### Using Transformations to Evaluate Integrals on ParallelogramsIn this section, we will demonstrate how to use the transformation $ u = y - x $, $ v = y $ to evaluate the given integral on a parallelogram. #### Problem StatementEvaluate the integral:\[\iint_R (y^2 - xy) \, dA\]where $ R $ is the region inside the parallelogram with vertices (0, 0), (1, 0), (2, 1), and (1, 1).#### Steps:1. **Transformation:** - Let $ u = y - x $ - Let $ v = y $2. **Illustration of the Region $ R $:** In the provided figure, we have a parallelogram plotted on the $ xy $-coordinate system. - The vertices of the parallelogram are: - (0, 0) - (1, 0) - (2, 1) - (1, 1) - The blue shaded area represents the region $ R $ over which the integral will be evaluated.3. **Description of the Graph:** - The $ x $-axis ranges from 0 to 2, while the $ y $-axis ranges from 0 to 1. - The parallelogram, illustrated in blue, is oriented such that it has a skewed shape with its sides parallel to the line $ y = x $. 4. **Jacobian Determinant:** Applying the transformation requires calculating the Jacobian determinant to adjust for the change of variables from $ (x,y) $ to $ (u,v) $.#### Conclusion:By transforming the region $ R $ using the given substitutions, we can convert the integral into a form that might be easier to evaluate. This is especially useful when dealing with complex regions, such as a non-rectangular parallelogram.### Illustration\[ \iint_R (y^2 - xy) \, dA \]- **Graph**: A parallelogram on the $ xy $-plane, vertices at (0, 0), (1, 0), (2, 1), and (1, 1). The shaded region denotes the area of integration.This method demonstrates the power of using

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Answered: Use the transformation u = y - x, v =…

Use the transformation u = y - x, v = y, to evaluate the integral on the parallelogram R of vertices (0, 0), (1, 0), (2, 1), and (1, 1) shown in the figure. (y2 - xy) dA

Calculus: Early Transcendentals

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ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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### Using Transformations to Evaluate Integrals on Parallelograms

In this section, we will demonstrate how to use the transformation $ u = y - x $, $ v = y $ to evaluate the given integral on a parallelogram.

#### Problem Statement
Evaluate the integral:

\[
\iint_R (y^2 - xy) \, dA
\]

where $ R $ is the region inside the parallelogram with vertices (0, 0), (1, 0), (2, 1), and (1, 1).

#### Steps:

1. **Transformation:**
- Let $ u = y - x $
- Let $ v = y $

2. **Illustration of the Region $ R $:**
In the provided figure, we have a parallelogram plotted on the $ xy $-coordinate system.

- The vertices of the parallelogram are:
- (0, 0)
- (1, 0)
- (2, 1)
- (1, 1)

- The blue shaded area represents the region $ R $ over which the integral will be evaluated.

3. **Description of the Graph:**
- The $ x $-axis ranges from 0 to 2, while the $ y $-axis ranges from 0 to 1.
- The parallelogram, illustrated in blue, is oriented such that it has a skewed shape with its sides parallel to the line $ y = x $.

4. **Jacobian Determinant:**
Applying the transformation requires calculating the Jacobian determinant to adjust for the change of variables from $ (x,y) $ to $ (u,v) $.

#### Conclusion:
By transforming the region $ R $ using the given substitutions, we can convert the integral into a form that might be easier to evaluate. This is especially useful when dealing with complex regions, such as a non-rectangular parallelogram.

### Illustration
\[ \iint_R (y^2 - xy) \, dA \]

- **Graph**: A parallelogram on the $ xy $-plane, vertices at (0, 0), (1, 0), (2, 1), and (1, 1). The shaded region denotes the area of integration.

This method demonstrates the power of using

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