Explanation Given: Equations of the solids: z = 25 e − ( x 2 + y 2 ) 4 , z = 0 , x 2 + y 2 = 16 Calculation: First convert the double integral into polar coordinates by substituting: x = r cos θ y = r sin θ d x d y = r d r d θ The corresponding limits in polar coordinates will be: 0 ≤ θ ≤ 2 π 0 ≤ r ≤ 4 Writing the double integral in polar coordinates, we can find the volume of the solid: V = ∫ 0 2 π ∫ 0 4 25 e − r 2 4 r d r d θ V = ∫ 0 2 π [ − 50 e − r 2 4 ] 0 4 d θ V = ∫ 0 2 π − 50 ( e − 4 − 1 ) d θ V = [ 50 ( 1 − e − 4 ) ] 0 2 π V = 100 π ( 1 − e − 4 ) V = 308.4052 Let d be the diameter of the hole that is removed from the solid

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ISBN: 9780131569898

Author: C. Henry Edwards, David E. Penney

Publisher: PEARSON

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Chapter 14.3, Problem 57E

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To calculate: The diameter of a hole

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Chapter 14.3, Problem 56E

Chapter 14.3, Problem 58E

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