Problem 36 - Calculus Early Transcendentals

Question

Find the average squared distance between the points of \(R=\\{(x, y):-2 \leq x \leq 2,0 \leq y \leq 2\\}\) and the origin.

Step-by-Step Solution

Answer

Answer: The average squared distance between the points in the region R and the origin is \(\frac{16}{3}\).

Step 1: 1. Define the squared distance function

In this case, we are finding the average squared distance between the points (x, y) in R and the origin (0, 0). The squared distance from the origin to a point (x, y) is given by: \(d(x, y) = x^2 + y^2\)

Step 2: 2. Set up the double integral

We use double integral to find the average of the squared distance function over the given region R. To find the average we divide the value obtained from the integration by the total area of the region. The region R has dimensions 4x2, so the total area is 8. The double integral can be set up as follows: $$ \frac{1}{8} \int_{-2}^{2} \int_{0}^{2} (x^2 + y^2) dy\, dx $$

Step 3: 3. Solve the inner integral

We first evaluate the inner integral with respect to y. The inner integral is: $$ \int_{0}^{2} (x^2 + y^2) dy $$ Evaluating the inner integral: $$ \int_{0}^{2} (x^2 + y^2) dy = x^2y + \frac{y^3}{3} \Big|_0^{2} = 2x^2 + \frac{8}{3} $$

Step 4: 4. Solve the outer integral

Now, we can substitute the result of the inner integral into the outer integral and evaluate: $$ \frac{1}{8}\int_{-2}^{2}(2x^2 + \frac{8}{3}) dx $$ Now we solve the outer integral: $$ \frac{1}{8}(\frac{2x^3}{3} + \frac{8x}{3}) \Big|_{-2}^{2} = \frac{1}{8}\left[\frac{16}{3} + \frac{16}{3}\right] - \frac{1}{8}\left[-\frac{16}{3} - \frac{16}{3}\right] = \frac{16}{3} $$

Step 5: 5. Find the average squared distance

The average squared distance from the origin to the points in the region R is: $$ \text{Average squared distance} = \frac{16}{3} $$ The average squared distance between the points in the region R and the origin is \(\frac{16}{3}\).