Problem 14 - Advanced Engineering Mathematics

Question

Evaluate the given integral by means of the indicated change of variables. $$ \begin{aligned} &\iint_{R} \frac{\cos \frac{1}{2}(x-y)}{3 x+y} d A, \text { where } R \text { is the region bounded by the }\\\ &\text { graphs of } y=x, y=x-\pi, y=-3 x+3, y=-3 x+6 \text { ; }\\\ &u=x-y, v=3 x+y \end{aligned} $$

Step-by-Step Solution

Answer

The integral evaluates to zero.

Step 1: Understanding the Region R

The region R is defined by the lines: \(y = x\), \(y = x - \pi\), \(y = -3x + 3\), and \(y = -3x + 6\). The intersection of these lines determines the boundaries of the region. Calculating intersections gives four points: \(A(0,0)\), \(B(\pi, \pi)\), \(C(0,6)\), and \(D(\pi-1, -3(\pi-1)+6)\). This forms a trapezoidal area on the xy-plane.

Step 2: Define the Transformation

Using the change of variables \(u = x - y\) and \(v = 3x + y\), solve for \(x\) and \(y\) in terms of \(u\) and \(v\). Solving the system gives: \(x = \frac{3v + u}{10}\) and \(y = \frac{3v - 9u}{10}\). This transformation will simplify the integral.

Step 3: Calculate the Jacobian

The Jacobian \(J\) of the transformation is used to adjust the differential area element when changing variables. Calculate \(abla(x,y)/(u,v)\) as follows: \[abla(x,y)/(u,v) = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix} = \begin{vmatrix} \frac{1}{10} & \frac{3}{10} \ -\frac{9}{10} & \frac{3}{10} \end{vmatrix}\] Evaluating, \(J = \frac{3}{10} + \frac{9}{100} = \frac{12}{100} = \frac{3}{25}\). Thus, \(dA = \left| \frac{3}{25} \right| du dv\).

Step 4: Transform Region R in uv-plane

Determine how the lines defining R transform in terms of \(u\) and \(v\). For example, \(y = x\) becomes \(u = 0\), \(y = x - \pi\) becomes \(u = -\pi\), \(y = -3x + 3\) becomes \(v = 3\), and \(y = -3x + 6\) becomes \(v = 6\). This gives a rectangle in the uv-plane with corners: \((0,3), (-\pi,3), (0,6), (-\pi,6)\).

Step 5: Solve the Integral in uv-plane

Rewrite the integral in terms of \(u\) and \(v\), and include the Jacobian: \[\int_{-\pi}^{0} \int_{3}^{6} \frac{\cos \frac{1}{2} u}{v} \frac{3}{25} \, dv \, du\]. Separate the integrals and solve: integrate with respect to \(v\) first and then \(u\). The integral becomes: \[\frac{3}{25} \int_{-\pi}^{0} \cos \frac{1}{2} u \, du \int_{3}^{6} \frac{1}{v} \, dv\]. Calculate the \(v\)-integral as \([\ln v]_{3}^{6}\) and the \(u\)-integral as \([2\sin(\frac{1}{2} u)]_{-\pi}^{0}\). Evaluating gives: \([\ln(6/3)] = \ln(2)\) and \(-2\sin(\frac{\pi}{2}) + 2\sin(\frac{-\pi}{2}) = 0\). Thus, the integral evaluates to zero.

Key Concepts

Double Integral

A double integral is an extension of a single integral and is used to accumulate quantities over a two-dimensional area. It allows us to compute areas, volumes, and other quantities that vary across a plane. In mathematics, the double integral of a function \( f(x, y) \) over a region \( R \) is denoted as \( \iint_{R} f(x, y) \, dA \), where \( dA \) is the infinitesimal area element of the region \( R \).
When dealing with double integrals, we often integrate over regions bounded by curves or lines. These boundaries determine the limits of the integral. A typical procedure involves setting up the integral with respect to one variable while treating the other variable as a constant, and then integrating successively.

• Double integrals can be used to find area. For instance, if \( f(x, y) = 1 \), the double integral gives the area of region \( R \).
• They can also measure volumes beneath surfaces or in solid objects by integrating over the relevant region.

In the context of the given exercise, the double integral involves a complex function over a trapezoidal region, necessitating the use of additional techniques like substitution to simplify the computation.

Change of Variables

The change of variables is a powerful technique used in calculus to simplify the evaluation of integrals, especially in multivariable calculus. It involves substituting variables with new ones, making the integral easier to solve by transforming the region of integration.
The standard transformation uses variables \( u \) and \( v \) in place of the original variables \( x \) and \( y \). The intention is to turn the region of integration into one with straightforward boundaries, like a rectangle or a circle, and simplify the integrand.

• The old variables \( x \) and \( y \) are expressed in terms of new variables \( u \) and \( v \) by solving either a given substitution equation or derived equations.
• In the given exercise, the transformation uses \( u = x - y \) and \( v = 3x + y \). This choice turns the complex region into a simple rectangle in the \( uv \) plane.

After transforming variables, the integral usually becomes easier to evaluate, reducing complicated boundaries and integrands.

Jacobian

The Jacobian is a crucial concept in multivariable calculus that arises when changing variables in double integrals. It is the determinant of the matrix of all first-order partial derivatives, giving the factor by which the differential area element is scaled during the transformation.
During a variable transformation \( (x, y) \rightarrow (u, v) \), the Jacobian \( J \) relates the differential areas \( du \, dv = J \, dx \, dy \). The formula for the Jacobian is given by:
\[abla(x,y)/(u,v) = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix}\]
Here, each term represents a partial derivative of \( x \) and \( y \) with respect to \( u \) and \( v \). Determining the Jacobian's value ensures that the transformed integral accurately accounts for the area change.

• It is essential to calculate this correctly to avoid mistakes in solving transformed integrals.
• In the example, the Jacobian is calculated to be \( \frac{3}{25} \), which scales the differential area in the transformed uv-plane.

With the Jacobian, you can confidently solve the integral over the newly defined region.

Trapezoidal Region

A trapezoidal region in geometry is a four-sided figure with one pair of parallel sides. It is a common type of region over which double integrals can be evaluated, thanks to its simple shape and well-defined boundaries.
In exercises like the one presented, defining the region accurately is vital for setting up the integral. In the Cartesian plane, a trapezoidal region can typically be identified by knowing the equations of its bounding lines. By solving for their intersections, you can pinpoint the vertices of the trapezoid.

• The current exercise describes a trapezoidal region defined by four lines: \( y = x \), \( y = x - \pi \), \( y = -3x + 3 \), and \( y = -3x + 6 \).
• Calculating their intersections yields points which form the vertices of the trapezoid.

This allows creating a bounded region suitable for double integration. After transforming variables, this region can be reshaped into simpler forms like a rectangle, easing the evaluation process. Understanding the geometric nature of the region assists in visualizing the area covered by the integral.