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1.3 Integral Calculus (iv) 1 ZA 1 (v) (ii) (vi) FIGURE 1.29 1 So much for the left side of the divergence theorem. To evaluate the surface integral we must consider separately the six faces of the cube: (i) (ii) (iii) (iv) (v) (vi) So the total flux is: 1 1 [ v · da = f' f' y²dy dz = } [ v. da v.da [ v · da = L 1 1 - S² S² y² dy dz = − } . [ = Sv. $ S v.da L 1 1 (2x + z²) dx dz = 3. 1 - S' S² ² dx d² = − } . L 1 1 [ v-da = L'S' dz S' S² 2y dx dy = 1. Sv da = - S' S L L · ½ 1 0dx dy = 0. v da = -+-+1+0=2, as expected. 33 33 Example 1.10. Check the divergence theorem using the function v = y²â+ (2xy + z²) ŷ + (2yz) z and a unit cube at the origin (Fig. 1.29). Solution In this case V.v=2(x + y), and 1 {2(x + y) dt = 2 f f f (x + y)dx dy dz. 1 S' L' = (x + y) dx = + y, = 1+y, f² (4 + y) dy = 1, [ '1dz = 1. 0 Thus, V.vdt = 2.

1.3 Integral Calculus (iv) 1 ZA 1 (v) (ii) (vi) FIGURE 1.29 1 So much for the left side of the divergence theorem. To evaluate the surface integral we must consider separately the six faces of the cube: (i) (ii) (iii) (iv) (v) (vi) So the total flux is: 1 1 [ v · da = f' f' y²dy dz = } [ v. da v.da [ v · da = L 1 1 - S² S² y² dy dz = − } . [ = Sv. $ S v.da L 1 1 (2x + z²) dx dz = 3. 1 - S' S² ² dx d² = − } . L 1 1 [ v-da = L'S' dz S' S² 2y dx dy = 1. Sv da = - S' S L L · ½ 1 0dx dy = 0. v da = -+-+1+0=2, as expected. 33 33 Example 1.10. Check the divergence theorem using the function v = y²â+ (2xy + z²) ŷ + (2yz) z and a unit cube at the origin (Fig. 1.29). Solution In this case V.v=2(x + y), and 1 {2(x + y) dt = 2 f f f (x + y)dx dy dz. 1 S' L' = (x + y) dx = + y, = 1+y, f² (4 + y) dy = 1, [ '1dz = 1. 0 Thus, V.vdt = 2.

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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The example below comes out of griffiths intro to electrodynamics. I was wondering if you can make a spherical and cylindrical problem like this example, one of each and use to divergence theorem to prove both sides of integrals from gauss divergence theorem. Hope this makes sense and thank 

1.3 Integral Calculus (iv) 1 ZA 1 (v) (ii) (vi) FIGURE 1.29 1 So much for the left side of the divergence theorem. To evaluate the surface integral we must consider separately the six faces of the cube: (i) (ii) (iii) (iv) (v) (vi) So the total flux is: 1 1 [ v · da = f' f' y²dy dz = } [ v. da v.da [ v · da = L 1 1 - S² S² y² dy dz = − } . [ = Sv. $ S v.da L 1 1 (2x + z²) dx dz = 3. 1 - S' S² ² dx d² = − } . L 1 1 [ v-da = L'S' dz S' S² 2y dx dy = 1. Sv da = - S' S L L · ½ 1 0dx dy = 0. v da = -+-+1+0=2, as expected. 33 33
Transcribed Image Text:1.3 Integral Calculus (iv) 1 ZA 1 (v) (ii) (vi) FIGURE 1.29 1 So much for the left side of the divergence theorem. To evaluate the surface integral we must consider separately the six faces of the cube: (i) (ii) (iii) (iv) (v) (vi) So the total flux is: 1 1 [ v · da = f' f' y²dy dz = } [ v. da v.da [ v · da = L 1 1 - S² S² y² dy dz = − } . [ = Sv. $ S v.da L 1 1 (2x + z²) dx dz = 3. 1 - S' S² ² dx d² = − } . L 1 1 [ v-da = L'S' dz S' S² 2y dx dy = 1. Sv da = - S' S L L · ½ 1 0dx dy = 0. v da = -+-+1+0=2, as expected. 33 33
Example 1.10. Check the divergence theorem using the function v = y²â+ (2xy + z²) ŷ + (2yz) z and a unit cube at the origin (Fig. 1.29). Solution In this case V.v=2(x + y), and 1 {2(x + y) dt = 2 f f f (x + y)dx dy dz. 1 S' L' = (x + y) dx = + y, = 1+y, f² (4 + y) dy = 1, [ '1dz = 1. 0 Thus, V.vdt = 2.
Transcribed Image Text:Example 1.10. Check the divergence theorem using the function v = y²â+ (2xy + z²) ŷ + (2yz) z and a unit cube at the origin (Fig. 1.29). Solution In this case V.v=2(x + y), and 1 {2(x + y) dt = 2 f f f (x + y)dx dy dz. 1 S' L' = (x + y) dx = + y, = 1+y, f² (4 + y) dy = 1, [ '1dz = 1. 0 Thus, V.vdt = 2.
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