Double Integral Over A Set: Step-by-Step Solution

by ADMIN 50 views

Integrating functions over specific sets can be a tricky business, especially when the region of integration is defined by inequalities. Let's dive into a fascinating problem involving integration over a set and unravel the steps to ensure we arrive at the correct solution. In this guide, we'll tackle the integral of a function over a region defined by inequalities, pinpoint common pitfalls, and equip you with the knowledge to conquer such challenges.

Defining the Region Ω

Our journey begins with a clear definition of the region over which we'll be integrating. The set Ω is defined as follows:

Ω = (x, y) ∈ ℝÂČ 0 < x < y < 2x < 2

This set lives in the two-dimensional real plane (ℝÂČ) and is bounded by a series of inequalities. To truly grasp the essence of this region, let's break down these inequalities:

  1. 0 < x: This inequality dictates that x must be a positive number, confining our region to the right side of the y-axis.
  2. x < y: This tells us that y must be greater than x, placing our region above the line y = x.
  3. y < 2x: Here, y must be less than 2x, bounding our region below the line y = 2x.
  4. 2x < 2: This inequality simplifies to x < 1, further restricting our region to the left of the vertical line x = 1.

Visualizing this region is key. Imagine the plane carved out by these lines. We have a triangular-shaped region nestled between y = x and y = 2x, with its boundaries further defined by the lines x = 0 and x = 1. Accurately defining and visualizing this region is the first step to setting up the integral correctly. Understanding the region Ω is crucial for setting up the iterated integral properly. The inequalities defining Ω dictate the bounds of integration, and any misinterpretation here will lead to an incorrect result. The region is a triangle bounded by the lines y = x, y = 2x, and x = 1 in the first quadrant. It’s helpful to sketch this region to visualize the limits of integration. The sketch of the region of integration will help determine the order of integration (whether to integrate with respect to x first or y first) and the appropriate limits for each variable. A clear picture of Ω ensures that the integral accurately sums the function over the intended domain. Failing to properly define Ω can lead to integrating over an incorrect area, resulting in a flawed calculation. Therefore, investing time in understanding the boundaries and shape of Ω is essential for a correct solution. The geometry of Ω directly influences the mathematical setup of the integral, so it's not just a preliminary step but a fundamental component of the integration process. Remember, the visual representation of the integration region is the compass that guides you through the integral's solution. Without it, you’re navigating in the dark, increasing the risk of errors and misinterpretations. So, take a moment, sketch the region, and let the geometry illuminate your path to the correct answer. This step is paramount in ensuring that the double integral correctly accounts for the function's behavior across the specified area. A well-defined integration domain is the cornerstone of accurate integration.

The Integral in Question

Now that we've painted a vivid picture of our region, let's turn our attention to the integral we're tasked with solving:

∏Ω y dx dy

This is a double integral, instructing us to integrate the function y over the region Ω. The notation dx dy indicates that we'll be integrating with respect to x first, followed by y. However, the order of integration can be switched, and we'll explore this flexibility later. The key here is understanding what this integral represents. It's essentially summing up the values of the function y over the entire region Ω. Imagine dividing Ω into infinitesimally small rectangles, multiplying the value of y at each rectangle by its area (dx dy), and then adding up all these products. That's the essence of the double integral. The integral of y over Ω represents the weighted average of the y-coordinate across the region. Understanding this geometric interpretation helps in visualizing the outcome of the integration. The double integral breaks down the problem into a series of smaller, manageable integrations. The function y acts as a weight at each point in the region, contributing to the overall sum based on its value. This weighted sum is what the integral calculates. It’s not merely about finding an area; it’s about accumulating the product of y and the infinitesimal area elements (dx dy) across the entire domain. The integration process essentially sweeps across the region, adding up these contributions. The result of the integral gives a single value that encapsulates the overall distribution of y within Ω. The act of integrating y over this region is like taking a census of the y-values and calculating a sort of average, but one that’s weighted by the infinitesimal area. This concept is fundamental in many areas of physics and engineering, where integrals are used to calculate things like mass, charge, or moments of inertia. The mathematical operation of integration is a powerful tool for summarizing continuous quantities over a region, providing insights that discrete sums cannot. Therefore, grasping the essence of what the integral is doing—accumulating the weighted values of y—is crucial for both setting it up correctly and interpreting the final answer. The conceptual understanding of the integral as a sum is what bridges the gap between the abstract notation and the concrete result. This lays the groundwork for solving the definite integral with confidence and accuracy.

Setting Up the Iterated Integral

The real magic happens when we translate this double integral into an iterated integral, which is something we can actually compute. This involves determining the limits of integration for both x and y. Remember, the order of integration matters, and we'll start by integrating with respect to x first. Looking back at our region Ω, we see that for a fixed value of y, x ranges from y/2 to y. This is because the left boundary of our region is the line y = 2x (or x = y/2), and the right boundary is the line y = x. Next, we need to determine the bounds for y. From our inequalities, we know that 0 < x < 1 and x < y < 2x. The maximum value of 2x within our region is 2 (when x = 1), and since y < 2x, the upper bound for y is also effectively limited. The intersection of y = 2x and x = 1 gives us y = 2, but the intersection of y = x and x = 1 gives us y = 1. So, y ranges from 0 to 1. Therefore, our iterated integral is set up as follows:

∫(from 0 to 1) ∫(from y/2 to y) y dx dy

This iterated integral is the key to unlocking the solution. It breaks down the double integral into two single integrals, which we can evaluate step-by-step. The setup of the iterated integral is the most critical step. It translates the geometric definition of the region into algebraic limits that can be used in the integration process. The inner integral (with respect to x) is evaluated first, treating y as a constant. The limits of integration for x (y/2 to y) are determined by the boundaries of the region along the x-direction for a fixed y. These boundaries are the lines that define the