91Ó°ÊÓ

When solving double integrals, it's often useful to simplify the region of integration by transforming variables. This technique is called the transformation of variables and it helps in changing the coordinates to make complex integration regions simpler.

For instance, in our exercise, we transformed variables by setting:

• \( u = x - y \)
• \( v = x + y \)

Using these transformations allows us to describe the integration region more easily. The original square, defined with vertices, is transformed into a parallelogram in the new coordinate system.

The transformation makes it easier to handle the integration because the complexity of the region's shape is reduced.

In transforming coordinate systems, the Jacobian determinant is crucial because it accounts for the change in area (or volume, in higher dimensions) resulting from the transformation.

For the transformation in our exercise, where

• \( u = x - y \)
• \( v = x + y \)

we calculate the Jacobian as the determinant of the matrix formed by the partial derivatives of the new variables with respect to the old ones:\[J = \left| \begin{array}{cc} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{array} \right| = \left| \begin{array}{cc} \frac{1}{2} & \frac{1}{2} \-\frac{1}{2} & \frac{1}{2} \end{array} \right| = \frac{1}{2}.\]The Jacobian value of \(\frac{1}{2}\) indicates how the area is scaled in the transformation. This factor is essential in altering the differential area from \(dx \times dy\) to \(du \times dv\).

Determining the limits of integration is essential when transferring the integral to a new set of variables. After transforming to the \(uv\) plane using the transformation, the original square region is mapped onto a parallelogram.

To find the new limits:

• The variable \(v\) has the limits of 2 to 6, determined by the transformed equations of the square's boundaries.
• The variable \(u\) is bound between \(-v + 2\) and \(v - 2\).

Setting these correct limits ensures you're integrating over the correct area in the new coordinate plane.

Properly understanding and determining these limits is fundamental when using a transformation of variables.

A definite integral calculates the exact area under a curve within specified bounds. In double integration, you are finding the volume under a surface. After transforming the variables and setting the new limits, the original problem becomes easier to solve.

The integrand in this example is converted to \(\frac{u^2}{v^2}\) due to the variable transformation. We first integrate with respect to \(u\), and then with respect to \(v\). For the inner integral:\[\int_{-v+2}^{v-2} \frac{1}{2} \cdot \frac{u^2}{v^2} \, du = \frac{1}{6v^2}[(v-2)^3 - (-v+2)^3]\]This result is then used in the outer integral:\[\int_{2}^{6} \frac{1}{6v^2}[(v-2)^3 - (-v+2)^3] \, dv\]Finally, you compute the definite integral to find the solution.

In our problem, symmetry in the region yields a final integral value of zero, which simplifies interpretation.