Let R be the region in the first quadrant of the xy-plane bounded by the hyperbolas xy = 1, u xy = 4, and the lines y = x, y = 16x. Use the transformation x=y= uv with u> 0 and v> 0 to rewrite the integral below over an appropriate region G in the uv-plane. Then evaluate the uv-integral over G. |xy|dxdy [[[√ + √x) d | dxdy = | (Type an exact answer.)
Let R be the region in the first quadrant of the xy-plane bounded by the hyperbolas xy = 1, u xy = 4, and the lines y = x, y = 16x. Use the transformation x=y= uv with u> 0 and v> 0 to rewrite the integral below over an appropriate region G in the uv-plane. Then evaluate the uv-integral over G. |xy|dxdy [[[√ + √x) d | dxdy = | (Type an exact answer.)
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
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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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![### Transforming Integrals Using Coordinate Transformations in the First Quadrant
**Problem Description**
Consider the region \( R \) in the first quadrant of the xy-plane, bounded by the following boundaries:
- The hyperbolas \( xy = 1 \) and \( xy = 4 \).
- The lines \( y = x \) and \( y = 16x \).
You are required to use the transformation \( x = \frac{u}{v} \) and \( y = uv \) where \( u > 0 \) and \( v > 0 \) to:
1. Rewrite the integral over an appropriate region \( G \) in the uv-plane.
2. Evaluate the uv-integral over \( G \).
**Integral to be Transformed**
\[
\iint_{R} \left( \frac{y}{x} + \sqrt{xy} \right) \,dx\,dy
\]
---
### Steps in the Solution
1. **Transformation of Coordinates:**
- Given the transformation \( x = \frac{u}{v} \) and \( y = uv \), these new coordinates will help simplify the integration process.
2. **Boundaries in the uv-plane:**
- The hyperbolas \( xy = 1 \) and \( xy = 4 \) transform to \( u^2 = 1 \) and \( u^2 = 4 \), so \( u = 1 \) and \( u = 2 \) respectively (considering \( u > 0 \)).
- The lines \( y = x \) and \( y = 16x \) transform to \( uv = \frac{u}{v} \) and \( uv = \frac{16u}{v} \), simplifying to \( v^2 = 1 \) and \( v^2 = 16 \) respectively (considering \( v > 0 \)), which means \( v = 1 \) and \( v = 4 \).
3. **Jacobian Determinant Calculation:**
- When transforming from \( (x, y) \) to \( (u, v) \), the area element \( dx\,dy \) changes to \( |J| \, du\,dv \), where \( J \) is the Jacobian of the transformation.
- The Jacobian \( J \) for \( x](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffd49838c-9fe9-4c2b-b877-1f4ac641c60c%2Ff2b530d2-e298-455a-90b5-2685662cfc22%2Fiwu1ioh_processed.png&w=3840&q=75)
Transcribed Image Text:### Transforming Integrals Using Coordinate Transformations in the First Quadrant
**Problem Description**
Consider the region \( R \) in the first quadrant of the xy-plane, bounded by the following boundaries:
- The hyperbolas \( xy = 1 \) and \( xy = 4 \).
- The lines \( y = x \) and \( y = 16x \).
You are required to use the transformation \( x = \frac{u}{v} \) and \( y = uv \) where \( u > 0 \) and \( v > 0 \) to:
1. Rewrite the integral over an appropriate region \( G \) in the uv-plane.
2. Evaluate the uv-integral over \( G \).
**Integral to be Transformed**
\[
\iint_{R} \left( \frac{y}{x} + \sqrt{xy} \right) \,dx\,dy
\]
---
### Steps in the Solution
1. **Transformation of Coordinates:**
- Given the transformation \( x = \frac{u}{v} \) and \( y = uv \), these new coordinates will help simplify the integration process.
2. **Boundaries in the uv-plane:**
- The hyperbolas \( xy = 1 \) and \( xy = 4 \) transform to \( u^2 = 1 \) and \( u^2 = 4 \), so \( u = 1 \) and \( u = 2 \) respectively (considering \( u > 0 \)).
- The lines \( y = x \) and \( y = 16x \) transform to \( uv = \frac{u}{v} \) and \( uv = \frac{16u}{v} \), simplifying to \( v^2 = 1 \) and \( v^2 = 16 \) respectively (considering \( v > 0 \)), which means \( v = 1 \) and \( v = 4 \).
3. **Jacobian Determinant Calculation:**
- When transforming from \( (x, y) \) to \( (u, v) \), the area element \( dx\,dy \) changes to \( |J| \, du\,dv \), where \( J \) is the Jacobian of the transformation.
- The Jacobian \( J \) for \( x
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