Explanation Given: Thesphere x 2 + y 2 + z 2 = 4 and below by the paraboloid 3 z = x 2 + y 2 . Calculation: Consider a sphere x 2 + y 2 + z 2 = 4 and below by the paraboloid 3 z = x 2 + y 2 . The objective is to find the volume V of the solid bounded above by sphere and below by paraboloid. The intersection point of these two shapes is as; x 2 + y 2 + z 2 = 4 3 z + z 2 = 4 [ x 2 + y 2 = 3 z ] z 2 + 3 z − 4 = 0 z = 1 , − 4 Since solid is above of the x y -plane so z = 1 , z ≠ − 4 . Therefore, the intersection is a circle z = 1 which is x 2 + y 2 = 3 . So the volume of solid is; V = ∫ − 3 3 ∫ − 3 − x 2 3 − x 2 { 4 − x 2 − y 2 − 1 3 ( x 2 + y 2 ) } d y d x Now change this into polar form as x = r cos θ , y = r sin θ and x 2 + y 2 = r 2 So, r varies from θ from 0 to 2 π . Therefore, V = ∫ 0 2 π ∫ 0 3 { 4 − r 2 − 1 3 ( r 2 ) d r } d θ = ∫ 0 2 π ∫ 0 3 { r 4 − r 2 d r − 1 3 ( r 3 ) d r } d θ Consider the integral ∫ r 4 − r 2 d r and 4 − r 2 = t which gives − 2 r d r = d t so integral becomes; ∫ r 4 − r 2 d r = − 1 2 ∫ t d t = − 1 2

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ISBN: 9781465208880

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

Publisher: Kendall Hunt Publishing

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Chapter 12.3, Problem 52PS

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To find: The volume of the solid region.

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Chapter 12.3, Problem 51PS

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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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The volume of the solid region.

Solution Summary: The author calculates the volume of the solid bounded by paraboloid. The intersection point of these two shapes is a circle z=1.

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To find: The volume of the solid region.

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Chapter 12 Solutions