Polar Coordinate Transformation And Integral Evaluation Of ∫∫√(x² + Y²) DA

In this article, we will delve into the process of transforming a double integral from Cartesian coordinates to polar coordinates and subsequently evaluating it. Specifically, we will focus on the integral ∫₋ₐᵃ ∫₀ᵥ(ₐ² ₋ ₓ²) √(x² + y²) dy dx. This type of problem is common in multivariable calculus and requires a strong understanding of coordinate transformations and integration techniques. The transformation to polar coordinates is particularly useful when dealing with integrals over regions with circular symmetry, as it simplifies the integrand and the limits of integration. By converting to polar coordinates, we replace the variables x and y with r and θ, where r represents the distance from the origin and θ represents the angle with respect to the positive x-axis. This transformation often makes the integral more manageable and easier to solve.

Coordinate transformation is a powerful tool in calculus that allows us to simplify integrals by changing the coordinate system. The original integral in Cartesian coordinates, ∫₋ₐᵃ ∫₀ᵥ(ₐ² ₋ ₓ²) √(x² + y²) dy dx, represents the integration of the function √(x² + y²) over a specific region in the xy-plane. The region of integration is defined by the limits of the integrals: x ranges from -a to a, and for each x, y ranges from 0 to √(a² - x²). This region is the upper half of a circle with radius a centered at the origin. The integrand √(x² + y²) represents the distance from the origin to the point (x, y). Integrating this function over the region gives the weighted area of the region, where the weight is determined by the distance from the origin. However, integrating directly in Cartesian coordinates can be cumbersome due to the complexity of the integrand and the limits of integration. The square root term √(x² + y²) and the upper limit of the inner integral √(a² - x²) both suggest that a transformation to polar coordinates might simplify the problem. Polar coordinates are particularly well-suited for problems with circular symmetry, as they naturally capture the radial and angular aspects of the region. In polar coordinates, the circle is described by a simple equation r = a, and the angle θ ranges from 0 to π for the upper half of the circle. The integrand √(x² + y²) becomes simply r, which is much easier to handle. The differential area element dx dy transforms to r dr dθ, which accounts for the change in area when transforming from Cartesian to polar coordinates. Thus, transforming to polar coordinates not only simplifies the integrand but also makes the limits of integration more straightforward, leading to a more manageable integral.

To effectively transform the given integral, ∫₋ₐᵃ ∫₀ᵥ(ₐ² ₋ ₓ²) √(x² + y²) dy dx, into polar coordinates, we must first understand the relationships between Cartesian and polar coordinates. The transformation equations are: x = r cos θ, y = r sin θ. These equations relate the Cartesian coordinates (x, y) to the polar coordinates (r, θ), where r is the radial distance from the origin and θ is the angle measured counterclockwise from the positive x-axis. The expression √(x² + y²) in Cartesian coordinates becomes r in polar coordinates, which simplifies the integrand significantly. The differential area element dy dx in Cartesian coordinates transforms to r dr dθ in polar coordinates. This factor of r is crucial and arises from the Jacobian determinant of the transformation, which accounts for the change in area when transforming between the coordinate systems. The region of integration in the xy-plane is the upper half of a circle with radius a centered at the origin. In polar coordinates, this region is described by the inequalities 0 ≤ r ≤ a and 0 ≤ θ ≤ π. The radial distance r ranges from 0 to a, covering the entire radius of the circle, and the angle θ ranges from 0 to π, covering the upper half of the circle. With these transformations in mind, we can rewrite the integral in polar coordinates. The integral ∫₋ₐᵃ ∫₀ᵥ(ₐ² ₋ ₓ²) √(x² + y²) dy dx becomes ∫₀<binary data, 1 bytes><binary data, 1 bytes><binary data, 1 bytes> ∫₀ᵃ r * r dr dθ. The integrand √(x² + y²) is replaced by r, and the differential area element dy dx is replaced by r dr dθ. The limits of integration are changed to reflect the polar coordinate system: θ ranges from 0 to π, and r ranges from 0 to a. This transformed integral is now in a form that is much easier to evaluate.

