Calculate Double Integrals Over Triangular Regions Step-by-Step Guide

In the realm of multivariable calculus, double integrals play a crucial role in calculating the volume under a surface, the area of a region in the plane, and various other physical quantities. This comprehensive guide delves into the process of evaluating a double integral over a triangular region, providing a detailed, step-by-step approach that will equip you with the knowledge and skills to tackle similar problems with confidence. Our specific focus will be on the double integral ∬_R (x + y) dA, where R represents a triangular region bounded by the lines y = -x + 1, y = x + 1, and y = 3. We will explore the geometric interpretation of this integral, the methods for setting up the limits of integration, and the techniques for evaluating the integral itself. This guide is designed to be accessible to students and professionals alike, with clear explanations, illustrative diagrams, and practical examples to reinforce your understanding.

Before we delve into the specifics of calculating double integrals over triangular regions, let's first establish a solid understanding of the concept of double integrals themselves. A double integral is a mathematical tool used to calculate the integral of a function over a two-dimensional region. In simpler terms, it's the extension of the single integral concept to functions of two variables. Imagine a surface defined by a function f(x, y) over a region R in the xy-plane. The double integral of f(x, y) over R represents the volume between the surface and the xy-plane. If f(x, y) is always positive, this volume is above the plane; if f(x, y) is sometimes negative, the integral represents the net volume, considering regions above the plane as positive and regions below as negative. Double integrals have a wide range of applications in various fields, including physics, engineering, economics, and computer graphics. They are used to calculate areas, volumes, mass, moments of inertia, and probabilities, among other things. The notation ∬_R f(x, y) dA represents the double integral of the function f(x, y) over the region R, where dA represents an infinitesimal area element in the region. The key to evaluating a double integral lies in setting up the correct limits of integration, which define the boundaries of the region R and allow us to break down the integral into a series of single integrals that can be evaluated using standard calculus techniques.

To effectively calculate a double integral, accurately defining the region of integration is paramount. This region, denoted as R, serves as the foundation for setting up the limits of integration, which in turn dictate the boundaries over which the integral is evaluated. In our specific problem, the region R is defined as a triangle bounded by three lines: y = -x + 1, y = x + 1, and y = 3. Visualizing this region is a crucial first step. The lines y = -x + 1 and y = x + 1 are both straight lines with slopes of -1 and 1, respectively, and they intersect at the point (0, 1). The line y = 3 is a horizontal line that intersects the other two lines. To determine the vertices of the triangle, we need to find the points where these lines intersect. The intersection of y = -x + 1 and y = 3 occurs when 3 = -x + 1, which gives x = -2. So, one vertex is (-2, 3). Similarly, the intersection of y = x + 1 and y = 3 occurs when 3 = x + 1, which gives x = 2. Thus, another vertex is (2, 3). As mentioned earlier, the intersection of y = -x + 1 and y = x + 1 occurs at (0, 1), which is the third vertex. Now that we have the vertices of the triangle, we can sketch the region R in the xy-plane. This visual representation will be invaluable in determining the limits of integration. Understanding the geometry of the region is essential for choosing the most convenient order of integration (either integrating with respect to x first or y first) and for expressing the limits of integration in terms of the appropriate variables. In this case, we have a triangular region, which presents us with two primary approaches for setting up the limits of integration: integrating with respect to x first (horizontal strips) or integrating with respect to y first (vertical strips). We will explore both methods in the subsequent sections to illustrate the nuances of each approach.

With a clear understanding of the region of integration, the next crucial step is to establish the limits of integration. This process involves expressing the boundaries of the region R in terms of the variables x and y, which will allow us to transform the double integral into an iterated integral – a sequence of single integrals that can be evaluated using standard calculus techniques. There are two primary ways to set up the limits of integration for a double integral: integrating with respect to x first and then y (dx dy), or integrating with respect to y first and then x (dy dx). The choice between these two approaches often depends on the shape of the region R and the complexity of the integrand. For our triangular region bounded by y = -x + 1, y = x + 1, and y = 3, let's first consider integrating with respect to x first (dx dy). This means we will be considering horizontal strips within the region. For a given value of y, the x-values range from the line y = -x + 1 (which can be rewritten as x = 1 - y) to the line y = x + 1 (which can be rewritten as x = y - 1). Thus, the inner limits of integration for x are from 1 - y to y - 1. The outer limits of integration for y are determined by the lowest and highest y-values in the region, which are 1 and 3, respectively. Therefore, when integrating with respect to x first, our limits of integration are: 1 - y ≤ x ≤ y - 1 and 1 ≤ y ≤ 3. Now, let's consider integrating with respect to y first (dy dx). This means we will be considering vertical strips within the region. For a given value of x, the y-values range from the line y = -x + 1 to the line y = x + 1 for the section of the triangle below the intersection of the lines y = -x + 1 and y = x + 1 with the line y = 3. However, for the section of the triangle above this intersection, the y-values range from the line y = -x + 1 to the line y = 3 and from the line y = x + 1 to the line y = 3. This means we would need to split the integral into two parts. The x-values range from -2 to 0 for the region bounded by y = -x + 1 and y = 3 and from 0 to 2 for the region bounded by y = x + 1 and y = 3. Therefore, integrating with respect to x first (dx dy) appears to be the simpler approach in this case, as it avoids splitting the integral into multiple parts. The choice of integration order can significantly impact the complexity of the calculation, so carefully analyzing the region and the integrand is crucial for efficient problem-solving. In the next section, we will use these limits of integration to set up the iterated integral and evaluate it.

