Explanation Given information: The integral ∭ D ( x 2 + y 2 + z 2 ) d x d y d z and the region D is surface x 2 + y 2 + z 2 ≤ 2 . Formula used: Coordinates in rectangular form are denoted by ( x , y , z ) and coordinates in cylindrical form are denoted by ( r , θ , z ) . To convert the rectangular coordinates to cylindrical coordinates use the following conversions: r = x 2 + y 2 tan θ = y x z = z To convert the cylindrical coordinates to rectangular coordinates use the following conversions: x = r cos θ y = r sin θ z = z Calculation: Consider the integral ∭ D ( x 2 + y 2 + z 2 ) d x d y d z and the region D x 2 + y 2 + z 2 ≤ 2 , these are in rectangular coordinates ( x , y , z ) , to convert it in cylindrical coordinates.Use the relation r = x 2 + y 2 and square both sides of it, r = x 2 + y 2 r 2 = x 2 + y 2 The region D is converted to the cylindrical form as, x 2 + y 2 + z 2 ≤ 2 r 2 + z 2 ≤ 2 z 2 ≤ 2 − r 2 − 2 − r 2 ≤ z ≤ 2 − r 2 The range of r, is evaluated by substituting z = 0 in the equation x 2 + y 2 + z 2 ≤ 2 , x 2 + y 2 + z 2 ≤ 2 r 2 ≤ 2 r ≤ 2 We neglect the negative sign since r is positive. Thus, the region D can be described with help of coordinates in cylindrical form as, 0 ≤ r ≤ 2 , − 2 − r 2 ≤ z ≤ 2 − r 2 , 0 ≤ θ ≤ 2 π and d x d y d z = r d z d r d θ The integral is computed as, ∭ D ( x 2 + y 2 + z 2 ) d x d y d z = ∭ D ( r 2 + z 2 ) r d z d r d θ = ∫ 0 2 π ∫ 0 2 ∫ − 2 − r 2 2 − r 2 ( r 2 + z 2 ) r d z d r d θ = ∫ 0 2 π ∫ 0 2 ∫ − 2 − r 2 2 − r 2 ( r 3 + r z 2 ) d z d r d θ Now, integrate with respect to z , ∭ D ( x 2 + y 2 + z 2 ) d x d y d z = ∫ 0 2 π ∫ 0 2 ( r 3 z + r z 3 3 ) − 2 − r 2 2 − r 2 d r d θ = ∫ 0 2 π ∫ 0 2 ( r 3 ( 2 − r 2 ) + r ( 2 − r 2 ) 3 3 − r 3 ( − 2 − r 2 ) − r ( − 2 − r 2 ) 3 3 ) d r d θ = ∫ 0 2 π ∫ 0 2 ( 2 r 3 ( 2 − r 2 ) + 2 r ( 2 − r 2 ) ( 2 − r 2 ) 3 ) d r d θ = ∫ 0 2 π ∫ 0 2 ( 4 r 3 ( 2 − r 2 ) + 4 r 3

Calculus: Special Edition: Chapters 1-5 (w/ WebAssign)

6th Edition

ISBN: 9781524908102

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

Publisher: Kendall Hunt Publishing

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Chapter 12.7, Problem 49PS

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To calculate: The integral ∭D( x 2+ y 2+ z 2)dx dy dz and the region D is surface x2+y2+z2≤2 .

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Chapter 12.7, Problem 48PS

Chapter 12.7, Problem 50PS

Chapter 12 Solutions

Calculus: Special Edition: Chapters 1-5 (w/ WebAssign)

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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The integral ∭ D ( x 2 + y 2 + z 2 ) d x d y d z and the region D is surface x 2 + y 2 + z 2 ≤ 2 .

Solution Summary: The author calculates the value of the integral displaystyleundersetDiiint.

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To calculate: The integral ∭D( x 2+ y 2+ z 2)dx dy dz and the region D is surface x2+y2+z2≤2 .

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Chapter 12 Solutions