Question

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

Short answer

Transform and integrate using Jacobian; solve resulting integrals.

Step-by-step solution

Analyze the Region

The region \( R \) is a triangle with vertices \( (1,0), (4,0), (4,3) \). This triangle is in the Cartesian plane, with two sides along the x and y axes.

Set Up the Transformation

Choose a transformation to map the triangular region into a simpler region. For this exercise, a suitable transformation can be \( u = x + y \) and \( v = x - y \). This maps the equations of the sides to simpler boundaries.

Find the Bounds in New Variables

Using the vertices of \( R \):- At \( (1, 0) \): \( u = 1 + 0 = 1 \), \( v = 1 - 0 = 1 \).- At \( (4, 0) \): \( u = 4 + 0 = 4 \), \( v = 4 - 0 = 4 \).- At \( (4, 3) \): \( u = 4 + 3 = 7 \), \( v = 4 - 3 = 1 \).Thus, the bounds in \((u, v)\) are \(1 \leq v \leq 4\), and for each fixed \(v\), \(v+2 \leq u \leq v+3\).

Compute the Jacobian

The transformation equations are \( u = x + y \) and \( v = x - y \). To find the Jacobian \( J \) of the transformation: \[ J = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix} \]Find \( x = \frac{u + v}{2} \) and \( y = \frac{u - v}{2} \), then calculate the partial derivatives and determinant:\[ J = \begin{vmatrix} \frac{1}{2} & \frac{1}{2} \ \frac{1}{2} & -\frac{1}{2} \end{vmatrix} = -\frac{1}{2} \]},

Set Up the Integral

Transform the integral using \( u \) and \( v \). The original integral \( \iint_{R} \sqrt{\frac{x+y}{x-y}} \, dA \) becomes:\[ \int_{1}^{4} \int_{v+2}^{v+3} \sqrt{\frac{u}{v}} \cdot \left| -\frac{1}{2} \right| \, du \, dv \]\[ = \frac{1}{2} \int_{1}^{4} \int_{v+2}^{v+3} \sqrt{\frac{u}{v}} \, du \, dv \]

Evaluate the Inner Integral

Evaluating the inner integral \( \int_{v+2}^{v+3} \sqrt{\frac{u}{v}} \, du \):\[ \int \sqrt{\frac{u}{v}} \, du = \frac{2}{3} \left( \frac{u}{v} \right)^{3/2} \cdot v \]Evaluate it from \( u = v+2 \) to \( u = v+3 \).

Evaluate the Outer Integral

Substitute the evaluation of the inner integral into the outer integral:\[ \frac{1}{2} \int_{1}^{4} \left( \frac{2}{3} \left( (v+3)^{3/2} - (v+2)^{3/2} \right) \right) \, dv \]Simplify and compute this definite integral where necessary.

Simplify and Solve

Finally, integrate and simplify the expression to find the value of the original integral over the region \( R \). Carry out the computation carefully to find the exact solution to the evaluation within the given limits.

Key concepts

Coordinate Transformation

When solving double integrals over complex regions, a coordinate transformation can significantly simplify the problem. The goal of a transformation is to convert coordinates from one system to another, often making the region of integration simpler in the process. In our exercise, we transformed the Cartesian coordinates to a new system using

• \( u = x + y \)
• \( v = x - y \)

This choice allows us to express the given triangular region with new boundary lines that are easier to work with. Transformations like these are essential in calculus, as they often turn a difficult problem into a manageable one. Remember, the key in choosing a transformation is aligning it to simplify the region of integration.

Jacobian Determinant

The Jacobian determinant is a fundamental tool when changing variables for integration. It is necessary because it accounts for how area scales between the coordinate systems. Essentially, it tells us how much an infinitesimal piece of the original region stretches or shrinks in the new coordinates.For the transformation with

• \( u = x + y \)
• \( v = x - y \)

we find the inverse transformations:

• \( x = \frac{u + v}{2} \)
• \( y = \frac{u - v}{2} \)

With these, the Jacobian determinant \( J \) is:\[ J = \begin{vmatrix} \frac{1}{2} & \frac{1}{2} \ \frac{1}{2} & -\frac{1}{2} \end{vmatrix} = -\frac{1}{2} \]A negative Jacobian indicates a change in orientation, but what matters primarily is the absolute value, used to adjust the integrand.

Integral Bounds

Integral bounds define the limits within which we evaluate the integral. They are essentially the boundaries of the region in the new coordinates.In our case, after the transformation, we calculated the integral bounds in terms of \( u \) and \( v \):

• For \( v \), the bounds are from 1 to 4.
• For each fixed \( v \), \( u \) ranges from \( v+2 \) to \( v+3 \).

These bounds correspond to the transformed region, mapping the original triangle to a;relatively simple rectangular region in the \( u \)-\( v \) plane. When setting these bounds, ensure they correctly represent the shape and limits of the original region in the transformed coordinate system.

Region of Integration

Identifying the region of integration correctly is crucial before performing any integral transformation. Our starting point is a triangular region in the Cartesian plane, with vertices at

• \((1,0)\)
• \((4,0)\)
• \((4,3)\)

To transform this, we needed a new set of coordinates that maintain the integrity of the region's shape while simplifying its description.By converting this into the \( u \)-\( v \) plane, the complex shape becomes a simpler rectangle defined by the bounds we calculated. This makes the work of evaluating the integral much easier.Remember, visualizing the region and its bounds in both the original and transformed systems helps in understanding the underlying geometry and ensuring accurate integration.