Evaluating The Double Integral ∫₀¹ ∫₀²ˣ (x² + Y²) Dy Dx A Step-by-Step Guide
Introduction to Double Integrals
In the realm of calculus, double integrals play a pivotal role in determining volumes, areas, and other essential properties of two-dimensional regions. This article delves into the computation of a specific double integral, providing a step-by-step solution and highlighting the underlying concepts. Let's explore the value of the double integral ∫₀¹ ∫₀²ˣ (x² + y²) dy dx. Understanding double integrals is crucial in various fields, including physics, engineering, and computer graphics, where they are used to calculate areas, volumes, moments of inertia, and other quantities. The integral ∫₀¹ ∫₀²ˣ (x² + y²) dy dx represents the volume under the surface z = x² + y² over the region in the xy-plane defined by 0 ≤ x ≤ 1 and 0 ≤ y ≤ 2x. This region is a triangle bounded by the lines y = 0, x = 1, and y = 2x. Evaluating this double integral involves performing two successive integrations. First, we integrate the function x² + y² with respect to y, treating x as a constant, over the interval 0 to 2x. Then, we integrate the resulting expression with respect to x over the interval 0 to 1. This process allows us to find the exact value of the double integral, which represents the volume under the surface. Double integrals are also used in probability and statistics to calculate probabilities over two-dimensional regions. For example, if we have a joint probability density function f(x, y), the probability that a random variable (X, Y) falls within a region R is given by the double integral of f(x, y) over R. In fluid dynamics, double integrals can be used to calculate the flux of a fluid across a surface. The flux represents the amount of fluid flowing through the surface per unit time. In thermodynamics, double integrals can be used to calculate the work done by a system in a two-dimensional process. The work is given by the integral of the pressure with respect to volume over the process path.
Problem Statement: Evaluating ∫₀¹ ∫₀²ˣ (x² + y²) dy dx
The core of this discussion lies in evaluating the double integral: ∫₀¹ ∫₀²ˣ (x² + y²) dy dx. This integral represents the volume under the surface defined by the function f(x, y) = x² + y² over a specific region in the xy-plane. The limits of integration, 0 to 1 for x and 0 to 2x for y, define this region. To solve this, we will follow a systematic approach involving two steps: first, integrating with respect to y while treating x as a constant, and second, integrating the result with respect to x. This process will yield a numerical value, representing the definite integral's solution. Understanding the problem statement is the first crucial step in solving any mathematical problem. In this case, we are asked to evaluate a double integral, which means we need to find the value of the integral of a function over a two-dimensional region. The function we are integrating is f(x, y) = x² + y², which is a paraboloid opening upwards. The region of integration is defined by the limits 0 ≤ x ≤ 1 and 0 ≤ y ≤ 2x. This region is a triangle in the xy-plane with vertices at (0, 0), (1, 0), and (1, 2). The double integral ∫₀¹ ∫₀²ˣ (x² + y²) dy dx represents the volume under the surface z = x² + y² and above this triangular region. To evaluate this double integral, we will first integrate with respect to y, treating x as a constant. This will give us a function of x, which we will then integrate with respect to x. The order of integration is important, and in this case, integrating with respect to y first is the most straightforward approach. We could also integrate with respect to x first, but that would require splitting the region of integration into two parts, which would make the calculation more complex. The problem statement clearly defines the function, the region of integration, and the order of integration, which makes it easier to proceed with the solution. It is important to carefully read and understand the problem statement before attempting to solve it.
