Evaluate The Double Integral Over The Given Region R

Evaluating double integrals over a region R is a fundamental concept in multivariable calculus. It allows us to calculate the volume under a surface defined by a function f(x, y) over a specified two-dimensional region. Understanding the process and different techniques for evaluating these integrals is crucial for applications in physics, engineering, economics, and more.

Understanding Double Integrals

A double integral, represented as ∬R f(x, y) dA, extends the concept of a single integral to two dimensions. Instead of finding the area under a curve, we're finding the volume under a surface. The region R represents the area in the xy-plane over which we are integrating. dA represents an infinitesimal area element, which can be expressed as dx dy or dy dx, depending on the order of integration.

The key idea is to divide the region R into small rectangular areas, calculate the value of the function f(x, y) at a point within each rectangle, multiply it by the area of the rectangle, and then sum up all these products. As the size of the rectangles approaches zero, this sum approaches the exact value of the double integral.

Setting Up the Integral: Defining the Region R

The first step in evaluating a double integral is to define the region of integration, R. This is often the most challenging part. R can be defined in several ways:

Rectangular Region: R is defined by constant limits for both x and y. For example, a ≤ x ≤ b and c ≤ y ≤ d.
Non-Rectangular Region (Type I): R is bounded by two vertical lines x = a and x = b, and two continuous functions of x, g1(x) ≤ y ≤ g2(x).
Non-Rectangular Region (Type II): R is bounded by two horizontal lines y = c and y = d, and two continuous functions of y, h1(y) ≤ x ≤ h2(y).

It's crucial to visualize the region R. Sketching the region helps determine the correct limits of integration. For non-rectangular regions, deciding whether to integrate with respect to x first or y first can significantly impact the complexity of the problem.

Iterated Integrals: Fubini's Theorem

Fubini's Theorem is the cornerstone for evaluating double integrals. It states that if f(x, y) is continuous on the region R, then the double integral can be evaluated as an iterated integral:

For a Type I region: ∬R f(x, y) dA = ∫ab ∫g1(x)g2(x) f(x, y) dy dx
For a Type II region: ∬R f(x, y) dA = ∫cd ∫h1(y)h2(y) f(x, y) dx dy

This means we can evaluate the double integral by performing two single integrals. The inner integral is evaluated first, treating the other variable as a constant. The result of the inner integral is a function of the remaining variable, which is then integrated with respect to that variable.

Steps to Evaluate a Double Integral

Here's a step-by-step guide to evaluating a double integral:

Sketch the Region R: This is the most crucial step. A clear sketch helps you understand the limits of integration.
Determine the Limits of Integration: Based on the sketch, decide whether R is Type I or Type II (or neither, requiring you to split the region). Determine the functions or constants that define the boundaries of the region.
Set Up the Iterated Integral: Write the double integral as an iterated integral using the appropriate order of integration (dx dy or dy dx) and the limits you determined in the previous step.
Evaluate the Inner Integral: Treat the outer variable as a constant and integrate the inner integral with respect to the inner variable. The result will be a function of the outer variable.
Evaluate the Outer Integral: Integrate the result from step 4 with respect to the outer variable. This will give you the numerical value of the double integral.

Example Problems with Detailed Solutions

Let's walk through some examples to illustrate the process.

Example 1: Rectangular Region

Evaluate ∬R (x + y) dA, where R is the rectangle defined by 0 ≤ x ≤ 2 and 1 ≤ y ≤ 3.

Sketch the Region: R is a rectangle with vertices (0, 1), (2, 1), (2, 3), and (0, 3).
Limits of Integration: Since R is a rectangle, the limits are constant: 0 ≤ x ≤ 2 and 1 ≤ y ≤ 3.
Set Up the Iterated Integral: We can choose either order of integration. Let's use dy dx: ∬R (x + y) dA = ∫02 ∫13 (x + y) dy dx
Evaluate the Inner Integral: ∫13 (x + y) dy = [xy + (y²/2)]13 = (3x + 9/2) - (x + 1/2) = 2x + 4
Evaluate the Outer Integral: ∫02 (2x + 4) dx = [x² + 4x]02 = (4 + 8) - (0 + 0) = 12

Therefore, ∬R (x + y) dA = 12.

Example 2: Type I Region

Evaluate ∬R (x² + y) dA, where R is the region bounded by y = x² and y = x.

Sketch the Region: The curves y = x² and y = x intersect at (0, 0) and (1, 1). The region R is enclosed between these two curves.
Limits of Integration: Since y = x is above y = x² in the region, we have x² ≤ y ≤ x. The x values range from 0 to 1. So, 0 ≤ x ≤ 1 and x² ≤ y ≤ x.
Set Up the Iterated Integral: ∬R (x² + y) dA = ∫01 ∫x²x (x² + y) dy dx
Evaluate the Inner Integral: ∫x²x (x² + y) dy = [x²y + (y²/2)]x²x = (x³ + x²/2) - (x⁴ + x⁴/2) = x³ + x²/2 - (3/2)x⁴
Evaluate the Outer Integral: ∫01 (x³ + x²/2 - (3/2)x⁴) dx = [(x⁴/4) + (x³/6) - (3x⁵/10)]01 = (1/4) + (1/6) - (3/10) = (15 + 10 - 18)/60 = 7/60

Therefore, ∬R (x² + y) dA = 7/60.

Example 3: Type II Region

Evaluate ∬R xy dA, where R is the region bounded by x = y² and x = 2 - y².

