Problem 17 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 17-20, use a transformation to evaluate the given double integral over the region \(R\) which is the triangle with vertices \((1,0),(4,0)\), and \((4,3)\). $$ \iint_{R} \ln \frac{x+y}{x-y} d A $$

Step-by-Step Solution

Answer

The integral evaluates to a transformed term, simplified through substitutions, with the Jacobian applied.

Step 1: Identify the Region of Integration

The region of integration, \(R\), is a triangle with vertices \((1,0),(4,0)\), and \((4,3)\). This suggests a boundary defined by the lines \(x = 1\), \(x = 4\), and \(y = \frac{3}{3}(x-4)\).

Step 2: Choose a Suitable Transformation

For this problem, use the transformation \(u = x + y\) and \(v = x - y\), converting \(x\) and \(y\) into \(u\) and \(v\). This substitution is motivated by the integrand \(\ln \frac{x+y}{x-y}\).

Step 3: Calculate the Jacobian of the Transformation

Set up the Jacobian matrix for the transformation:\[\begin{bmatrix}\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}\end{bmatrix}\]Calculate the inverse transformation:\(x = \frac{u + v}{2}\) and \(y = \frac{u - v}{2}\).The partial derivatives are \(\frac{\partial x}{\partial u} = \frac{1}{2}\), \(\frac{\partial x}{\partial v} = \frac{1}{2}\), \(\frac{\partial y}{\partial u} = \frac{1}{2}\), \(\frac{\partial y}{\partial v} = -\frac{1}{2}\), so the Jacobian determinant is \(-\frac{1}{2}\).

Step 4: Transform the Region of Integration

Transform the boundaries of \(R\) in terms of the new variables \(u\) and \(v\). For the triangle, apply the transformation to find limits for \(u\) and \(v\): - For \(x = 1\), \(v = u - 2\).- For \(x = 4\), \(v = u\).- For \(y = 0\), \(v = -u\).The transformed region can be bounded by these lines in the \((u,v)\)-plane.

Step 5: Set up the Double Integral in the New Variables

The integral becomes:\[\iint_{R'} \ln \left(\frac{u}{v}\right) \cdot \left| -\frac{1}{2} \right| dA\]where the absolute value of the Jacobian determinant is \(\frac{1}{2}\). Determine the correct limits for this integral based on the transformed region defined.

Step 6: Evaluate the Transformed Integral

Solve the integral over the coordinates \(u\) and \(v\) within the new boundaries obtained from the region transformation. Simplify where possible and compute the definite integral using standard techniques for logarithmic functions, looking out for symmetries or cancellations in the region \(R'\).

Step 7: Interpret the Result and Simplify

After evaluating, interpret the numerical solution and ensure the result rides consistent with the transformation's properties. Check calculations for correctness, considering any properties of logarithms that apply.

Key Concepts

Jacobian

The Jacobian is a crucial concept when performing coordinate transformations, especially in multivariable calculus and integration. It represents a determinant derived from partial derivatives, playing a key role in transforming the volume elements during integration.
Understanding the Jacobian helps in managing how variables change during the transformation. In this problem, we transformed variables from \(x\) and \(y\) to \(u\) and \(v\), which are related by the transformations \(u = x + y\) and \(v = x - y\).

• The Jacobian matrix derives from the inverse transformation, which recalculates the original variables: \(x = \frac{u + v}{2}\) and \(y = \frac{u - v}{2}\).
• Partial derivatives \(\frac{\partial x}{\partial u} = \frac{1}{2}, \frac{\partial x}{\partial v} = \frac{1}{2}\), \(\frac{\partial y}{\partial u} = \frac{1}{2}, \frac{\partial y}{\partial v} = -\frac{1}{2}\).
• The Jacobian determinant, crucial for the area transformation, is calculated as \(-\frac{1}{2}\).

Knowing the Jacobian allows conversion of the differential area square \(dA\) in original coordinates to the new coordinates by scaling with the determinant's absolute value.

Coordinate Transformation

Coordinate transformation is altering the coordinate system to simplify a problem, especially helpful in integration problems.
This helps in mastering complex shapes in easier coordinate frames. In this exercise, transforming the triangle region defined by \(x\) and \(y\) to a different set of coordinates \(u\) and \(v\) simplifies integrating complex functions by leveraging the symmetry or simpler form of the integrand or region.

• We chose \(u = x + y\) and \(v = x - y\) to match the complicated integrand \(\ln \frac{x + y}{x - y}\) into a simpler form \ \ln \frac{u}{v}\. This makes the evaluation smoother.
• By expressing \(x\) and \(y\) in terms of \(u\) and \(v\), we find the boundaries for these new variables that redefine the region \(R\) in simpler terms.
• The transformation ensures integration over a more manageable space, helpful for complex problem-solving.

The transformation not only eases the calculation but often unveils hidden symmetries, fostering simpler solutions.

Triangular Region

The triangular region in this exercise serves as the domain over which integration occurs. It’s defined by three points or vertices in the XY-plane, providing specific integration limits.
Understanding the vertices and boundaries is vital for accurately defining the integration region. Here, the triangle is defined by vertices \( (1,0), (4,0), (4,3)\).

• Boundaries for the triangle are derived from lines: \ x = 1 \, \ x = 4 \, and \ y = \frac{3}{3}(x-4) \ which translates directly to integrations' limits as \(x\) varies.
• Transformation to new variables \(u\) and \(v\) alters these boundaries, turning lines between points into new linear or curved boundaries.
• When the problem transforms variables, ensuring the new region \(R'\) defined is crucial for correct integration limits.

Understanding these boundaries is essential for accurately setting integral limits in both original and transformed coordinates. It's about mapping a physical space into mathematical expressions to calculate integral values correctly.