Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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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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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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