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

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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