a. Explanation Given : W is the region bounded by y + z = 2 , 2 x = y , x = 0 and z = 0 . Calculation : Step 1: The upper face z = 2 − y intersects the first quadrant of the xy -plane ( z = 0 ) in the line y = 2 . Therefore the projection of W onto the xy -plane is the triangle D defined by D : 0 ≤ x ≤ 1 , 2 x ≤ y ≤ 2 Therefore, W is the following region: 0 ≤ x ≤ 1 , 2 x ≤ y ≤ 2 , 0 ≤ z ≤ 2 – y We obtain the following iterated integral: ∭ W ( 2 x − 4 y + 6 z ) d V = ∫ 0 1 ∫ 2 x 2 ∫ 0 2 − y ( 2 x − 4 y + 6 z ) d z d y d x = ∫ 0 1 ∫ 2 x 2 2 x z − 4 y z + 3 z 2 | z = 0 2 − y d y d x b. To determine To calculate : The value of the triple integral f ( x , y , z ) = 2 x − 4 y + 6 z by projecting W onto the yz plane. c. To determine To calculate : The value of the triple integral f ( x , y , z ) = 2 x − 4 y + 6 z by projecting W onto the xz plane.

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ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 16.3, Problem 30E

To determine

To calculate: The value of the triple integral f(x,y,z)=2x−4y+6z by projecting W onto the xy plane.

To determine

To calculate: The value of the triple integral f(x,y,z)=2x−4y+6z by projecting W onto the yz plane.

To determine

To calculate: The value of the triple integral f(x,y,z)=2x−4y+6z by projecting W onto the xz plane.

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Chapter 16.3, Problem 29E

Chapter 16.3, Problem 31E

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Chapter 16 Solutions

CALCULUS (CLOTH)

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The value of the triple integral f ( x , y , z ) = 2 x − 4 y + 6 z by projecting W onto the xy plane.

Solution Summary: The author calculates the triple integral f(x,y,z)=2x-4y+6z by projecting W onto the xy plane.

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To calculate: The value of the triple integral f(x,y,z)=2x−4y+6z by projecting W onto the xy plane.

To calculate: The value of the triple integral f(x,y,z)=2x−4y+6z by projecting W onto the yz plane.

To calculate: The value of the triple integral f(x,y,z)=2x−4y+6z by projecting W onto the xz plane.

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Chapter 16 Solutions