Defining Polar Coordinates

When dealing with integrals that involve circular or radial symmetry, converting to polar coordinates can greatly simplify the problem. Polar coordinates provide an alternative way to describe points in a plane using a radial distance r from the origin and an angle θ measured from the positive x-axis. The relationships between Cartesian coordinates (x, y) and polar coordinates (r, θ) are given by: x = rcos(θ), y = rsin(θ). The radial distance r is calculated as r = √(x² + y²), and the angle θ can be found using θ = arctan(y/x). In the context of double integrals, the differential area element dA in Cartesian coordinates (dx dy) transforms to r dr dθ in polar coordinates. This transformation factor r arises from the Jacobian determinant of the coordinate transformation and is essential for correctly evaluating integrals in polar coordinates. Understanding these transformations is crucial for setting up the integral in polar coordinates and determining the appropriate limits of integration. The region of integration in the original Cartesian integral often dictates the limits of integration in polar coordinates. For instance, if the region is a circle or a sector of a circle, the limits for r and θ can be easily determined. The radial distance r typically ranges from 0 to the radius of the circle, while the angle θ ranges from 0 to 2π for a full circle or a fraction thereof for a sector. By correctly converting to polar coordinates and establishing the appropriate limits of integration, we can often simplify complex integrals into more manageable forms.

Setting up the Integral in Polar Form

To set up the integral in polar form, we must first identify the region of integration and express it in terms of polar coordinates. The original integral, ∫₋ₐᵃ ∫₀ᵥ(ₐ² ₋ ₓ²) √(x² + y²) dy dx, is defined over a region in the xy-plane. The limits of integration for y, 0 ≤ y ≤ √(a² - x²), indicate that we are considering the upper half of a circle centered at the origin with radius a. The limits of integration for x, -a ≤ x ≤ a, confirm that we are integrating over the entire width of this semicircle. In polar coordinates, this region is described by 0 ≤ r ≤ a and 0 ≤ θ ≤ π. The radial distance r ranges from 0 to the radius a, and the angle θ ranges from 0 to π, covering the upper half of the circle. Next, we need to transform the integrand √(x² + y²) into polar coordinates. Using the transformation equations, we know that x² + y² = r², so √(x² + y²) = r. Additionally, the differential area element dy dx in Cartesian coordinates transforms to r dr dθ in polar coordinates. This factor of r is crucial and arises from the Jacobian determinant of the transformation. Now we can rewrite the integral in polar coordinates. The integral ∫₋ₐᵃ ∫₀ᵥ(ₐ² ₋ ₓ²) √(x² + y²) dy dx becomes ∫₀<binary data, 1 bytes><binary data, 1 bytes><binary data, 1 bytes> ∫₀ᵃ r * r dr dθ, which simplifies to ∫₀<binary data, 1 bytes><binary data, 1 bytes><binary data, 1 bytes> ∫₀ᵃ r² dr dθ. This integral is now in a form that is much easier to evaluate. The integrand is a simple power of r, and the limits of integration are constant, making the integration process straightforward.

Now that we have transformed the integral into polar coordinates, we can proceed with evaluating it. The integral in polar form is ∫₀<binary data, 1 bytes><binary data, 1 bytes><binary data, 1 bytes> ∫₀ᵃ r² dr dθ. To evaluate this double integral, we first integrate with respect to r, treating θ as a constant. The inner integral is ∫₀ᵃ r² dr. The antiderivative of r² with respect to r is (1/3)r³, so we evaluate this from r = 0 to r = a: [(1/3)a³] - [(1/3)(0)³] = (1/3)a³. Now we substitute this result back into the outer integral, which becomes ∫₀<binary data, 1 bytes><binary data, 1 bytes><binary data, 1 bytes> (1/3)a³ dθ. This is now a single integral with respect to θ. The integrand (1/3)a³ is a constant with respect to θ, so we can simply multiply it by the length of the interval of integration, which is π - 0 = π. Thus, the integral evaluates to (1/3)a³ * π = (πa³)/3. Therefore, the value of the original double integral ∫₋ₐᵃ ∫₀ᵥ(ₐ² ₋ ₓ²) √(x² + y²) dy dx is (πa³)/3. This result represents the weighted area of the upper half of the circle with radius a, where the weight at each point is proportional to its distance from the origin. Evaluating the integral in polar coordinates significantly simplifies the process compared to attempting to evaluate the original integral in Cartesian coordinates. The transformation to polar coordinates allows us to take advantage of the circular symmetry of the region and the integrand, resulting in a more manageable integral.

