Explanation Given : , the region D bounded by y = x , y = 4 x − x 2 and y = 0 . Calculation : Step 1: The region D between y = x and y = 4 x − x 2 is a vertically simple region. Step 2: To find the inequalities defining the region D as a vertically simple region, we must first compute the x-coordinates of the points of intersection of the two curves by solving the following equation, x = 4 x − x 2 ⇒ x 2 − 3 x = x ( x − 3 ) = 0 ⇒ x 1 = 0 , x 2 = 3 Thus, the following are the inequalities defined for D: 0 ≤ x ≤ 3 , x ≤ y ≤ 4 x − x 2 Step 3: Evaluate the double integral f ( x , y ) = x is over D by computing the following iterated integral: ∫ ∫ D x d A = ∫ 0 3 ∫ x 4 x − x 2 x d y d x = ∫ 0 3 x y | y = x 4 x − x 2 d x = ∫ 0 3 ( 4 x 2 − x 3 − x 2 ) d x = ∫ 0 3 ( 3 x 2 − x 3 ) d x = x 3 − x 4 4 | 0 3 = 27 − 81 4 = 27 4 Step 4: The region shown in the below figure is the union of two horizontally simple regions. To determine the inequalities defining this region, we must first find the y-coordinate of A. As we found that the x-coordinate of this point is x=3 in Step 2, hence y=3. We now, solve the equation of the right curve y = 4 x − x 2 for x in terms of y: y = 4 x − x 2 x 2 − 4 x + y = 0 x 1 , 2 = 2 ± 4 − y In this part of boundary x ≤ 3 , hence the negative root must be taken. That is, x = 2 − 4 − y We obtain the following inequalities for the domain D: 0 ≤ y ≤ 3 , y ≤ x ≤ 2 + 4 − y The other part of the domain in question will have to be 3 ≤ y ≤ 4 , 2 − 4 − y ≤ x ≤ 2 + 4 − y

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ISBN: 9781319323394

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 16.2, Problem 16E

To determine

The value of the double integral of $f (x, y) = x$ over the region D bounded by $y = x, y = 4 x - x^{2} and y = 0$ as a vertically simple region and as a horizontally simple region.

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Chapter 16.2, Problem 15E

Chapter 16.2, Problem 17E

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Chapter 16 Solutions

CALCULUS W/SAPLING ACCESS >IC<

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The value of the double integral of f ( x , y ) = x over the region D bounded by y = x , y = 4 x − x 2 and y = 0 as a vertically simple region and as a horizontally simple region.

Solution Summary: The author calculates the value of the double integral of f (x, y ) = x over the region D as a vertically and horizontally simple region.

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The value of the double integral of $f (x, y) = x$ over the region D bounded by $y = x, y = 4 x - x^{2} and y = 0$ as a vertically simple region and as a horizontally simple region.

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Chapter 16 Solutions

The value of the double integral of f ( x , y ) = x over the region D bounded by y = x , y = 4 x − x 2 and y = 0 as a vertically simple region and as a horizontally simple region.

Solution Summary: The author calculates the value of the double integral of f (x, y ) = x over the region D as a vertically and horizontally simple region.

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The value of the double integral of f(x,y)=x over the region D bounded by y=x,y=4x−x2and y=0 as a vertically simple region and as a horizontally simple region.

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Chapter 16 Solutions

The value of the double integral of $f (x, y) = x$ over the region D bounded by $y = x, y = 4 x - x^{2} and y = 0$ as a vertically simple region and as a horizontally simple region.