Problem 45 - Calculus Early Transcendentals

Question

Find the following average values. The average distance between points of the disk \(\\{(r, \theta): 0 \leq r \leq a\\}\) and the origin

Step-by-Step Solution

Answer

Answer: The average distance between points in the disk and the origin is \(\frac{2}{3}a\).

Step 1: Set up the integral

We want to find the average distance between the points on the disk and the origin. First, we have to set up the integral in terms of polar coordinates. Since the disk extends from \(0 \leq r \leq a\), the integral will be taken over the area of the disk. The distance between a point \((r, \theta)\) and the origin is simply \(r\). We'll integrate with respect to \(r\) and \(\theta\): $$\text{Average distance} = \frac{\int_{0}^{2\pi}\int_{0}^{a} r^2 dr d\theta}{\int_{0}^{2\pi}\int_{0}^{a} r dr d\theta}$$

Step 2: Evaluate the integrals

To find the average distance, we have to evaluate the two integrals separately. First, let's evaluate the integral at the numerator: $$\int_{0}^{2\pi}\int_{0}^{a} r^2 dr d\theta = \int_{0}^{2\pi} \left[\frac{1}{3}r^3 \right]_0^a d\theta = \int_{0}^{2\pi} \frac{1}{3}a^3 d\theta = \left[\frac{1}{3}a^3 \theta \right]_0^{2\pi} = \frac{2\pi}{3}a^3$$ Now, let's evaluate the integral at the denominator: $$\int_{0}^{2\pi}\int_{0}^{a} r dr d\theta = \int_{0}^{2\pi} \left[\frac{1}{2}r^2 \right]_0^a d\theta = \int_{0}^{2\pi} \frac{1}{2}a^2 d\theta = \left[\frac{1}{2}a^2 \theta \right]_0^{2\pi} = \pi a^2$$ Finally, let's find the average distance by dividing the two results: $$\text{Average distance} = \frac{\frac{2\pi}{3}a^3}{\pi a^2} = \frac{2}{3}a$$

Step 3: Final Answer

The average distance between points in the disk \(\\{(r, \theta): 0 \leq r \leq a\\}\) and the origin is \(\frac{2}{3}a\).