Performing the Inner Integration

To evaluate the double integral in polar coordinates, ∫₀<binary data, 1 bytes><binary data, 1 bytes><binary data, 1 bytes> ∫₀ᵃ r² dr dθ, we begin by performing the inner integration with respect to r. This means we treat θ as a constant and integrate the function r² with respect to r over the interval [0, a]. The integral ∫₀ᵃ r² dr is a standard power rule integration. The antiderivative of r² with respect to r is (1/3)r³. We evaluate this antiderivative at the limits of integration, a and 0, and subtract the values: [(1/3)a³] - [(1/3)(0)³] = (1/3)a³. This result represents the value of the inner integral. Now we have a single integral remaining: ∫₀<binary data, 1 bytes><binary data, 1 bytes><binary data, 1 bytes> (1/3)a³ dθ. This integral is much simpler to evaluate because the integrand (1/3)a³ is a constant with respect to θ. Performing the inner integration is a crucial step in evaluating double integrals. By integrating with respect to one variable first, we reduce the double integral to a single integral, which is often easier to solve. In this case, the inner integration with respect to r allowed us to simplify the integrand and obtain a constant value that could then be integrated with respect to θ. This technique of performing inner integration first is a common strategy in multivariable calculus for evaluating double and triple integrals.

Completing the Outer Integration

After performing the inner integration, we are left with the outer integral, which is ∫₀<binary data, 1 bytes><binary data, 1 bytes><binary data, 1 bytes> (1/3)a³ dθ. This integral is with respect to θ, and the integrand (1/3)a³ is a constant. To evaluate this integral, we can simply multiply the constant by the length of the interval of integration. The interval for θ is [0, π], so the length of the interval is π - 0 = π. Therefore, the integral evaluates to (1/3)a³ * π = (πa³)/3. This is the final result of the double integral. The value (πa³)/3 represents the volume under the surface z = √(x² + y²) over the region defined by the upper half of the circle with radius a. The constant a determines the size of the region, and the result scales accordingly. Completing the outer integration is the final step in evaluating the double integral. By integrating with respect to the remaining variable, we obtain a numerical value that represents the overall result of the integration. In this case, the outer integration was straightforward because the integrand was a constant. However, in other cases, the integrand may be a function of the remaining variable, requiring further integration techniques. The process of evaluating a double integral involves careful attention to the limits of integration and the order of integration. By breaking the integral into inner and outer integrals, we can systematically evaluate the integral and obtain the desired result.

In conclusion, we have successfully transformed the integral ∫₋ₐᵃ ∫₀ᵥ(ₐ² ₋ ₓ²) √(x² + y²) dy dx into polar coordinates and evaluated it. The transformation to polar coordinates simplified the integral by replacing the Cartesian coordinates (x, y) with polar coordinates (r, θ), where x = r cos θ, y = r sin θ, and √(x² + y²) = r. The differential area element dy dx was replaced by r dr dθ. The limits of integration were also transformed to reflect the polar coordinate system, with r ranging from 0 to a and θ ranging from 0 to π, representing the upper half of a circle with radius a centered at the origin. The transformed integral in polar coordinates was ∫₀<binary data, 1 bytes><binary data, 1 bytes><binary data, 1 bytes> ∫₀ᵃ r² dr dθ. We first evaluated the inner integral with respect to r, which resulted in (1/3)a³. Then, we evaluated the outer integral with respect to θ, which yielded the final result of (πa³)/3. This result represents the value of the original double integral and provides insight into the weighted area of the region of integration. The process of transforming to polar coordinates is a powerful technique for simplifying integrals that involve circular symmetry. By converting to polar coordinates, we can often reduce complex integrals to more manageable forms, making them easier to evaluate. This technique is widely used in multivariable calculus and has applications in various fields, including physics, engineering, and computer graphics.