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
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
8th Edition
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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2e331d2d-c7b3-4724-8206-f82883f9d427%2F6e14c757-2526-4272-a497-ceead1a16a3b%2Ft8fnf7k_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### 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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