Explanation Given: The solid region bounded by the graphs z = 25 − x 2 − y 2 , x 2 + y 2 = 16 and z = 0 . Formula used: Volume in polar coordinate is given as V = ∫ θ 1 θ 2 ∫ r 1 r 2 f ( r , θ ) ⋅ r d r d θ . In polar coordinates, d A = r d r d θ , x = r cos θ and y = r sin θ ; The relationship between rectangular and polar coordinate is given as x 2 + y 2 = r 2 . Use the integration formula, ∫ x n d x = x n + 1 n + 1 Calculation: Convert the equation z = 25 − x 2 − y 2 in polar coordinates, z = 25 − x 2 − y 2 f ( x , y ) = 25 − ( x 2 + y 2 ) f ( r , θ ) = 25 − r 2 Find the limit of r . Take the provided equation into consideration, x 2 + y 2 = 16 As x 2 + y 2 = r 2 , thus, r 2 = 16 r = 4 Therefore, upper limit of r is 4, i.e., bounds of r is 0 ≤ r ≤ 4

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ISBN: 9781337275576

Author: Larson, Ron, Edwards, Bruce

Publisher: Cengage Learning,

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Chapter 14, Problem 28RE

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To calculate: The volume of the solid region bounded by the graphs of the provided equations using double integrals in polar coordinates.

z=25−x2−y2, x2+y2=16 and z=0

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Chapter 14, Problem 27RE

Chapter 14, Problem 29RE

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The volume of the solid region bounded by the graphs of the provided equations using double integrals in polar coordinates. z = 25 − x 2 − y 2 , x 2 + y 2 = 16 and z = 0

Solution Summary: The author calculates the volume of the solid region bounded by the graphs using double integrals in polar coordinates.

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To calculate: The volume of the solid region bounded by the graphs of the provided equations using double integrals in polar coordinates.

z=25−x2−y2, x2+y2=16 and z=0

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Chapter 14 Solutions

The volume of the solid region bounded by the graphs of the provided equations using double integrals in polar coordinates. z = 25 − x 2 − y 2 , x 2 + y 2 = 16 and z = 0

Solution Summary: The author calculates the volume of the solid region bounded by the graphs using double integrals in polar coordinates.

Videos

To calculate: The volume of the solid region bounded by the graphs of the provided equations using double integrals in polar coordinates. z=25−x2−y2, x2+y2=16 and z=0

Want to see the full answer?

Chapter 14 Solutions

To calculate: The volume of the solid region bounded by the graphs of the provided equations using double integrals in polar coordinates.

z=25−x2−y2, x2+y2=16 and z=0