Evaluating The Double Integral Of Cos(x² + Y²) Over A Semicircular Region

In the realm of calculus, double integrals play a pivotal role in calculating volumes, areas, and other important quantities in two-dimensional spaces. This article delves into the evaluation of a specific double integral, focusing on the integral of the cosine function with a quadratic argument, cos(x² + y²), over a semicircular region denoted by S. The region S is defined as the set of all points (x, y) that satisfy two conditions: x² + y² ≤ 4, which represents a circle centered at the origin with a radius of 2, and x ≥ 0, which restricts the region to the right half-plane. This problem elegantly combines concepts from multivariable calculus and coordinate geometry, offering a rich learning experience.

When confronted with double integrals over circular or semicircular regions, a strategic approach is to transition from Cartesian coordinates (x, y) to polar coordinates (ρ, θ). This transformation often simplifies the integral significantly, as it allows us to leverage the inherent symmetry of the region. In polar coordinates, x and y are expressed as x = ρ cos θ and y = ρ sin θ, respectively, where ρ represents the radial distance from the origin and θ denotes the angle with respect to the positive x-axis. The element of area, dxdy, in Cartesian coordinates transforms to ρ dρ dθ in polar coordinates. This change of variables is crucial for effectively evaluating integrals over circular domains.

In this exploration, we will meticulously walk through the process of transforming the given double integral into polar coordinates, carefully defining the limits of integration based on the semicircular region S. We will then proceed to evaluate the integral step-by-step, applying appropriate integration techniques and leveraging the properties of trigonometric functions. By the end of this discussion, you will gain a deeper understanding of how to handle double integrals over non-rectangular regions and appreciate the power of coordinate transformations in simplifying complex calculations. This problem serves as an excellent example of the interplay between different mathematical concepts and highlights the importance of strategic problem-solving in calculus.

We are tasked with evaluating the double integral ∬S cos(x² + y²) dxdy, where S represents the region defined by S = {(x, y) | x² + y² ≤ 4 and x ≥ 0}. This region is a semicircle centered at the origin with a radius of 2, lying in the right half-plane. The integrand, cos(x² + y²), is a function that depends on the square of the distance from the origin, making it a natural candidate for transformation into polar coordinates. The challenge lies in correctly setting up the integral in polar coordinates and then evaluating it accurately.

To effectively evaluate the double integral, we transition from Cartesian coordinates (x, y) to polar coordinates (ρ, θ). In polar coordinates, x = ρ cos θ and y = ρ sin θ. The region S, defined by x² + y² ≤ 4 and x ≥ 0, transforms into a simpler representation in polar coordinates. The inequality x² + y² ≤ 4 becomes ρ² ≤ 4, implying that 0 ≤ ρ ≤ 2. The condition x ≥ 0 translates to ρ cos θ ≥ 0. Since ρ is always non-negative, this implies that cos θ ≥ 0. The angles θ for which cos θ is non-negative lie in the range -π/2 ≤ θ ≤ π/2. However, since we are considering the entire semicircle in the right half-plane, the range of θ is 0 ≤ θ ≤ π/2.

The integrand, cos(x² + y²), becomes cos(ρ²), as x² + y² = ρ² in polar coordinates. The element of area, dxdy, transforms into ρ dρ dθ. Therefore, the double integral in polar coordinates is expressed as ∬S cos(ρ²) ρ dρ dθ. The limits of integration for ρ are from 0 to 2, and the limits of integration for θ are from 0 to π/2. Thus, the integral becomes ∫0π/2 ∫02 cos(ρ²) ρ dρ dθ. This transformation significantly simplifies the integral, making it more amenable to evaluation.

Now, we proceed to evaluate the double integral in polar coordinates: ∫0π/2 ∫02 cos(ρ²) ρ dρ dθ. We first evaluate the inner integral with respect to ρ. To do this, we use a substitution method. Let u = ρ², then du = 2ρ dρ, so ρ dρ = (1/2) du. When ρ = 0, u = 0, and when ρ = 2, u = 4. The inner integral becomes ∫04 cos(u) (1/2) du = (1/2) ∫04 cos(u) du. The antiderivative of cos(u) is sin(u), so we have (1/2) [sin(u)]04 = (1/2) (sin(4) - sin(0)) = (1/2) sin(4).

Next, we evaluate the outer integral with respect to θ. The integral is ∫0π/2 (1/2) sin(4) dθ. Since (1/2) sin(4) is a constant with respect to θ, the integral simplifies to (1/2) sin(4) ∫0π/2 dθ = (1/2) sin(4) [θ]0π/2 = (1/2) sin(4) (π/2 - 0) = (π/4) sin(4). Thus, the value of the double integral is (π/4) sin(4). This result demonstrates the effectiveness of transforming to polar coordinates and using appropriate integration techniques to solve complex integrals.

Let's analyze the provided alternative representations of the double integral to identify the correct one and understand why the others are incorrect. The original integral is ∬S cos(x² + y²) dxdy, where S = {(x, y) | x² + y² ≤ 4 and x ≥ 0}. We have already established that the correct transformation to polar coordinates yields ∫0π/2 ∫02 cos(ρ²) ρ dρ dθ.

1. ∫02 ∫0π/2 cos(ρ²) dρ dθ: This option is incorrect because it is missing the crucial factor of ρ in the integrand. When transforming from Cartesian to polar coordinates, the area element dxdy becomes ρ dρ dθ. Omitting the ρ would lead to an incorrect result.
2. ∫02 ∫0π/2 ρ cos(ρ²) dρ dθ: This option correctly includes the ρ factor but has the integration limits reversed. The correct order of integration is first with respect to ρ (from 0 to 2) and then with respect to θ (from 0 to π/2). Reversing the limits would change the region of integration and thus the value of the integral.
3. ∫02 ∫-√(4-x²)√(4-x²) ρ cos(ρ²) dy dx: While this option attempts to represent the integral in Cartesian coordinates, it is unnecessarily complex and prone to errors. The limits of integration for y are correct for the semicircular region, but the presence of ρ cos(ρ²) within the Cartesian integral is confusing and incorrect. It is much simpler to evaluate the integral in polar coordinates.
4. ∫-π/2π/2 ∫02 ρ cos(ρ²) dρ dθ: This option is the closest to the correct form but has a slight error in the limits of integration for θ. The correct range for θ is 0 to π/2, representing the right half of the circle. Integrating from -π/2 to π/2 would cover the entire circle, not just the semicircle.

The importance of the Jacobian determinant (ρ in this case) in coordinate transformations cannot be overstated. It accounts for the change in area when mapping from one coordinate system to another. Failing to include the Jacobian leads to incorrect results, as it distorts the area element and consequently the integral value. Similarly, the limits of integration must accurately reflect the region of integration. Incorrect limits will lead to evaluating the integral over a different region, resulting in an erroneous answer. The correct choice ensures that the integral is evaluated over the intended semicircular region.

In summary, the double integral ∬S cos(x² + y²) dxdy, where S is the semicircular region defined by x² + y² ≤ 4 and x ≥ 0, can be effectively evaluated by transforming to polar coordinates. The correct transformation yields the integral ∫0π/2 ∫02 cos(ρ²) ρ dρ dθ, which evaluates to (π/4) sin(4). This problem underscores the importance of choosing the appropriate coordinate system to simplify integration and highlights the role of the Jacobian determinant in ensuring accurate transformations. Understanding these concepts is crucial for mastering multivariable calculus and tackling a wide range of problems involving integration over non-rectangular regions. The ability to strategically apply coordinate transformations and integration techniques is a valuable skill for any student of mathematics and related fields.