Explanation Given: A solid bounded above by the paraboloid z = x 2 + y 2 and below by the square region 0 ≤ x ≤ 1 , 0 ≤ y ≤ 1 and z = 0 Concept Used: A double integral ∬ R f ( x , y ) d A over the range R : a ≤ x ≤ b , g 1 ( x ) ≤ y ≤ g 2 ( x ) is defined as ∬ R f ( x , y ) d A = ∫ a b ∫ g 1 ( x ) g 2 ( x ) f ( x , y ) d y d x And A double integral ∬ R f ( x , y ) d A over the range R : m 1 ( x ) ≤ x ≤ m 2 ( x ) , c ≤ y ≤ d is defined as ∬ R f ( x , y ) d A = ∫ c d ∫ m 1 ( x ) m 2 ( x ) f ( x , y ) d x d y Calculation: For thesolid bounded above by the paraboloid z = x 2 + y 2 and below by the square region 0 ≤ x ≤ 1 , 0 ≤ y ≤ 1 and z = 0 The volume is given by V = ∫ y = 0 1 ∫ x = 0 1 ( x 2 + y 2 − 0 ) d x d y V = ∫ y = 0 1 ∫ x = 0 1 ( x 2 + y 2 ) d x d y The square region D bounded by 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1 is

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ISBN: 9781524916817

Author: SMITH KARL J, STRAUSS MONTY J, TODA MAGDALENA DANIELE

Publisher: Kendall Hunt Publishing

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Chapter 12, Problem 57SP

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ToCalculateThe volume of the solid that is bounded above by the paraboloid z=x2+y2 and below by the square region 0≤x≤1 , 0≤y≤1 and z=0 using double integral.

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Chapter 12, Problem 56SP

Chapter 12, Problem 58SP

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ToCalculateThe volume of the solid that is bounded above by the paraboloid z = x 2 + y 2 and below by the square region 0 ≤ x ≤ 1 , 0 ≤ y ≤ 1 and z = 0 using double integral.

Solution Summary: The author calculates the volume of the solid that is bounded above by the paraboloid z=x2+y2.

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ToCalculateThe volume of the solid that is bounded above by the paraboloid z=x2+y2 and below by the square region 0≤x≤1 , 0≤y≤1 and z=0 using double integral.

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