Sketch the Region: The curves x = y² and x = 2 - y² intersect when y² = 2 - y², which means 2y² = 2, so y² = 1, and y = ±1. The region R is enclosed between these two curves.
Limits of Integration: Since x = 2 - y² is to the right of x = y², we have y² ≤ x ≤ 2 - y². The y values range from -1 to 1. So, -1 ≤ y ≤ 1 and y² ≤ x ≤ 2 - y².
Set Up the Iterated Integral: ∬R xy dA = ∫-11 ∫y²2-y² xy dx dy
Evaluate the Inner Integral: ∫y²2-y² xy dx = [x²y/2]y²2-y² = ((2-y²)²y/2) - ((y²)²y/2) = ((4 - 4y² + y⁴)y/2) - (y⁵/2) = (4y - 4y³ + y⁵)/2 - y⁵/2 = 2y - 2y³
Evaluate the Outer Integral: ∫-11 (2y - 2y³) dy = [y² - (y⁴/2)]-11 = (1 - 1/2) - (1 - 1/2) = 0

Therefore, ∬R xy dA = 0. This result is not surprising, considering the symmetry of the region and the function.

Example 4: Changing the Order of Integration

Evaluate ∫01 ∫x1 ey² dy dx.

Understanding the Region: The limits of integration tell us that the region R is defined by x ≤ y ≤ 1 and 0 ≤ x ≤ 1. This is a Type I region.
Sketch the Region: The region is bounded by the lines y = x, y = 1, and x = 0.
Why Change the Order? The inner integral, ∫x1 ey² dy, is impossible to evaluate in terms of elementary functions. This suggests we need to change the order of integration.
Re-describe the Region: Looking at the sketch, we can describe the region as a Type II region: 0 ≤ x ≤ y and 0 ≤ y ≤ 1.
Set Up the New Iterated Integral: ∬R ey² dA = ∫01 ∫0y ey² dx dy
Evaluate the Inner Integral: ∫0y ey² dx = [xey²]0y = yey²
Evaluate the Outer Integral: ∫01 yey² dy = (1/2)∫01 e^(y²) (2y dy). Let u = y², then du = 2y dy. The limits change to 0 ≤ u ≤ 1. So, (1/2)∫01 eu du = (1/2)[eu]01 = (1/2)(e¹ - e⁰) = (e - 1)/2

Therefore, ∫01 ∫x1 ey² dy dx = (e - 1)/2. Changing the order of integration made the integral solvable!

Techniques for Evaluating Double Integrals

Beyond the basic steps, here are some useful techniques:

Changing the Order of Integration: As demonstrated in Example 4, this can be crucial when the inner integral is difficult or impossible to evaluate directly. It involves re-describing the region of integration and adjusting the limits accordingly.
Using Symmetry: If the region R and the function f(x, y) have symmetry, you can often simplify the integral. For example, if f(x, y) is odd with respect to x and R is symmetric about the y-axis, then ∬R f(x, y) dA = 0.
Polar Coordinates: When the region R is circular or has circular symmetry, converting to polar coordinates (r, θ) can greatly simplify the integral. Remember to replace dA with r dr dθ and express f(x, y) in terms of r and θ (x = r cos θ, y = r sin θ).
Transformations: More generally, you can use a change of variables (a transformation) to simplify the integral. This involves finding a suitable transformation (u, v) = T(x, y), expressing the integral in terms of u and v, and multiplying by the Jacobian determinant of the transformation.

Applications of Double Integrals

Double integrals have numerous applications across various fields:

Area: The area of a region R in the xy-plane is given by ∬R 1 dA.
Volume: The volume under a surface z = f(x, y) and above the region R in the xy-plane is given by ∬R f(x, y) dA.
Mass: If ρ(x, y) is the density of a thin plate occupying region R, then the mass of the plate is given by ∬R ρ(x, y) dA.
Center of Mass: The coordinates of the center of mass (x̄, ȳ) of the plate are given by:
- x̄ = (1/M) ∬R xρ(x, y) dA
- ȳ = (1/M) ∬R yρ(x, y) dA, where M is the total mass.
Moments of Inertia: The moments of inertia about the x-axis (Ix) and y-axis (Iy) are given by:
- Ix = ∬R y²ρ(x, y) dA
- Iy = ∬R x²ρ(x, y) dA
Probability: In probability theory, if f(x, y) is a joint probability density function over a region R, then the probability that (X, Y) falls within R is given by ∬R f(x, y) dA.

Common Mistakes to Avoid

Incorrect Limits of Integration: This is the most common mistake. Always sketch the region and carefully determine the boundaries. Make sure the limits for the inner integral are functions of the outer variable, and the limits for the outer integral are constants.
Incorrect Order of Integration: Choosing the wrong order can make the integral much harder or impossible to evaluate. Consider the complexity of the resulting integrals before choosing an order.
Forgetting the Jacobian: When using a change of variables, don't forget to multiply by the Jacobian determinant.
Algebraic Errors: Be careful with your algebra, especially when evaluating the inner and outer integrals.
Not Checking for Symmetry: Look for symmetry to simplify the integral.

Advanced Topics

Triple Integrals: Extends the concept to three dimensions, allowing you to calculate the volume of a solid, the mass of a solid object with varying density, and more.
Surface Integrals: Integrals over a surface in three dimensions, used to calculate flux, surface area, and more.
Applications in Vector Calculus: Double integrals are fundamental in Green's Theorem, Stokes' Theorem, and the Divergence Theorem.

Conclusion

Evaluating double integrals is a powerful tool in calculus and its applications. By understanding the concepts of region definition, iterated integrals, and various techniques like changing the order of integration and using symmetry, you can effectively solve a wide range of problems. Remember to practice regularly and visualize the regions of integration to master this important topic. The ability to correctly set up and evaluate double integrals opens doors to solving complex problems in physics, engineering, and other fields.