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.
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
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
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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