Evaluating The Double Integral ∫₀⁴ ∫₀^(y²/4) Y/(x² + Y²) Dx Dy

In this article, we will delve into the process of evaluating the double integral ∫₀⁴ ∫₀^(y²/4) y/(x² + y²) dx dy. This integral presents an interesting challenge due to the form of the integrand and the limits of integration. We will systematically break down the problem, first tackling the inner integral and then proceeding to the outer integral. This step-by-step approach will allow us to understand each stage of the calculation and arrive at the final solution. Double integrals are a fundamental concept in multivariable calculus, with applications ranging from calculating areas and volumes to solving problems in physics and engineering. The ability to evaluate these integrals accurately is a crucial skill for anyone working in these fields. The given integral specifically involves a rational function within the integrand, which often requires careful consideration of the integration techniques used. Moreover, the limits of integration, particularly the upper limit of the inner integral being a function of y, add another layer of complexity that needs to be addressed methodically. Therefore, a thorough understanding of both the theoretical underpinnings and the practical steps involved is essential for mastering the evaluation of such integrals. This article aims to provide a comprehensive guide to solving this particular double integral, while also reinforcing the general principles applicable to a wider range of similar problems.

Step 1: Evaluating the Inner Integral

Let's start by focusing on the inner integral, which is ∫₀^(y²/4) y/(x² + y²) dx. To evaluate this, we recognize that the integrand resembles the derivative of the arctangent function. Specifically, we recall that the derivative of arctan(x/a) with respect to x is a/(x² + a²). In our case, we have y/(x² + y²), which matches this form with a = y. Therefore, we can use this knowledge to find the antiderivative. The inner integral, ∫₀^(y²/4) y/(x² + y²) dx, requires a strategic approach due to the form of the integrand. Recognizing the similarity to the derivative of the arctangent function is crucial here. The function y/(x² + y²) closely resembles the form a/(x² + a²), which is the derivative of arctan(x/a) with respect to x. By identifying 'y' as the constant 'a' in this context, we can proceed with the integration. The antiderivative of y/(x² + y²) with respect to x is arctan(x/y). This step involves a direct application of the known integral of a specific form, which is a common technique in calculus. The ability to recognize such patterns and apply the corresponding antiderivatives is a key skill in integral calculus. Once we have the antiderivative, the next step is to evaluate it at the limits of integration, which in this case are 0 and y²/4. This will give us a function in terms of 'y' that we can then use in the outer integral. This process highlights the fundamental theorem of calculus, which connects differentiation and integration and allows us to evaluate definite integrals by finding antiderivatives. The methodical approach of breaking down the double integral into inner and outer integrals, and then tackling each one systematically, is a standard technique that simplifies the problem and makes it more manageable.

Applying this, the antiderivative of y/(x² + y²) with respect to x is arctan(x/y). Now, we evaluate this antiderivative at the limits of integration, 0 and y²/4:

[arctan(x/y)] from 0 to y²/4 = arctan((y²/4)/y) - arctan(0/y) = arctan(y/4) - arctan(0) = arctan(y/4)

Step 2: Evaluating the Outer Integral

Now we move on to the outer integral, which becomes ∫₀⁴ arctan(y/4) dy. This integral is a bit more challenging and requires integration by parts. Integration by parts is a technique used to integrate the product of two functions. The formula for integration by parts is ∫u dv = uv - ∫v du. To apply integration by parts to ∫₀⁴ arctan(y/4) dy, we need to choose suitable functions for u and dv. A common strategy is to choose u as the function that becomes simpler when differentiated and dv as the rest of the integrand. In this case, a natural choice for u is arctan(y/4), since its derivative is a rational function, which is generally easier to work with. The remaining part of the integrand, dy, will be our dv. This choice sets the stage for applying the integration by parts formula and simplifying the integral. The success of integration by parts often hinges on the correct choice of u and dv, and this comes with practice and familiarity with different types of integrals. The goal is to transform the original integral into a simpler one that can be evaluated more easily. In this particular case, the choice of u = arctan(y/4) and dv = dy leads to a new integral that involves a rational function, which can be solved using standard techniques. This step is a crucial part of the overall process of evaluating the double integral, as it allows us to progress from a complex integral involving an inverse trigonometric function to a more manageable form. The careful application of integration by parts is a testament to the power of this technique in solving a wide range of integrals.

