91Ó°ÊÓ

When dealing with integrals over complex regions, transforming variables can simplify calculations. In this exercise, we were presented with a double integral over a parallelogram. Instead of using the Cartesian \((x, y)\) coordinates directly, we applied a transformation that maps these coordinates to \((u, v)\) coordinates.
The transformation formulas are given by:\[ x = \frac{1}{4}(u+v), \quad y = \frac{1}{4}(v - 3u) \]This substitution is aimed at simplifying the region of integration. By using the mapped vertices from the original coordinates, we can redefine the limits for a simpler geometric representation.
This approach often reduces the complexity of the integral calculation and allows us to compute the integral over a transformed region in a straightforward way.

The Jacobian determinant plays a crucial role when transforming coordinates. It helps account for how areas scale as they transform from one set of variables to another. In our exercise, after expressing \(x\) and \(y\) in terms of \(u\) and \(v\), the Jacobian determinant \(J\) was calculated through the partial derivatives:

• \(\frac{\partial x}{\partial u} = \frac{1}{4}, \frac{\partial x}{\partial v} = \frac{1}{4}\)
• \(\frac{\partial y}{\partial u} = -\frac{3}{4}, \frac{\partial y}{\partial v} = \frac{1}{4}\)

The Jacobian determinant is given by:
\[J = \begin{vmatrix} \frac{1}{4} & \frac{1}{4} \ -\frac{3}{4} & \frac{1}{4} \end{vmatrix} = \left( \frac{1}{4} \cdot \frac{1}{4} \right) - \left( \frac{1}{4} \cdot \left( -\frac{3}{4} \right) \right) = \frac{1}{4}\]
The determinant helps account for the change in area when switching between the \((x, y)\) and \((u, v)\) coordinate systems, ensuring integral results are consistent.

A double integral computes the volume under a surface in a defined region, usually in 2D space. Here, our double integral involves a function over the parallelogram region \(R\). Using the change of variables and the Jacobian, the integral becomes easier to solve.
Once transformed, the original integrand \(4x + 8y\) in terms of \(u\) and \(v\) becomes \(-5u + 2v\). Combined with the Jacobian determinant \(\frac{1}{4}\), our integral is defined as:
\[\iint_{R} (4x + 8y) \, dA = \int_{0}^{4}\int_{4-u}^{8} (-5u + 2v) \cdot \frac{1}{4} \, dv \, du\]
The next step requires evaluating this expression by integrating with respect to each variable within the specified limits, leading us to find the value of the integral over the transformed region.

Parametrization is a method of turning geometrical shapes into algebraic expressions. In our exercise, it involves changing from the \((x, y)\) plane to a \((u, v)\) plane using determined transformations. This step helps redefine the boundaries of the region of interest.
First, we identified new bounds from the mapping of vertices for the \(uv\)-plane, which are:

• \(0 \leq u \leq 4\)
• For each fixed \(u\), \(4 - u \leq v \leq 8\)

These constraints define where we compute the integral and ensure that we correctly cover the original parallelogram. Parametrizing in this manner simplifies both the region of integration and often the integrand itself.