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Chapter 14: Problem 35

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.

Short Answer

Expert verified
Answer: The average squared distance between the points in region R and the origin is \(\frac{10}{3}\).

Step by step solution

01

Understand the problem

We need to find the average squared distance between points in the region R and the origin. The region R is defined by the inequalities: $$-2 \leq x \leq 2, 0 \leq y \leq 2$$
02

Calculate the area of the region R

In order to find the average squared distance, we need to divide the sum of squared distances by the area of the region. The area of the region R can be calculated as follows: $$Area = (2-(-2))\times(2-0)=4\times2=8$$
03

Set up the integral

The squared distance between any point \((x, y)\) and the origin can be calculated using the formula: $$D^2 = x^2 + y^2$$ Now, we need to integrate the squared distance over the range of x and y values in the region R. The integral we need to calculate is: $$\int\int_R (x^2+y^2) dA$$
04

Integrate to find the sum of squared distances

Now, we need to integrate the expression for squared distance over the region R. The integral can be split into two parts, one for each coordinate: $$\int_{-2}^2 \int_{0}^2 (x^2+y^2)dy dx$$ Now, let's first integrate with respect to y and then with respect to x: $$\int_{-2}^2 \left[ \int_{0}^2 (x^2+y^2)dy \right] dx$$ $$\int_{-2}^2 \left[ x^2y+\frac{1}{3}y^3 \right]_{0}^{2} dx$$ $$\int_{-2}^2 (2x^2 + \frac{8}{3}) dx$$ Now, let's integrate with respect to x: $$\left[\frac{2}{3}x^3+\frac{8}{3}x\right]_{-2}^{2}$$ $$\left(\frac{16}{3}+8\right)-\left(\frac{-16}{3}+8\right)$$ $$\frac{32}{3}+16$$ $$\frac{80}{3}$$
05

Find the average squared distance

Now, to calculate the average squared distance, we divide the sum of squared distances by the total area: $$\frac{\frac{80}{3}}{8} = \frac{80}{3\times8}=\frac{80}{24}=\frac{10}{3}$$ The average squared distance between the points in region R and the origin is \(\frac{10}{3}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Double Integration
Double integration is a method used to compute the integral of a function over a two-dimensional area. It involves integrating a function twice, once with respect to one variable and then again with respect to another. In the context of calculating average squared distances, we use double integration to sum up all the contributions of squared distances across the region.The general form of a double integral is \[ \int \int_R f(x,y) \, dA \] where - \(R\) is the region of integration,- \(f(x,y)\) is the function to be integrated, and- \(dA\) represents an infinitesimally small area element in the region.By performing double integration, we compute the total of these contributions to find quantities like area or total squared distance. This approach is essential when dealing with problems involving a function over a region with two variables, like the average squared distance problem.
Region of Integration
The region of integration is the specific area over which the integration takes place. For problems involving double integration, defining the region of integration is crucial as it determines the limits of the integrals.In our exercise, the region \(R\) is defined by the inequalities:- \(-2 \leq x \leq 2\)- \(0 \leq y \leq 2\)This forms a rectangular region on the Cartesian plane, where:- The x-values range from -2 to 2 - The y-values range from 0 to 2Understanding the boundaries of this region helps us set the proper limits when setting up our integrals. These limits ensure that the function is evaluated over the correct area, capturing all necessary values to find the total or average in scenarios like squared distance calculations.
Squared Distance Formula
The squared distance formula is used to find the squared distance from a point to another, often from a point to the origin. It's practically used since it avoids the necessity of computing square roots which can complicate calculus operations.For any point \((x,y)\) in the plane, the squared distance \(D^2\) from this point to the origin \((0,0)\) is given by:\[ D^2 = x^2 + y^2 \]This formula comes from the Pythagorean theorem, where the distance between two points in a plane involves the hypotenuse of a right triangle, with sides given by the differences in x and y coordinates.In the context of the original exercise, this formula is applied across many points in the region to calculate the total squared distance by integrating \(x^2 + y^2\) over the defined region \(R\). This total is then used to find the average squared distance.
Area Calculation
Calculating the area of a region is a fundamental step when dealing with double integration problems, particularly when determining averages.For the region \(R\), which is a rectangle defined by the points:- \(-2 \leq x \leq 2\)- \(0 \leq y \leq 2\) The area can be found by considering the differences in the boundaries of x and y:- The length along the x-axis is \(2 - (-2) = 4\)- The length along the y-axis is \(2 - 0 = 2\)Multiplying these lengths together gives the total area of the region:\[ \text{Area} = 4 \times 2 = 8 \text{ square units} \]Understanding and calculating this area is necessary to get the average squared distance. After obtaining the total squared distance through integration, divide by this area to find the average, as shown in the original problem solutions.

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Most popular questions from this chapter

Use integration to find the volume of the following solids. In each case, choose a convenient coordinate system, find equations for the bounding surfaces, set up a triple integral, and evaluate the integral. Assume that \(a, b, c, r, R,\) and h are positive constants. Find the volume of a solid right circular cone with height \(h\) and base radius \(r\).

Triangle medians A triangular region has a base that connects the vertices (0,0) and \((b, 0),\) and a third vertex at \((a, h),\) where \(a > 0, b > 0,\) and \(h > 0\) a. Show that the centroid of the triangle is \(\left(\frac{a+b}{3}, \frac{h}{3}\right)\) b. Note that the three medians of a triangle extend from each vertex to the midpoint of the opposite side. Knowing that the medians of a triangle intersect in a point \(M\) and that each median bisects the triangle, conclude that the centroid of the triangle is \(M\)

Let \(D\) be the region bounded by the ellipsoid \(x^{2} / a^{2}+y^{2} / b^{2}+z^{2} / c^{2}=1,\) where \(a > 0, b > 0\) and \(c>0\) are real numbers. Let \(T\) be the transformation \(x=a u, y=b v, z=c w\). Find the center of mass of the upper half of \(D(z \geq 0)\) assuming it has a constant density.

Evaluate the Jacobians \(J(u, v, w)\) for the following transformations. $$x=v w, y=u w, z=u^{2}-v^{2}$$

Determine whether the following statements are true and give an explanation or counterexample. a. A thin plate of constant density that is symmetric about the \(x\) -axis has a center of mass with an \(x\) -coordinate of zero. b. A thin plate of constant density that is symmetric about both the \(x\) -axis and the \(y\) -axis has its center of mass at the origin. c. The center of mass of a thin plate must lie on the plate. d. The center of mass of a connected solid region (all in one piece) must lie within the region.
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