Question

Find the following average values. The average distance squared between points on the unit disk \(\\{(r, \theta): 0 \leq r \leq 1\\}\) and the point (1,1)

Short answer

Based on the given solution, calculate the average distance squared between points on the unit disc and the point (1,1). Answer: The average distance squared between points on the unit disc and the point (1,1) is \(\frac{5}{2}\).

Step-by-step solution

Convert the given point (1,1) from Cartesian to Polar coordinates

In order to find the distance between points on the unit disc and the point (1,1), we need to convert the point (1,1) from Cartesian coordinates to polar coordinates. The polar coordinates can be found by using the following formulas: \(r = \sqrt{x^2 + y^2}\) \(\theta = \arctan(\frac{y}{x})\) For the point (1,1), we have: \(r = \sqrt{1^2 + 1^2} = \sqrt{2}\) \(\theta = \arctan(\frac{1}{1}) = \arctan(1) = \frac{\pi}{4}\) So, the point (1,1) in polar coordinates is \((\sqrt{2}, \frac{\pi}{4})\).

Find the distance squared between points on the unit disc and the point (1,1)

Now, we need to find the distance squared between points on the unit disc, \((r, \theta)\), and the point \((1,1)\) which corresponds to \((\sqrt{2}, \frac{\pi}{4})\) in polar coordinates. The distance between two points in polar coordinates can be found using the following formula: \(D^2 = r_1^2 + r_2^2 - 2r_1r_2\cos(\theta_1 - \theta_2)\) In our case, we have: \(D^2 = r_1^2 + (\sqrt{2})^2 - 2r \sqrt{2} \cos(\theta - \frac{\pi}{4}) = r^2 + 2 - 2r\sqrt{2} \cos(\theta - \frac{\pi}{4})\)

Integrate the distance squared over the unit disc

Now we need to integrate the distance squared over the unit disc (0 ≤ r ≤ 1) to compute the average squared distance. Since we are dealing with polar coordinates, the area element is given by \(r \, dr \, d\theta\), and we need to integrate both with respect to \(r\) and \(\theta\). The resulting integral is: \(\int_{0}^{1} \int_{0}^{2\pi} \left( r^2 + 2 - 2r\sqrt{2} \cos(\theta - \frac{\pi}{4})\right) r \, dr \, d\theta\) To find the average distance squared, we need to divide by the area of the unit disc, which is \(\pi\).

Compute the integral and average squared distance

Split the integral into three parts: \(A = \int_{0}^{1} \int_{0}^{2\pi} r^3\, dr \, d\theta\) \(B = 2 \int_{0}^{1} \int_{0}^{2\pi} r\, dr \, d\theta\) \(C = -2\sqrt{2} \int_{0}^{1} \int_{0}^{2\pi} r^2 \cos(\theta - \frac{\pi}{4})\, dr \, d\theta\) Then, compute each part: \(A = 2\pi \int_{0}^{1} r^3\, dr = 2\pi \left[\frac{1}{4}r^4\right]_0^1 = \frac{1}{2}\pi \) \(B = 4\pi \int_{0}^{1} r\, dr = 4\pi \left[\frac{1}{2}r^2\right]_0^1 = 2\pi\) \(C = -2\sqrt{2} \int_{0}^{1} \int_{0}^{2\pi} r^2 \cos(\theta - \frac{\pi}{4})\, dr \, d\theta\) Since \(\cos(\theta - \frac{\pi}{4})\) is an odd function and we are integrating over a full period, \(C = 0\) Now, sum the parts: \(A + B + C = \frac{1}{2}\pi + 2\pi + 0 = \frac{5}{2}\pi\) Finally, to find the average squared distance, divide by the area of the unit disc: \(\frac{\frac{5}{2}\pi}{\pi} = \frac{5}{2}\). The average distance squared between points on the unit disc and the point (1,1) is \(\frac{5}{2}\).

Key concepts

Polar Coordinates

In calculus and other branches of mathematics, polar coordinates are an alternative to Cartesian coordinates that can make dealing with circular or rotationally symmetric figures a lot simpler. Instead of using horizontal (x) and vertical (y) coordinates to determine the position of a point, polar coordinates use a pair \(r, \theta\). \(
\) Here, \(r\) is the radial distance from a fixed central point (usually the origin), and \(\theta\) is the angular displacement from a fixed direction, typically the positive x-axis. This system is particularly useful when solving problems involving circles, spirals, or anything with rotational symmetry. \(
\)

• The radial coordinate \(r\) tells you how far away the point is from the origin.
• The angular coordinate \(\theta\) gives the direction of the point, measured in radians in a counterclockwise direction from the positive x-axis.

\(
\) To convert a point from Cartesian to Polar coordinates, we use the formulas: \(r = \sqrt{x^2 + y^2}\) and \(\theta = \arctan\left(\frac{y}{x}\right)\). This conversion step is crucial, especially when tackling geometry problems involving circles or arcs, as it transforms how we consider distances and angles.

Integration in Polar Coordinates

When calculating the integral in polar coordinates, we approach it slightly differently than in Cartesian coordinates. In polar coordinates, the area element or differential area needs to take the circular nature of the coordinate system into account. \(
\)

• The differential area element in polar coordinates is defined as \(r \, dr \, d\theta\).
• This comes from the structure of how the polar system stretches the radial component by \(r\), as each 'slice' extends outward from the origin like a piece of pie.

\(
\) To integrate a function over an area defined by polar coordinates, we integrate with respect to \(r\) and \(\theta\). For example, to calculate the average distance squared, an integral is set up over both radial distance and angular displacement. This gives us the full picture of how a varying parameter, such as distance, changes over a circular area. The integration gives us a summation of all infinitesimally small pieces of the entire area under consideration.

Unit Disc Area

The unit disc is simply a disc in the plane where all points are within a unit distance from the central point or origin. It's defined in polar coordinates by the condition \(0 \leq r \leq 1\). This forms a clean circle with a radius of 1. The area of such a disc in the Cartesian plane is known to be \(\pi\), but when dealing with polar coordinates, it's important to reiterate how this area is calculated with integration. \(
\) In polar coordinates, to compute the total area of the unit disc, we integrate with respect to \(r\) and \(\theta\). The formula is: \(\int_{0}^{1} \int_{0}^{2\pi} r \, dr \, d\theta = \pi \)

• This is because the radius varies from 0 to 1, and \(\theta\) covers a full circle, from 0 to \(2\pi\) radians.
• The unit disc's radial symmetry makes it easy to integrate over, as each radial increment forms a ring segment with a proportional increase in circumference.

\(
\) This calculation is foundational in understanding how areas and distances, even average distances, come into play within the domain of polar coordinates.