Problem 45 - University Calculus: Early Transcendentals

Question

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 value of \(d^2\) is \(\frac{a^2}{2} + h^2\).

Step 1: Setting up the coordinate system

To simplify the problem, place the center of the circle at the origin of a coordinate system, thus the circle is centered at (0,0) with radius \(a\). Place \(P_{0}\) on the x-axis, so it has coordinates \((h, 0)\), where \(h\) is the distance from \(P_{0}\) to the center of the circle.

Step 2: Expressing the distance squared

The distance from any point \(P = (x, y)\) to \(P_{0} = (h,0)\) is given by the formula \(d^2 = (x - h)^2 + (y - 0)^2 = (x-h)^2 + y^2\). We want to find the average value of \(d^2\) over the entire disk with radius \(a\).

Step 3: Setting up the integral for average value

The average value of \(d^2\) over the circle is given by the double integral:\[ \frac{1}{\pi a^2} \int \int_{R} \left( (x-h)^2 + y^2 \right) \, dx \, dy \]where \(R\) is the region within the circle \(x^2 + y^2 \leq a^2\).

Step 4: Convert to polar coordinates

Using polar coordinates, where \(x = r\cos(\theta)\) and \(y = r\sin(\theta)\), the integral becomes:\[ \frac{1}{\pi a^2} \int_{0}^{2\pi} \int_{0}^{a} \left( (r\cos(\theta) - h)^2 + (r\sin(\theta))^2 \right) r \, dr \, d\theta \].

Step 5: Simplifying the integral

First, expand \( (r\cos(\theta) - h)^2 + r^2\sin^2(\theta) \) to get:\[ r^2\cos^2(\theta) - 2hr\cos(\theta) + h^2 + r^2\sin^2(\theta) \]Simplifying, we note that \( \cos^2(\theta) + \sin^2(\theta) = 1 \), giving:\[ r^2 - 2hr\cos(\theta) + h^2 \].

Step 6: Evaluate the integral over r

Separate the integral into three parts:\[ \frac{1}{\pi a^2} \left( \int_{0}^{2\pi} \int_{0}^{a} r^3 \, dr \, d\theta - 2h \int_{0}^{2\pi} \int_{0}^{a} r^2\cos(\theta) \, dr \, d\theta + h^2 \int_{0}^{2\pi} \int_{0}^{a} r \, dr \, d\theta \right) \].

Step 7: Solving the radial integrals

Compute each integral:1. \( \int_{0}^{a} r^3 \, dr = \frac{a^4}{4} \).2. \( \int_{0}^{a} r \, dr = \frac{a^2}{2} \).3. The integral \( \int_{0}^{a} r^2\cos(\theta) \, dr \) evaluates to \( \frac{a^3\cos(\theta)}{3} \).For the angular parts:1. \( \int_{0}^{2\pi} \cos(\theta) \, d\theta = 0 \), which results in the second integral being zero.

Step 8: Combine all parts

Substitute the evaluated integrals back:\[ \frac{1}{\pi a^2} \left( \frac{a^4}{4} \cdot 2\pi + h^2 \cdot \frac{a^2}{2} \cdot 2\pi \right) = \frac{1}{2}a^2 + h^2 \].

Step 9: Final result

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

Key Concepts

Polar Coordinates

Polar coordinates are an alternative to the Cartesian coordinate system, often useful in problems involving circular or rotational symmetry. Instead of using coordinates

• \((x, y)\)

it represents points on a plane using

• \((r, \theta)\)

where:

• \(r\) is the distance from the origin to the point.
• \(\theta\) is the angle between the positive x-axis and the line connecting the origin to the point.

This transformation simplifies calculations when dealing with circular shapes, such as circles or disks, since it aligns with the radial symmetry of these objects. By replacing the Cartesian coordinates with polar coordinates in integrals, the limits of integration over a circular region become more straightforward. In our exercise, transforming to polar coordinates allows us to efficiently integrate over the circle precisely because it diminishes the complexity of the bounds

• \(r = 0\) to \(r = a\)
• \(\theta = 0\) to \(\theta = 2\pi\)

replacing \((x-h)^2 + y^2\) with a polar expression. By capitalizing on the geometry of the problem, we streamline our integration using the simpler bounds of the polar coordinate system.

Double Integral

A double integral is a type of integral used to calculate over two-dimensional areas, such as the one enclosed by a circle in this exercise. It extends the idea of a single integral, which finds the area under a curve, to finding the volume under a surface. When we set up a double integral, it is expressed as follows:

• \(\int \int_R\) over a region \(R\)

Here, it is used to find the average value of the distance squared

• \(d^2 = (x-h)^2 + y^2\)

for every point inside the circle. Double integrals can be quite sophisticated, but are easier to manage when symmetry or transformation can simplify the computations. Polar coordinates are particularly advantageous here since the circular symmetry of the problem correlates directly to the natural format of a polar double integral.

Coordinate Transformation

Coordinate transformation is the process of changing from one coordinate system to another, and it serves as a vital tool in tackling complex integrations. When dealing with problems with inherent symmetry like a circle, transforming to different coordinates can significantly simplify the mathematics involved. For this exercise, switching from Cartesian coordinates

• \((x, y)\)

to polar coordinates

• \((r, \theta)\)

made the problem easier to solve by aligning the coordinate system with the shape of the region we are integrating over. A transformation like this does two main things:

• It simplifies the integral's limits.
• It can make complicated expressions manageable.

Specifically, in this exercise, the transformation is key to rewriting the expression for distance squared in terms of the polar axes, allowing the original double integral over

• \((x, y)\) space

to be expressed over a simpler form in

• \((r, \theta)\) space

thereby simplifying the solution, and ultimately calculating the average value of the squared distance more intuitively.