With the limits of integration meticulously established, we now embark on the crucial task of evaluating the double integral. This process involves transforming the double integral into an iterated integral, which is essentially a sequence of single integrals that can be solved using the fundamental theorem of calculus. Recall that our double integral is ∬R (x + y) dA, and we have determined that integrating with respect to x first (dx dy) is the more straightforward approach. This gives us the following iterated integral: ∫1^3 ∫(y-1)^(1-y) (x + y) dx dy. The inner integral, ∫(y-1)^(1-y) (x + y) dx, is evaluated with respect to x, treating y as a constant. Applying the power rule of integration, we find the antiderivative of x + y with respect to x to be (1/2)x^2 + xy. Evaluating this antiderivative at the limits of integration (y - 1) and (1 - y) and subtracting, we get: [(1/2)(1-y)^2 + y(1-y)] - [(1/2)(y-1)^2 + y(y-1)] = (1/2)(1 - 2y + y^2) + y - y^2 - (1/2)(y^2 - 2y + 1) - y^2 + y = (1/2) - y + (1/2)y^2 + y - y^2 - (1/2)y^2 + y - (1/2) - y^2 + y = -2y^2 + 2y. Now we have reduced the double integral to a single integral: ∫_1^3 (-2y^2 + 2y) dy. We evaluate this integral with respect to y. Applying the power rule again, the antiderivative of -2y^2 + 2y with respect to y is (-2/3)y^3 + y^2. Evaluating this antiderivative at the limits of integration 3 and 1 and subtracting, we get: [(-2/3)(3)^3 + (3)^2] - [(-2/3)(1)^3 + (1)^2] = [-18 + 9] - [-2/3 + 1] = -9 - [1/3] = -28/3. Therefore, the value of the double integral ∬_R (x + y) dA over the triangular region R is -28/3. This result represents the net signed volume between the surface z = x + y and the xy-plane over the region R. The negative sign indicates that a larger portion of the volume lies below the xy-plane. In this section, we have demonstrated the step-by-step process of evaluating a double integral, from setting up the limits of integration to applying the fundamental theorem of calculus. This methodical approach can be applied to a wide range of double integral problems, enabling you to calculate areas, volumes, and other important quantities in various contexts.

In this comprehensive guide, we have explored the intricacies of calculating double integrals over triangular regions. We began by establishing a solid foundation in the concept of double integrals, highlighting their significance in multivariable calculus and their applications in diverse fields. We then delved into the crucial step of defining the region of integration, emphasizing the importance of visualizing the region and determining its boundaries. The process of setting up the limits of integration was discussed in detail, with a focus on the two primary approaches: integrating with respect to x first and then y (dx dy), and integrating with respect to y first and then x (dy dx). We illustrated how the choice of integration order can impact the complexity of the calculation and how to strategically select the most efficient approach. The core of our exploration centered on the evaluation of the double integral ∬_R (x + y) dA over a specific triangular region bounded by the lines y = -x + 1, y = x + 1, and y = 3. We meticulously demonstrated the step-by-step process of transforming the double integral into an iterated integral, evaluating the inner integral, and then evaluating the outer integral. Our calculations led us to the result of -28/3, which represents the net signed volume between the surface z = x + y and the xy-plane over the triangular region R. This result underscores the importance of understanding the geometric interpretation of double integrals and the ability to accurately set up and evaluate them. The techniques and strategies presented in this guide provide a solid foundation for tackling a wide range of double integral problems, empowering you to confidently apply these concepts in your studies and professional endeavors. Whether you are a student grappling with multivariable calculus or a professional applying these techniques in your field, the knowledge and skills gained from this guide will undoubtedly enhance your problem-solving abilities and deepen your understanding of the power and versatility of double integrals.

Correct answer

The correct answer is (D) 28/3.