Problem 31 - Thomas Calculus

Question

Average value If \(f\) is continuous, the average value of the polar coordinate \(r\) over the curve \(r=f(\theta), \alpha \leq \theta \leq \beta,\) with respect to \(\theta\) is given by the formula $$r_{\mathrm{av}}=\frac{1}{\beta-\alpha} \int_{\alpha}^{\beta} f(\theta) d \theta$$ Use this formula to find the average value of \(r\) with respect to \(\theta\) over the following curves \((a>0)\) $$ \begin{array}{l}{\text { a. The cardioid } r=a(1-\cos \theta)} \\ {\text { b. The circle } r=a} \\ {\text { c. The circle } r=a \cos \theta, \quad-\pi / 2 \leq \theta \leq \pi / 2}\end{array} $$

Step-by-Step Solution

Answer

a. Average value is \(a\), b. Average value is \(a\), c. Average value is \(\frac{2a}{\pi}\).

Step 1: Understand the Problem

We are given a problem to find the average value of the polar coordinate \( r \) over specific curves using the formula for the average value of \( r \) in terms of \( \theta \). This involves evaluating the integral of the function \( f(\theta) \) over a specified interval.

Step 2: Set Up the Formula for Each Curve

For each part, we will set up the formula \( r_{\text{av}} = \frac{1}{\beta - \alpha} \int_{\alpha}^{\beta} f(\theta) \, d\theta \), where \( f(\theta) \), \( \alpha \), and \( \beta \) are specified by the given curves.

Step 3: Solve Part (a) The Cardioid

The function for the cardioid is \( r = a(1-\cos\theta) \) over the interval \( [0, 2\pi] \). Substituting into the formula:\[ r_{\text{av}} = \frac{1}{2\pi - 0} \int_{0}^{2\pi} a(1-\cos\theta) \, d\theta = \frac{a}{2\pi} \left( 2\pi - \int_{0}^{2\pi} \cos\theta \, d\theta \right) \]Calculate the integral of \( \cos\theta \):\[ \int \cos\theta \, d\theta = \sin\theta \text{ evaluated from } 0 \text{ to } 2\pi \text{ is } 0 \]Thus,\[ r_{\text{av}} = \frac{a}{2\pi} (2\pi) = a \]

Step 4: Solve Part (b) The Circle

For the circle \( r = a \), notice that the function is constant, so:\[ r_{\text{av}} = \frac{1}{2\pi - 0} \int_{0}^{2\pi} a \, d\theta = \frac{a}{2\pi} [a\theta]_{0}^{2\pi} = \frac{a}{2\pi}(a(2\pi) - a(0)) = a \]

Step 5: Solve Part (c) Another Circle

For the circle \( r = a \cos\theta \) over the interval \( [-\pi/2, \pi/2] \):\[ r_{\text{av}} = \frac{1}{\pi - (-\pi)} \int_{-\pi/2}^{\pi/2} a\cos\theta \, d\theta = \frac{a}{\pi} \left[ \sin\theta \right]_{-\pi/2}^{\pi/2} \]The evaluated integral:\[ \sin(\pi/2) - \sin(-\pi/2) = 1 - (-1) = 2 \]Thus,\[ r_{\text{av}} = \frac{a}{\pi} \times 2 = \frac{2a}{\pi} \]

Key Concepts

Average Value Theorem

The Average Value Theorem provides an interesting and helpful way to find the "average" size of a function over a certain interval. For functions defined in polar coordinates, the challenge is to compute this by integrating a function of \( \theta \). The average value \( r_{\text{av}} \) of a polar function \( r = f(\theta) \) over an interval \([\alpha, \beta]\) can be defined as:
\[ r_{\text{av}} = \frac{1}{\beta - \alpha} \int_{\alpha}^{\beta} f(\theta) \, d\theta \]
This formula essentially finds the mean height of the curve described by the function over the specified angle interval. Here, it means summing up tiny slices of the area beneath the curve and dividing by the length of the interval. Such an average is particularly insightful in revealing general characteristics of the curve's geometry.

Integral Calculus

Integral Calculus is concerned with the concept of an integral, which, essentially, gives us a way to accumulate quantities over a space or interval. Specifically, in the context of polar coordinates as considered in the problem examples, it allows us to calculate areas under curves defined by functions of \( \theta \).

Integrating a function in polar coordinates involves setting up the integral with respect to \( \theta \), ensuring to account for any changes in angle that define the curve. When dealing with curves like cardioids or circles, utilizing integral calculus helps us compute the average value over specified intervals of \( \theta \) so that we can understand these shapes better in terms of their spatial orientation and scale.

Cardioid

A cardioid is a heart-shaped curve, which is a special type of limaçon, and it has interesting properties in polar coordinates. The math behind it is simple yet intriguing: it can be described as \( r = a(1 - \cos\theta) \).

In the given problem, evaluating the integral of a cardioid from 0 to \( 2\pi \), while considering how \( \cos\theta \) fluctuates over this range, shows that the cosine function completes a full period over the interval, resulting in a net zero contribution due to symmetry when integrated. Thus, computing:
\[ r_{\text{av}} = \frac{a}{2\pi} \times (2\pi) = a \]
This tells us that the average radius across one complete cardioid loop is just \( a \), illustrating how the shape balances its spread around the pole.

Circle

In polar coordinates, a circle has a simple representation and provides a distinct case as explored in the exercise. For a fixed radius, we have an equation like \( r = a \). When integral calculus is applied here, it shows that regardless of the interval, since \( r \) is a constant throughout, its average over any complete rotation is simply the radius \( a \).

The intriguing case arises with equations like \( r = a \cos\theta \). For such a circle, assessed over \( [-\pi/2, \pi/2] \), the function places the circle in a half-plane, and the integral contributes a distinct result of

\[ r_{\text{av}} = \frac{2a}{\pi} \]
This calculation highlights how the circle's position and direction can vary dramatically within polar systems depending on the bounds and trigonometric influence on \( r \).