Explanation Concept used: The divergence theorem states that, Let S be a smooth, orientable surface that encloses a solid region R in ℝ 3 . If F → is a continuous vector field whose components have continuous partial derivatives in an open set containing R , then ∬ S F → ⋅ N → d s = ∭ R div F → d V ; where N → is the outward unit normal field for the surface S . Consider the vector function ∬ S F → ⋅ N → d s for F → = x 3 i ^ + y 3 j ^ + 3 a 2 z k ^ (constant a > 0 ); and the surface bounded by the cylinder S : x 2 + y 2 = a 2 and the planes z = 0 and z = 1

7th Edition

ISBN: 9781524916817

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

Publisher: Kendall Hunt Publishing

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Chapter 13.7, Problem 39PS

To determine

To calculate: The value of the surface integral ∬SF→⋅N→ds for F→=x3i^+y3j^+3a2zk^ (constant a>0 ); and the surface bounded by the cylinder S:x2+y2=a2 and the planes z=0 and z=1 .

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Chapter 13.7, Problem 38PS

Chapter 13.7, Problem 40PS

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The value of the surface integral ∬ S F → ⋅ N → d s for F → = x 3 i ^ + y 3 j ^ + 3 a 2 z k ^ (constant a > 0 ); and the surface bounded by the cylinder S : x 2 + y 2 = a 2 and the planes z = 0 and z = 1 .

Solution Summary: The author explains the divergence theorem, where S is a smooth, orientable surface whose components have continuous partial derivatives in an open set.

Concept explainers

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To calculate: The value of the surface integral ∬SF→⋅N→ds for F→=x3i^+y3j^+3a2zk^ (constant a>0 ); and the surface bounded by the cylinder S:x2+y2=a2 and the planes z=0 and z=1 .

Want to see the full answer?

Chapter 13 Solutions

The value of the surface integral ∬ S F → ⋅ N → d s for F → = x 3 i ^ + y 3 j ^ + 3 a 2 z k ^ (constant a &gt; 0 ); and the surface bounded by the cylinder S : x 2 + y 2 = a 2 and the planes z = 0 and z = 1 .

Solution Summary: The author explains the divergence theorem, where S is a smooth, orientable surface whose components have continuous partial derivatives in an open set.

Concept explainers

Videos

To calculate: The value of the surface integral ∬SF→⋅N→ds for F→=x3i^+y3j^+3a2zk^ (constant a>0 ); and the surface bounded by the cylinder S:x2+y2=a2 and the planes z=0 and z=1 .

Want to see the full answer?

Chapter 13 Solutions

The value of the surface integral ∬ S F → ⋅ N → d s for F → = x 3 i ^ + y 3 j ^ + 3 a 2 z k ^ (constant a > 0 ); and the surface bounded by the cylinder S : x 2 + y 2 = a 2 and the planes z = 0 and z = 1 .