Explanation Given: The surface is z = x y ( x 2 + y 2 ) 2 and rectangular region as 1 ≤ x ≤ 2 , 1 ≤ y ≤ 4 Approach: We can also use double integral to find volume under the given surface and over the rectangular region. If there is a surface z define as z = f ( x , y ) and region R as a ≤ x ≤ b , c ≤ y ≤ d Then volume ( V ) under given surface and over the rectangular region R is V = ∬ R f ( x , y ) d x d y = ∫ c d ∫ a b f ( x , y ) d x d y Calculation: Given surface is z = x y ( x 2 + y 2 ) 2 and region is 1 ≤ x ≤ 2 , 1 ≤ y ≤ 4 Volume under given surface and over rectangular region is V = ∫ 1 4 ∫ 1 2 ( x y ( x 2 + y 2 ) 2 ) d x d y = ∫ 1 4 ∫ 1 2 ( y x ( x 2 + y 2 ) − 2 ) d x d y = ∫ 1 4 y [ 1 2 ×

2nd Edition

ISBN: 9780321964038

Author: GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.

Publisher: Pearson Addison Wesley,

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Chapter 9.5, Problem 37E

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The surface is z=xy(x2+y2)2 and rectangular region as 1≤x≤2,1≤y≤4

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Chapter 9.5, Problem 36E

Chapter 9.5, Problem 38E

Chapter 9 Solutions

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The surface is z = x y ( x 2 + y 2 ) 2 and rectangular region as 1 ≤ x ≤ 2 , 1 ≤ y ≤ 4

Solution Summary: The author explains that we can also use double integral to find volume under the given surface and over the rectangular region.

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To find: The surface is z=xy(x2+y2)2 and rectangular region as 1≤x≤2,1≤y≤4

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Chapter 9 Solutions

To find:

The surface is z=xy(x2+y2)2 and rectangular region as 1≤x≤2,1≤y≤4