Explanation A Triple integral of a function f ( x , y , z ) over a solid D which is bounded above by a surface z = u ( x , y ) and bounded below by z = v ( x , y ) that project onto a region A & given as follows: ∭ D f ( x , y , z ) = ∬ A ∫ u ( x , y ) v ( x , y ) f ( x , y , z ) d z d A The objective is to find the volume of a right pyramid with height H and a square base of side S . Consider a solid right pyramid with square vertices as ( S 2 , 0 , 0 ) , ( 0 , S 2 , 0 ) , ( − S 2 , 0 , 0 ) , ( 0 , − S 2 , 0 ) and top as ( 0 , 0 , H ) . Evaluate the volume of the solid in the first octant and then multiply the result by 4. First, we require the equation of the plane passing through ( S 2 , 0 , 0 ) , ( 0 , S 2 , 0 ) , ( 0 , 0 , H ) Let: P = ( S 2 , 0 , 0 ) , Q = ( 0 , S 2 , 0 ) , R = ( 0 , 0 , H ) Hence, P Q = ( − S 2 , S 2 , 0 ) , P R = ( − S 2 , 0 , H ) Let n be a vector normal to the plane in the first octant. Thus, n = P Q × P R n = | i j k − S 2 S 2 0 − S 2 0 H | n = i ( H S 2 ) + j ( H S 2 ) + k ( S 2 2 ) Hence, the equation of the plane in the first octant is: H S 2 ( x − S 2 ) + H S 2 ( y ) + S 2 2 ( z ) = 0 Thus, set up the integral as follows: ∭ D d V = ∫ 0 S 2 ∫ 0 ( S 2 − x ) ∫ 0 2 H S ( S 2 − x</

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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 12.5, Problem 52PS

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Chapter 12 Solutions

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Ch. 12.1 - Prob. 1PS Ch. 12.1 - Prob. 2PS Ch. 12.1 - Prob. 3PS Ch. 12.1 - Prob. 4PS Ch. 12.1 - Prob. 5PS Ch. 12.1 - Prob. 6PS Ch. 12.1 - Prob. 7PS Ch. 12.1 - Prob. 8PS Ch. 12.1 - Prob. 9PS Ch. 12.1 - Prob. 10PS

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