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This triple integral can be broken into a product of integrals and evaluated, as follows. 27 (3r) dx dr de = 3 dr de Jo dx -4 Jo - 3(27)(

This triple integral can be broken into a product of integrals and evaluated, as follows. 27 (3r) dx dr de = 3 dr de Jo dx -4 Jo - 3(27)(

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Use the Divergence Theorem to calculate the surface integral F. dS; that is, calculate the flux of F across S. F(x, y, 2) = 3xy?i + xej + z°k, S is the surface of the solid bounded by the cylinder y? + z? = 4. and the planes x = -4 and x = 4 Step 1 If the surface S has positive orientation and bounds the simple solid E, then the Divergence Theorem tells us that F. ds = div F dV. For F(x, y, z) = 3xy?i + xej + z³k we have 3(,2 + z²) div F = 3y2 + 3:2 Step 2 Since S bounds the cylinder y2 + z2 = 4 between the planes x = -4 and x = 4, we will use cylindrical coordinates, with polar coordinates in the yz-plane. Therefore, y = r cos(0), z = r sin(8), and x = x. We, therefore, have the following. F. dS = div F dV 2 (3r2 cos?(0) + 3,2sin?( ?(0) dx dr de 3,2 sin² (0) | -4 3p3 dx dr de 3p.3 Step 3 This triple integral can be broken into a product of integrals and evaluated, as follows. 27 (2 27 3 de dr dx Jo 3(27)(
Transcribed Image Text:Use the Divergence Theorem to calculate the surface integral F. dS; that is, calculate the flux of F across S. F(x, y, 2) = 3xy?i + xej + z°k, S is the surface of the solid bounded by the cylinder y? + z? = 4. and the planes x = -4 and x = 4 Step 1 If the surface S has positive orientation and bounds the simple solid E, then the Divergence Theorem tells us that F. ds = div F dV. For F(x, y, z) = 3xy?i + xej + z³k we have 3(,2 + z²) div F = 3y2 + 3:2 Step 2 Since S bounds the cylinder y2 + z2 = 4 between the planes x = -4 and x = 4, we will use cylindrical coordinates, with polar coordinates in the yz-plane. Therefore, y = r cos(0), z = r sin(8), and x = x. We, therefore, have the following. F. dS = div F dV 2 (3r2 cos?(0) + 3,2sin?( ?(0) dx dr de 3,2 sin² (0) | -4 3p3 dx dr de 3p.3 Step 3 This triple integral can be broken into a product of integrals and evaluated, as follows. 27 (2 27 3 de dr dx Jo 3(27)(
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