Let u = arctan(y/4) and dv = dy. Then, du = (1/4)/(1 + (y/4)²) dy = 4/(16 + y²) dy and v = y. Applying integration by parts:

∫₀⁴ arctan(y/4) dy = [y * arctan(y/4)] from 0 to 4 - ∫₀⁴ y * (4/(16 + y²)) dy

Step 3: Solving the Remaining Integral

The first term, [y * arctan(y/4)] from 0 to 4, is straightforward to evaluate: 4 * arctan(1) - 0 * arctan(0) = 4 * (π/4) = π. Now, let's tackle the remaining integral, -∫₀⁴ y * (4/(16 + y²)) dy. This integral can be solved using a substitution method. The remaining integral, -∫₀⁴ y * (4/(16 + y²)) dy, presents a slightly different challenge compared to the previous steps. Here, the substitution method proves to be the most effective approach. The key to using substitution lies in identifying a part of the integrand whose derivative is also present in the integral. In this case, the expression 16 + y² in the denominator is a good candidate, as its derivative, 2y, is closely related to the 'y' term in the numerator. By substituting u = 16 + y², we can simplify the integral significantly. This substitution transforms the integral into a form that is much easier to evaluate, typically involving a simpler function or a standard integral. The process of substitution is a fundamental technique in integral calculus and is widely used to solve a variety of integrals. It involves changing the variable of integration to simplify the integrand, making it more amenable to standard integration formulas. The success of the substitution method depends on the correct identification of the substitution variable, and this often requires practice and a good understanding of the structure of the integrand. In this particular case, the substitution u = 16 + y² not only simplifies the integral but also highlights the elegance and power of this technique in transforming complex problems into simpler ones.

Let u = 16 + y², then du = 2y dy. So, the integral becomes:

-∫₀⁴ y * (4/(16 + y²)) dy = -2 ∫ (1/u) du

We need to change the limits of integration as well: when y = 0, u = 16, and when y = 4, u = 16 + 4² = 32. Thus, the integral becomes:

-2 ∫₁₆³² (1/u) du = -2 [ln|u|] from 16 to 32 = -2 (ln(32) - ln(16)) = -2 ln(32/16) = -2 ln(2)

Step 4: Combining the Results

Finally, we combine the results from the previous steps. The outer integral was split into two parts, and we evaluated each part separately. Now, we add them together to get the final answer. Combining the results from the previous steps is the final stage in evaluating the double integral. This step requires careful attention to the individual components that were calculated earlier and how they fit together to form the complete solution. We had split the outer integral into two parts: the term resulting from integration by parts and the remaining integral that was solved using substitution. Each of these parts has its own value, and now we need to add them together. The term from integration by parts gave us π, while the remaining integral evaluated to -2 ln(2). Adding these two values gives us the final result for the double integral. This step underscores the importance of a systematic and organized approach to problem-solving in mathematics. By breaking down a complex problem into smaller, more manageable parts, we can tackle each part individually and then combine the results to obtain the final solution. This approach not only simplifies the calculations but also reduces the chances of errors. The final result, π - 2 ln(2), represents the definite value of the double integral, and it is the culmination of all the steps and techniques applied throughout the process.

∫₀⁴ arctan(y/4) dy = π - 2 ln(2)

Therefore, the value of the original double integral is:

∫₀⁴ ∫₀^(y²/4) y/(x² + y²) dx dy = π - 2 ln(2)

In conclusion, we have successfully evaluated the double integral ∫₀⁴ ∫₀^(y²/4) y/(x² + y²) dx dy. This process involved several key steps, including evaluating the inner integral using the arctangent function, applying integration by parts to the outer integral, using substitution to solve the remaining integral, and finally, combining the results. Each step required a careful understanding of the relevant calculus techniques and a systematic approach to the calculations. The final result, π - 2 ln(2), represents the definite value of the integral. Evaluating double integrals like this one is a fundamental skill in calculus and has wide applications in various fields, including physics, engineering, and computer science. The techniques used here, such as integration by parts and substitution, are essential tools in the calculus toolkit and are applicable to a wide range of problems. The ability to break down a complex integral into simpler parts and solve each part methodically is crucial for success in integral calculus. This article has provided a detailed walkthrough of the solution process, highlighting the key steps and techniques involved. By understanding these steps and practicing similar problems, one can develop a strong foundation in evaluating double integrals and apply these skills to solve more complex problems in the future.