Step-by-Step Solution
1. Inner Integral Evaluation
We initiate the process by evaluating the inner integral with respect to y, treating x as a constant: ∫₀²ˣ (x² + y²) dy. This involves finding the antiderivative of x² + y² with respect to y, which is x²y + (1/3)y³. We then evaluate this expression at the limits of integration, 0 and 2x. This step is crucial as it transforms the double integral into a single integral, making the subsequent steps more manageable. When evaluating the inner integral, we treat x as a constant because we are integrating with respect to y. This means that x² is treated as a constant term, and its integral with respect to y is simply x²y. The integral of y² with respect to y is (1/3)y³. So, the antiderivative of x² + y² with respect to y is x²y + (1/3)y³. Next, we need to evaluate this antiderivative at the limits of integration, which are 0 and 2x. This means we substitute y = 2x and y = 0 into the expression x²y + (1/3)y³ and subtract the second result from the first. When y = 2x, the expression becomes x²(2x) + (1/3)(2x)³ = 2x³ + (8/3)x³. When y = 0, the expression becomes x²(0) + (1/3)(0)³ = 0. Therefore, the value of the inner integral is 2x³ + (8/3)x³ - 0 = (14/3)x³. This result is a function of x, which we will then integrate with respect to x in the next step. The inner integral effectively calculates the area under the curve z = x² + y² along the y-axis for a fixed value of x. This area is a function of x, and the outer integral will sum up these areas over the interval 0 ≤ x ≤ 1 to give us the total volume under the surface.
2. Substituting Limits and Simplifying
Substituting the limits of integration, we get [x²(2x) + (1/3)(2x)³] - [x²(0) + (1/3)(0)³] = 2x³ + (8/3)x³. This simplification results in an expression in terms of x only, which is (14/3)x³. This expression now becomes the integrand for the outer integral. The substitution of the limits of integration is a critical step in evaluating definite integrals. It involves replacing the variable of integration (in this case, y) with the upper and lower limits of integration and subtracting the result obtained from the lower limit from the result obtained from the upper limit. This process effectively calculates the change in the antiderivative over the interval of integration, which gives us the definite integral's value. In this case, substituting y = 2x into the antiderivative x²y + (1/3)y³ gives us x²(2x) + (1/3)(2x)³ = 2x³ + (8/3)x³. Substituting y = 0 into the antiderivative gives us x²(0) + (1/3)(0)³ = 0. Subtracting the second result from the first gives us 2x³ + (8/3)x³ - 0 = (14/3)x³. This simplified expression, (14/3)x³, represents the value of the inner integral, which is a function of x. This function will now be integrated with respect to x in the next step to find the value of the double integral. The simplification process is important because it reduces the complexity of the expression, making it easier to integrate in the next step. In this case, combining the terms 2x³ and (8/3)x³ into (14/3)x³ simplifies the expression and makes it more straightforward to integrate with respect to x.
3. Outer Integral Evaluation
Next, we evaluate the outer integral: ∫₀¹ (14/3)x³ dx. This involves finding the antiderivative of (14/3)x³ with respect to x, which is (14/3) * (1/4)x⁴ = (7/6)x⁴. We then evaluate this antiderivative at the limits of integration, 0 and 1. Evaluating the outer integral is the final step in calculating the value of the double integral. This step involves integrating the result obtained from the inner integral with respect to the remaining variable (in this case, x). The outer integral effectively sums up the values of the inner integral over the interval of integration, giving us the total value of the double integral. In this case, we are integrating the function (14/3)x³ with respect to x over the interval 0 ≤ x ≤ 1. The antiderivative of (14/3)x³ with respect to x is (14/3) * (1/4)x⁴ = (7/6)x⁴. This antiderivative represents the function whose derivative is (14/3)x³. To evaluate the definite integral, we need to find the difference between the values of the antiderivative at the upper and lower limits of integration. This means we substitute x = 1 and x = 0 into the expression (7/6)x⁴ and subtract the second result from the first. When x = 1, the expression becomes (7/6)(1)⁴ = 7/6. When x = 0, the expression becomes (7/6)(0)⁴ = 0. Therefore, the value of the outer integral is 7/6 - 0 = 7/6. This value represents the final answer to the double integral. The outer integral effectively calculates the volume under the surface z = x² + y² over the triangular region in the xy-plane. This volume is represented by the definite integral, which we have now evaluated to be 7/6.
