91Ó°ÊÓ

In calculus, the technique known as a change of variables is crucial for simplifying complex integral calculations. It is especially useful when dealing with complicated region shapes, such as parallelograms, where direct integration may be challenging. By choosing appropriate substitution variables, we can transform the original integral into an easier one.
In the given problem, the region is a parallelogram. The chosen variables are:

• \( u = x - y \)
• \( v = x + 4y \)

These substitutions are selected to simplify the function and region to more manageable forms, allowing for straightforward evaluation of the integral. The choice of substitutions depends on the geometry of the region and the complexity of the function involved.

A parallelogram is a four-sided polygon with opposite sides that are equal in length and parallel. This geometric shape is significant in calculus when setting up specific coordinate transformations, especially useful in the change of variables.
For the exercise, the region \( R \) bounded by the vertices \((0,0), (1,1), (5,0),\) and \((4,-1)\) forms a parallelogram. This shape suggests an appropriate transformation that maintains the structure and symmetries, aiding in simplification.

• The parallelogram's slant and orientation often guides the choice of substitution variables to linearize the boundaries.
• Once mapped onto a rectangle in the \( uv \)-plane, the integration limits become straightforward, easing the computation.

The Jacobian determinant plays a central role when performing change of variables in multiple integrals. It determines how volume elements transform under the new variables. You can think of it as a scaling factor for area or volume when integrating over transformed regions.
For the change of variables, if \( u = u(x, y) \) and \( v = v(x, y) \), the Jacobian determinant is expressed as: \[J = \begin{vmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{vmatrix}\]The solution step-by-step examines this matrix and calculates:

• \( J = \frac{1}{5} \)

This determinant reflects how the transformation scales the area of the region being integrated over.

Double integrals are used to compute volumes under surfaces or the total amount of some quantity over a two-dimensional region. When dealing with transformations, the double integral includes evaluating the absolute value of the Jacobian, ensuring the area differential is always positive.
The formulation starts with the original surface \( z = f(x, y) \) and requires integrating \( \int\int f(x, y) \, dA \). After employing the change of variables, this translates to:\[V = \int\int f(u, v) |J| \, du \, dv\]In this problem, the absolute value of the Jacobian \( |J| \) ensures the positivity of the transformed area element. The limits of integration are determined from the transformed parallelogram on the \( uv \)-plane, simplifying the computation considerably.

• \( 0 \leq u \leq 1 \)
• \( 0 \leq v \leq 5 \)