Explanation Given : Domain D : 0 ≤ x ≤ 1 , x 2 ≤ y ≤ 4 − x 2 , Double Integral ∫ ∫ D y d A Calculation : Step 1: The domain D is the shaded region in the figure D is a vertically simple region and the limits of integration 0 ≤ x ≤ 1 ︸ limits of outer integral x 2 ≤ y ≤ 4 − x 2 ︸ limits of inner integral Step 2: Set up the iterated integral. The iterated integral is ∫ ∫ D y d A = ∫ 0 1 ∫ x 2 4 − x 2 y d y d x Step 3: Evaluate the inner integral with respect to y

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Chapter 16.2, Problem 4E

To determine

The sketch of the domain $D : 0 \leq x \leq 1, x^{2} \leq y \leq 4 - x^{2}$ and the evaluated value of $\int \int_{D} y d A$ as an iterated integral.

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Chapter 16.2, Problem 3E

Chapter 16.2, Problem 5E

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Chapter 16 Solutions

Calculus - Standalone book

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The sketch of the domain D : 0 ≤ x ≤ 1 , x 2 ≤ y ≤ 4 − x 2 and the evaluated value of ∫ ∫ D y d A as an iterated integral.

Solution Summary: The author explains that the iterated integral for domain D is 6.67. The shaded region in the figure is a vertically simple region.

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The sketch of the domain $D : 0 \leq x \leq 1, x^{2} \leq y \leq 4 - x^{2}$ and the evaluated value of $\int \int_{D} y d A$ as an iterated integral.

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Chapter 16 Solutions

The sketch of the domain D : 0 ≤ x ≤ 1 , x 2 ≤ y ≤ 4 − x 2 and the evaluated value of ∫ ∫ D y d A as an iterated integral.

Solution Summary: The author explains that the iterated integral for domain D is 6.67. The shaded region in the figure is a vertically simple region.

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The sketch of the domain D:0≤x≤1,x2≤y≤4−x2 and the evaluated value of ∫∫DydA as an iterated integral.

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Chapter 16 Solutions

The sketch of the domain $D : 0 \leq x \leq 1, x^{2} \leq y \leq 4 - x^{2}$ and the evaluated value of $\int \int_{D} y d A$ as an iterated integral.