91Ó°ÊÓ

Calculate the squared distance between a point (x, y) and the origin. Using the distance formula, we find that the squared distance (d^2) is given by: \(d^2 = x^2 + y^2\)

We shall set up a double integral to find the average squared distance in the given region R. To find the average value of a function over a rectangular region, we can use the following formula: \(Average = \frac{1}{Area(R)} \iint_R f(x, y) dA\) In this case, the area of R is \((2 - (-2))(2 - 0) = 4 \cdot 2 = 8\) and the function f(x, y) = \(x^2 + y^2\). Therefore, we can write the double integral for the average squared distance as: \(Average = \frac{1}{8} \iint_R (x^2 + y^2) dA\)

We will now evaluate the double integral by first integrating with respect to y, followed by x. The limits of integration are -2 to 2 for x and 0 to 2 for y, as given in the problem. Thus, we have: \(Average = \frac{1}{8} \int_{-2}^2 \int_{0}^2 (x^2 + y^2) dy dx\) First, we will integrate with respect to y: \(Average = \frac{1}{8} \int_{-2}^2 \left[\int_{0}^2 (x^2 + y^2) dy\right] dx\) \(Average = \frac{1}{8} \int_{-2}^2 \left[x^2y + \frac{1}{3}y^3\right]_0^2 dx\) \(Average = \frac{1}{8} \int_{-2}^2 (2x^2 + \frac{8}{3}) dx\) Now, we will integrate with respect to x: \(Average = \frac{1}{8} \left[\left(\frac{2}{3}x^3 + \frac{8}{3}x\right)\right]_{-2}^2\) \(Average = \frac{1}{8} \left[\left(\frac{16}{3} + \frac{16}{3}\right) - \left(\frac{-16}{3} + \frac{-16}{3}\right)\right]\) \(Average = \frac{1}{8} \cdot \frac{64}{3}\)

Finally, multiply the fraction by the result of the double integral to find the average squared distance: \(Average = \frac{1}{8} \cdot \frac{64}{3} = \frac{64}{24} = \frac{8}{3}\) Therefore, the average squared distance between the points in the region R and the origin is \(\frac{8}{3}\).