Explanation Given : f ( x , y ) = ( ln y ) − 1 , Domain D bounded by y = e x and y = e x Calculation : Step 1: To express D as a horizontally simple region, we must first rewrite the equations of the curves y = e x and y = e x with x as a function of y. y = e x ⇒ x = ln y y = e x ⇒ x = ln y ⇒ x = ln 2 y We obtain the following inequalities: 1 ≤ y ≤ e , ln 2 y ≤ x ≤ ln y Step 2: Using Theorem 2, we compute the double integral of f ( x , y ) = ( ln y ) − 1 over D as the following iterated integral...

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ISBN: 9781319323394

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 16.2, Problem 31E

To determine

The value of integral of $f (x, y) = {(\ln y)}^{- 1}$ over the domain D bounded by $y = e^{x}$ and $y = e^{\sqrt{x}}$ .

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Chapter 16.2, Problem 30E

Chapter 16.2, Problem 32E

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Chapter 16 Solutions

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The value of integral of f ( x , y ) = ( ln y ) − 1 over the domain D bounded by y = e x and y = e x .

Solution Summary: The author explains how to determine the value of integral of f ( x, y ) = 1 over the domain D.

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The value of integral of $f (x, y) = {(\ln y)}^{- 1}$ over the domain D bounded by $y = e^{x}$ and $y = e^{\sqrt{x}}$ .

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