4. Final Calculation
Evaluating the antiderivative at the limits gives us [(7/6)(1)⁴] - [(7/6)(0)⁴] = 7/6. Therefore, the value of the double integral ∫₀¹ ∫₀²ˣ (x² + y²) dy dx is 7/6. The final calculation involves substituting the limits of integration into the antiderivative and simplifying the result. This step is crucial as it gives us the numerical value of the definite integral, which represents the area, volume, or other quantity that the integral is designed to calculate. In this case, we found the antiderivative of (14/3)x³ to be (7/6)x⁴. To evaluate the definite integral, we need to substitute the upper limit of integration (x = 1) and the lower limit of integration (x = 0) into the antiderivative and subtract the result obtained from the lower limit from the result obtained from the upper limit. Substituting x = 1 into the antiderivative gives us (7/6)(1)⁴ = 7/6. Substituting x = 0 into the antiderivative gives us (7/6)(0)⁴ = 0. Subtracting the second result from the first gives us 7/6 - 0 = 7/6. Therefore, the value of the definite integral ∫₀¹ (14/3)x³ dx is 7/6. This value represents the area under the curve y = (14/3)x³ between x = 0 and x = 1. It also represents the value of the outer integral, which, combined with the inner integral, gives us the value of the double integral ∫₀¹ ∫₀²ˣ (x² + y²) dy dx. The final calculation step ensures that we obtain the correct numerical value for the integral, which is essential for applications in various fields, such as physics, engineering, and economics.
Result: The Value of the Double Integral
Thus, the value of the double integral ∫₀¹ ∫₀²ˣ (x² + y²) dy dx is 7/6. This result represents the volume under the surface z = x² + y² over the region defined by 0 ≤ x ≤ 1 and 0 ≤ y ≤ 2x. This final result is the culmination of all the steps we have taken to evaluate the double integral. It represents the precise numerical value of the volume under the surface z = x² + y² over the specified region. This result is not just a number; it has a geometric interpretation and can be used in various applications. For example, if we were to build a solid object with the shape defined by the surface z = x² + y² and the region of integration, the volume of that object would be 7/6 cubic units. This result can also be used in physics to calculate the mass of an object with a density function given by x² + y² over the same region. In statistics, this result could represent the probability of an event occurring in a two-dimensional space. The value of the double integral, 7/6, is a precise and meaningful result that can be used in various contexts. It is a testament to the power of calculus and the ability to use integrals to solve real-world problems. The process of evaluating the double integral has not only given us the result but also provided us with a deeper understanding of the concepts and techniques involved in multivariable calculus.
Conclusion
In conclusion, evaluating the double integral ∫₀¹ ∫₀²ˣ (x² + y²) dy dx demonstrates the application of calculus in solving problems involving two variables. The step-by-step approach, from evaluating the inner integral to the final calculation, provides a clear understanding of the process. The result, 7/6, represents a specific volume and highlights the practical applications of double integrals in various fields. The process of evaluating a double integral involves careful attention to detail and a thorough understanding of the fundamental concepts of calculus. Each step, from setting up the integral to evaluating the limits of integration, plays a crucial role in obtaining the correct result. Double integrals are a powerful tool for solving problems in various fields, including physics, engineering, and economics. They allow us to calculate areas, volumes, moments of inertia, and other quantities in two-dimensional and three-dimensional spaces. The ability to evaluate double integrals is an essential skill for anyone working in these fields. In this article, we have demonstrated a step-by-step approach to evaluating a specific double integral, which can serve as a guide for solving other similar problems. The key is to break down the problem into smaller, manageable steps and to carefully apply the rules of calculus. The result we obtained, 7/6, is not just a number; it represents a specific quantity that can be interpreted and used in various contexts. This highlights the practical significance of calculus and its ability to provide solutions to real-world problems. The study of double integrals is an important part of multivariable calculus, which is a fundamental branch of mathematics with wide-ranging applications. By mastering the concepts and techniques of multivariable calculus, we can gain a deeper understanding of the world around us and solve complex problems in various fields.
