Explanation Given: The domain D : D = { ( x , y ) : 1 ≤ x + y ≤ 4 , − 4 ≤ y − 2 x ≤ 1 } The map F: u = x + y , v = y − 2 x G is inverse of map F . The double integral: ∬ D e x + y d x d y Equation 14: Jac ( G ) = Jac ( F ) − 1 Formula used: Jacobian = [ ∂ x ∂ u ∂ x ∂ v ∂ y ∂ u ∂ y ∂ v ] Calculation: We have been given map F as: u = x + y , v = y − 2 x Let us first sketch D by sketching the given set of curves: D = { ( x , y ) : 1 ≤ x + y ≤ 4 , − 4 ≤ y − 2 x ≤ 1 } To determine (b) To evaluate: The given integral

4th Edition

ISBN: 9781319323394

Author: Rogawski

Publisher: MAC HIGHER

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Coordinate Geometry

Coordinate geometry, also known as analytic geometry or Cartesian geometry in classical mathematics, is a type of geometry that is studied using a coordinate system. Coordinate geometry is a branch of mathematics that describes the position of points on …

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Chapter 16.6, Problem 40E

To determine

(a)

To compute:

The value of $J a c (G)$ .

To determine

(b)

To evaluate:

The given integral

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Chapter 16.6, Problem 39E

Chapter 16.6, Problem 41E

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Chapter 16 Solutions

CALCULUS W/SAPLING ACCESS >IC<

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(a) To compute: The value of J a c ( G ) .

Solution Summary: The author explains how to compute the value of J a c (G) using the equation 14.

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(a)

To compute:

The value of $J a c (G)$ .

(b)

To evaluate:

The given integral

Want to see the full answer?

Chapter 16 Solutions

(a) To compute: The value of J a c ( G ) .

Solution Summary: The author explains how to compute the value of J a c (G) using the equation 14.

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(a) To compute: The value of Jac(G).

(b) To evaluate: The given integral

Want to see the full answer?

Chapter 16 Solutions

(a)

To compute:

The value of $J a c (G)$ .

(b)

To evaluate:

The given integral