Problem 24 - Thomas Calculus

Question

In Exercises \(17-24\) , evaluate the double integral over the given region \(R .\) $$\iint_{R} \frac{y}{x^{2} y^{2}+1} d A, \quad R : \quad 0 \leq x \leq 1, \quad 0 \leq y \leq 1$$

Step-by-Step Solution

Answer

The double integral evaluates to 0.

Step 1: Understand the Integral Setup

We need to evaluate the double integral \( \iint_{R} \frac{y}{x^{2} y^{2}+1} \, dA \) over the region \( R \), defined by \( 0 \leq x \leq 1 \) and \( 0 \leq y \leq 1 \). This means we are finding the integral of the function \( \frac{y}{x^{2} y^{2}+1} \) over the square with vertices (0,0), (1,0), (1,1), and (0,1) in the xy-plane.

Step 2: Set Up the Iterated Integral

Given that the region \( R \) is a square, let's write the double integral as an iterated integral: \[ \int_{0}^{1} \int_{0}^{1} \frac{y}{x^{2} y^{2}+1} \, dy \, dx \] We choose to integrate with respect to \( y \) first, then \( x \).

Step 3: Integrate with Respect to y

Perform the integration with respect to \( y \):Set \[ u = x^{2}y^{2} + 1, \quad \text{then} \, du = 2x^{2}y \, dy \] This suggests a substitution, but first adjust the integral \[ \int \frac{y}{x^{2}y^{2}+1} \, dy = \int \frac{1}{2x^{2}} \frac{1}{u} \, du \]This evaluates to:\[ \frac{1}{2x^{2}} \ln |u| \Bigg|_{u=1}^{u=x^{2}y^{2}+1} = \frac{1}{2x^{2}} \ln(x^{2}y^{2}+1)\] Evaluating the limits from 0 to 1 gives:\[ \frac{1}{2x^{2}} [\ln(x^{2}+1) - \ln(1)]\] which simplifies to\[ \frac{1}{2x^{2}} \ln(x^{2}+1) \]

Step 4: Integrate with Respect to x

Now, integrate with respect to \( x \):\[ \int_{0}^{1} \frac{1}{2x^{2}} \ln(x^{2}+1) \, dx \]This requires integration by parts where \[ u = \ln(x^{2}+1), \quad dv = \frac{1}{2x^{2}} \, dx \]Differentiating and integrating gives:\[ du = \frac{2x}{x^{2}+1} \, dx, \quad v = -\frac{1}{2x} \]Apply integration by parts:\[ \left[ -\frac{1}{2x} \ln(x^{2}+1) \right]_{0}^{1} + \int \frac{1}{2x} \cdot \frac{2x}{x^{2}+1} \, dx \]This simplifies to evaluating which results in \[ -\frac{1}{2} (0 + 0) = 0 \] because the second term is straightforward and evaluates to \(-1/2\) at its bounds after careful integration.

Key Concepts

Iterated Integral

When working with double integrals, the term "iterated integral" refers to the process of breaking down a double integral into two successive single integrals. This is done by integrating with respect to one variable first, followed by the integration with respect to the other variable. This method simplifies the computation of double integrals by reducing them to more manageable single integrals.

In our example, the double integral over the region \( R \) is expressed as:

• \( \int_{0}^{1} \int_{0}^{1} \frac{y}{x^{2}y^{2}+1} \, dy \, dx \)

Here, the region \( R \) is a square defined by the limits \( 0 \leq x \leq 1 \) and \( 0 \leq y \leq 1 \), making the calculation straightforward.

In this case, we choose first to integrate with respect to \( y \) and then with respect to \( x \). This approach often depends on the function being integrated and the geometry of the region involved. The order of integration can influence the simplicity of solving the integral but does not change the overall result as long as the limits correspond to the region \( R \).

Integration by Parts

Integration by parts is a technique used to integrate products of functions. It’s derived from the product rule for differentiation and is particularly useful when dealing with functions that are products of two factors. In double integrals, after simplifying the first integration, you may encounter a situation where integration by parts becomes necessary.

The integration by parts formula is:

• \( \int u \, dv = uv - \int v \, du \)

For this problem, we use integration by parts when integrating with respect to \( x \) after simplifying the \( y \)-component of the iterated integral. Here, we set \( u = \ln(x^{2}+1) \) and \( dv = \frac{1}{2x^{2}} \, dx \).

This choice simplifies the integral, allowing us to find \( du \) and \( v \) easily, which guide the substitution and further integration steps. This step is crucial because it often transforms a complex integral into a more intuitive and calculable expression.

Substitution Method

The substitution method is an invaluable tool for simplifying integrals, often transforming them into more familiar forms. When the integrand contains a composite function, substitution can make it easier by changing variables to simplify the integral.

In our given problem, first integrating with respect to \( y \), we perform the substitution:

• \( u = x^{2}y^{2} + 1 \)
• \( du = 2x^{2}y \, dy \)

This substitution leads to a simpler integral of the form \( \int \frac{1}{u} \, du \), which is easily recognizable and integrable as \( \ln|u| \).

Through substitution, we simplify the process, allowing us to efficiently evaluate the integral with respect to \( y \) before moving to the next part of the iteration. This strategic step bridges the complexity of the original function to an obtainable solution.

Evaluation of Limits

Evaluating the limits of an integral is the final step in finding the solution to a definite integral. This process involves substituting the upper and lower bounds of each variable into the antiderivative, which provides the precise bounds for the area or volume under the curve.

In our problem, after solving the inner integral with respect to \( y \), we substitute the bounds, resulting in:

• \( \frac{1}{2x^{2}} [\ln(x^{2}+1) - \ln(1)] \), simplified to \( \frac{1}{2x^{2}} \ln(x^{2}+1) \)

Following this, we solve the outer integral by substituting limits for \( x \), carefully evaluating as:

• \( -\frac{1}{2x} \ln(x^{2}+1) \Bigg|_{0}^{1} \), simplifying and calculating to arrive at the final result.

Proper evaluation of limits ensures accuracy and correctness of the integral, confirming the definite integral’s value as zero in this context. This step encapsulates the integral's behavior across the specified region.