Problem 19 - Advanced Engineering Mathematics

Question

Evaluate the double integral over the region \(R\) that is bounded by the graphs of the given equations. Choose the most convenient order of integration. $$ \iint_{R} \frac{y}{1+x y} d A ; y=0, y=1, x=0, x=1 $$

Step-by-Step Solution

Answer

The value of the integral is \( 1 - \frac{\pi^2}{12} \).

Step 1: Understanding the Region of Integration

The region \( R \) is bounded by \( y=0 \), \( y=1 \), \( x=0 \), and \( x=1 \). This describes a square region in the xy-plane with vertices at \((0,0), (1,0), (1,1), (0,1)\).

Step 2: Choosing the Order of Integration

Since the region is a square, we can choose either order of integration conveniently. Here, we choose to integrate with respect to \( y \) first and then \( x \). So, the integral becomes \( \int_{0}^{1} \left( \int_{0}^{1} \frac{y}{1+xy} dy \right) dx \).

Step 3: Inner Integral with Respect to y

Evaluate the inner integral: \( \int_{0}^{1} \frac{y}{1+xy} dy \). Let \( v = 1 + xy \), then \( dv = x \, dy \), or \( dy = \frac{dv}{x} \). When \( y = 0 \), \( v = 1 \), and when \( y = 1 \), \( v = 1 + x \). Change the integral to \( \frac{1}{x} \int_{1}^{1+x} \frac{v-1}{v} dv \).

Step 4: Evaluating the Changed Integral

The integral \( \frac{1}{x} \int_{1}^{1+x} \left( 1 - \frac{1}{v} \right) dv \) becomes \( \frac{1}{x} \left( (v-1) - \ln|v| \right) \Big|_{1}^{1+x}\), which simplifies to \( \frac{1}{x} \left( x - \ln(1+x) \right) \). Therefore, the result of the inner integral is \( 1 - \frac{\ln(1+x)}{x} \).

Step 5: Outer Integral with Respect to x

Now, integrate the result of the inner integral with respect to \( x \): \( \int_{0}^{1} \left(1 - \frac{\ln(1+x)}{x} \right) dx \). This integral can be simplified to: \( \int_{0}^{1} 1 \, dx - \int_{0}^{1} \frac{\ln(1+x)}{x} dx \). The first term evaluates to \( 1 \), and the second integral is a known integral which evaluates to \( \frac{\pi^2}{12} \).

Step 6: Final Result

Subtract the evaluated integral of the second part from the first part: \( 1 - \frac{\pi^2}{12} \). This is the value of the given double integral.

Key Concepts

Region of Integration

In the process of evaluating a double integral, understanding the region of integration is a fundamental step. For this particular problem, the region \( R \) is defined by the bounds \( y=0 \), \( y=1 \), \( x=0 \), and \( x=1 \). This description indicates that the region is a simple square on the coordinate plane. The vertices of this square are at points \((0,0), (1,0), (1,1), (0,1)\).

This specific set-up makes the double integration task straightforward since the boundaries align along constant lines, forming a rectangle in Cartesian coordinates.
Understanding this layout is crucial as it helps to visualize the area over which the integration is performed.

The clarity of the region also simplifies the process of choosing integration orders and performing subsequent calculations.

Order of Integration

Choosing the order of integration can greatly influence the complexity of evaluating a double integral. In this example, since the defined region is a simple square, we have the flexibility to choose either order conveniently.

For simplification, it is often beneficial to integrate with respect to the variable that simplifies the inner integral more easily. Here, we choose to integrate with respect to \( y \) first, followed by \( x \).

By arranging the integral in this order, the expression becomes:

• \( \int_{0}^{1} \left( \int_{0}^{1} \frac{y}{1+xy} \, dy \right) \, dx \).

This decision simplifies the integration evaluation, breaking the task into manageable steps. Balancing ease of computation with the nature of the functions involved helps achieve a more efficient problem-solving process.

Integral Simplification

Simplifying integrals is an essential step in evaluating expressions within the bounds set by the region of integration. For the inner integral with respect to \( y \), simplification involves a substitution method to tackle the function \( \frac{y}{1+xy} \).
Using substitution:

• Let \( v = 1 + xy \).
• Then \( dv = x \, dy \), or \( dy = \frac{dv}{x} \).
• Change the limits of integration accordingly: when \( y = 0 \), \( v = 1 \), and when \( y = 1 \), \( v = 1 + x \).

These substitutions transform the integral into a more manageable form:

• \( \frac{1}{x} \int_{1}^{1+x} \frac{v-1}{v} \, dv \).

With this change, evaluation becomes simpler and leads directly to the simplified result, easing the way to calculate the outer integral.

Known Integrals

After simplifying expressions, using known integrals speeds up solving the problem. Once the change in variables simplifies the inner integral, the evaluated form can often be matched to known integrals from integral tables or mathematical references.

In this scenario, once the inner integral simplifies to \( 1 - \frac{\ln(1+x)}{x} \), we evaluate it as an outer integral from \( x=0 \) to \( x=1 \):

• \( \int_{0}^{1} \left(1 - \frac{\ln(1+x)}{x} \right) \, dx \)

Breaking it down further, the integration splits into two parts:

• \( \int_{0}^{1} 1 \, dx = 1 \)
• \( -\int_{0}^{1} \frac{\ln(1+x)}{x} \, dx \)

The first is straightforward, and the second term is a recognized integral which evaluates to \( \frac{\pi^2}{12} \). Utilizing these known integral results makes the integration both efficient and easily verifiable.