(a) Explanation Property used: If the given function f ( x , y , z ) is the joint density function then, it satisfies the equation ∬ ℝ 3 f ( x , y , z ) d A = 1 . Formula used: If g ( x ) is the function of x and h ( y ) is the function of y then, ∫ a b ∫ c d g ( x ) h ( y ) k ( z ) d z d y d x = ∫ a b g ( x ) d x ∫ c d h ( y ) d y ∫ e f k ( z ) d z (1) Given: The joint density function, f ( x , y , z ) = { Cxyz , if 0 ≤ x ≤ 2 , 0 ≤ y ≤ 2 , 0 ≤ z ≤ 2 0 , otherwise Calculation: ∬ ℝ 3 f ( x , y , z ) d A = ∫ − ∞ ∞ ∫ − ∞ ∞ ∫ − ∞ ∞ f ( x , y , z ) d A = ∫ − ∞ 0 ∫ − ∞ 0 ∫ − ∞ 0 f ( x , y , z ) d z d y d x + ∫ 0 2 ∫ 0 2 ∫ 0 2 f ( x , y , z ) d z d y d x + ∫ 2 ∞ ∫ 2 ∞ ∫ 2 ∞ f ( x , y , z ) d z d y d x = 0 + ∫ 0 2 ∫ 0 2 ∫ 0 2 f ( x , y , z ) d z d y d (b) To determine To find: The value of P ( X ≤ 1 , Y ≤ 1 , Z ≤ 1 ) . (c) To determine To find: The value of P ( X + Y + Z ≤ 1 ) .

8th Edition

ISBN: 9781285741550

Author: James Stewart

Publisher: Cengage Learning

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Question

Chapter 15.6, Problem 51E

(a)

To determine

To find: The value of a constant C.

(b)

To determine

To find: The value of P(X≤1,Y≤1,Z≤1).

(c)

To determine

To find: The value of P(X+Y+Z≤1).

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Chapter 15.6, Problem 48E

Chapter 15.6, Problem 52E

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Chapter 15 Solutions

Calculus: Early Transcendentals

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The value of a constant C .

Solution Summary: The author explains that the value of a constant is 18.

To find: The value of a constant C.

To find: The value of P(X≤1,Y≤1,Z≤1).

To find: The value of P(X+Y+Z≤1).

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Chapter 15 Solutions