum over and how to adjust for polar conversion, respectively. The extra \( r \) accounts for the stretching that occurs when transitioning from rectangular to polar coordinates.

Disk Area

Understanding the area of a disk is crucial when working with problems in polar coordinates. A disk is simply a circle with its center at a specific point and its boundary determined by a radius.
For a disk centered at the origin with radius \( a \), its area is found using the familiar formula for the area of a circle:

• **Disk Area Formula:**\[ \text{Area} (R) = \pi a^2 \]

This formula expresses that you have \( \pi \)—the constant that relates diameter to perimeter—multiplied by the square of the radius. It tells you the total space within the boundary, which is essential when calculating averages or setting integration bounds.

Polar Integration

Polar integration is a method used to integrate over regions that are better described in circular terms, like circles or sectors. This is common in physics and engineering when dealing with circular objects.
When integrating in polar coordinates, you replace \( dx \, dy \) with \( r \, dr \, d\theta \). Here is how it works:

• **Coordinate Conversion:** \[ x = r \cos(\theta), \quad y = r \sin(\theta) \]
• **Element of Area:** \[ dA = r \, dr \, d\theta \]

The limits for \( r \) and \( \theta \) depend on the specific region \( R \) you're considering. For a full circular region like a disk, \( r \) goes from 0 to some radius \( a \), and \( \theta \) spans from 0 to \( 2\pi \). This adaptation makes polar integration ideal for finding areas and averages in circular shapes.