Problem 14 - Thomas Calculus

Question

\begin{equation} \begin{array}{c}{\text { Use the transformation } x=u+(1 / 2) v, y=v \text { to evaluate the }} \\ {\text { integral }} \\ {\int_{0}^{2} \int_{y / 2}^{(y+4) / 2} y^{3}(2 x-y) e^{(2 x-y)^{2}} d x d y} \\ {\text { by first writing it as an integral over a region } G \text { in the uv-plane. }}\end{array} \end{equation}

Step-by-Step Solution

Answer

Transform the integral using the given coordinate transformation, evaluate each part successively, and determine the value.

Step 1: Understand the transformation

Recognize that a coordinate transformation is given by \( x = u + \frac{1}{2}v \) and \( y = v \). This transformation can change the integral's region and simplify the process of integration.

Step 2: Determine the new region in the uv-plane

Substitute the given bounds of integration in the xy-plane to convert them to the uv-plane. Initially, we have \( 0 \leq y \leq 2 \) and \( \frac{y}{2} \leq x \leq \frac{y+4}{2} \). Since \( y = v \), \( 0 \leq v \leq 2 \). For \( x \), substitute \( x = u + \frac{1}{2}v \), therefore, \( \frac{v}{2} \leq u + \frac{1}{2}v \leq \frac{v+4}{2} \), leading to \( 0 \leq u \leq 2 \).

Step 3: Express the integral in terms of u and v

Change the differential expression and the function to the new variables using the transformation. The function becomes \( v^3 \times 2u e^{(2u)^2} \) because \( 2x - y = 2u \). The differential \( dx \) becomes \( du \), and \( dy = dv \). Change the bounds according to the region \( 0 \leq v \leq 2 \) and \( 0 \leq u \leq 2 \).

Step 4: Evaluate the transformed integral

The integral in terms of \( u \) and \( v \) is \( \int_{0}^{2} \int_{0}^{2} v^3 (2u) e^{(2u)^2} du \, dv \). First, evaluate the integral \( \int_{0}^{2} (2u) e^{(2u)^2} du \), and then multiply the result by \( v^3 \) and integrate with respect to \( v \).

Step 5: Integrate over u

Integrate the expression \( 2u e^{(2u)^2} \) with respect to \( u \) from 0 to 2. Notice that this is an improper integral but can be approached by a method like substitution, or observe directly if the resulting expressions make it straightforward.

Step 6: Integrate over v

Integrate the resultant expression of \( v^3 \) from the previous integration over the range \( 0 \leq v \leq 2 \). After performing the calculations, this provides the result of the integral.

Key Concepts

Integral Calculus

Integral calculus is a fundamental branch of mathematics focused on accumulation and area under curves. It helps in solving problems related to the overall amount gathered by a function, such as total distance traveled or particular areas. This exercise involves evaluating a double integral, which calculates the volume under a surface defined in the xy-plane.

The double integral from the original exercise is given by:

• \( \int_{0}^{2} \int_{\frac{y}{2}}^{\frac{y+4}{2}} y^{3}(2x-y) e^{(2x-y)^{2}} \ dx \ dy \)

The purpose of the exercise involves transforming the region of integration using a new coordinate system to simplify this process.

The concepts of substitution and bounds are key in solving integrals more effectively, allowing specific transformations that convert complex integrals into manageable expressions.

UV-Plane Transformation

A uv-plane transformation is a technique used in integral calculus to simplify the integration process by changing the coordinate system. In the given exercise, we're using the transformation:

• \( x = u + \frac{1}{2}v \)
• \( y = v \)

This helps redefine the original problem to make the region of integration easier to handle. By transforming variables, we translate the problem into new axes, which often simplifies the arithmetic involved.

Transformation is fundamental when the geometry of the region is complex or when simplifying the function gripping the integral. Here, this approach permits clearing the relations between variables, often revealing symmetries.

Integration Bounds

Integration bounds are limits specifying where integration starts and ends. They define the region over which you calculate an integral. In this exercise, original bounds were:

• For \( y \): \( 0 \leq y \leq 2 \)
• For \( x \): \( \frac{y}{2} \leq x \leq \frac{y+4}{2} \)

Upon applying the uv-plane transformation, this becomes a rectangle with bounds:

• For \( v \): \( 0 \leq v \leq 2 \)
• For \( u \): \( 0 \leq u \leq 2 \)

Within the new system, the region becomes a straightforward square in the uv-coordinate plane, simplifying the later steps of integration.

Setting and comprehending the bounds is crucial as they inform us of the exact area or volume encompassing the function's influence.

Change of Variables

The change of variables is a pivotal concept in calculus used to convert a difficult integral into a more convenient one. This is done by applying a transformation to variables, which simplifies the evaluation puzzle. In our problem, we switched from the xy-plane to the uv-plane.

Under this transformation, not only do the limits change, but we also adjust the integrand itself, which here shifted from \( y^3(2x-y)e^{(2x-y)^2} \) to \( v^3(2u)e^{(2u)^2} \). The use of a new differential, often denoted as the Jacobian, ensures accurate translation between variable systems. For our simplified case, due to the nature of the transformation, the Jacobian is 1, indicating a direct translation during this specific operation.

Understanding and applying changes of variables provide an avenue toward solving integrals that initially seem unwieldy or too complex, making it an invaluable tool in calculus.