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[2020-02-06 Thu 17:18]
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Session 86: Proof of the Divergence Theorem | Part B: Flux and the Divergence Theorem | 4. Triple Integrals and Surface Integrals in 3-Space | Multivariable Calculus | Mathematics | MIT OpenCourseWare

1 Chalkboard

Figure 1: Proof of divergence theorem

Figure 1: Proof of divergence theorem

Figure 2: \(D\) is vertically simple

Figure 2: DD is vertically simple

Figure 3: Checking both sides has same outcome

Figure 3: Checking both sides has same outcome

Figure 4: Flux at top side

Figure 4: Flux at top side

Figure 5: Flux at bottom side

Figure 5: Flux at bottom side

Figure 6: QED and divide up complex regions

Figure 6: QED and divide up complex regions

2 How proof the divergence theorem

2.1 Front

How proof the divergence theorem

Describe only the steps

2.2 Back

  • Let \(\vb{F}\) be a vector field with only 1 component
    • You can proof separately and sum up then
  • Divide up the figure in cubes, the flux at shared sides will be 0
  • Set the Gauss theorem equation with this conditions
  • Expand the equations side by side, and compare

3 Draw a figure for proofing the divergence theorem

3.1 Front

Draw a figure for proofing the divergence theorem

Let \(\vb{F} = \ev{0,0,P(x,y,z)}\)

3.2 Back

  • SS is a closed surface assuming vertically simple
  • Lower surface: z=g(x,y)z = g(x,y)
  • Upper surface: z=h(x,y)z = h(x,y)

4 Set up the divergence theorem equation for proofing it

4.1 Front

Set up the divergence theorem equation for proofing it

Let \(\vb{F} = \ev{0,0,P}\), SS a closed surface vertically simple, where lower surface is z=g(x,y)z = g(x,y) and upper surface z=h(x,y)z = h(x,y)

And proof that has the same outcome

4.2 Back

\({\displaystyle \iint_S P \vu{k} \cdot \vu{n} \dd{S} = \iiint_D \pdv{P}{z} \dd{V}}\)

Right side:

\({\displaystyle \iiint_D \pdv{P}{z} \dd{V} = \iint_R \int_{g(x,y)}^{h(x,y)} \pdv{P}{z} \dd{z} \dd{x} \dd{y} = \iint_R (P(x,y,h) - P(x,y,g)) \dd{x} \dd{y}}\)

Left side:

  • Calculating \(\dd{\vb{S}}\) for top and bottom side
    • \(\dd{\vb{S}} = \pm \ev{-z_x, -z_y, 1} \dd{x} \dd{y}\)
    • Positive is for top side
    • Negative is for bottom side
  • Flux at lateral sides is 00 because \(\vb{F} = \ev{0,0,P}\) and it’s perpendicular to \(\dd{\vb{S}}\) of lateral sides
  • RR is the region shadow in the xy-planexy\text{-plane}
  • Top side
    • \({\displaystyle \iint_{\text{top}} P \vu{k} \cdot \dd{\vb{S}} = \iint_R P(x,y,z)\dd{x}\dd{y} = \iint_R P(x,y,h(x,y)) \dd{x}\dd{y}}\)
  • Bottom side
    • \({\displaystyle \iint_{\text{bottom}} P \vu{k} \cdot \dd{\vb{S}} = \iint_R -P(x,y,z)\dd{x}\dd{y} = \iint_R P(x,y,g(x,y)) \dd{x}\dd{y}}\)
  • Adding up all sides
    • \({\displaystyle \iint_S P \vu{k} \cdot \dd{\vb{S}} = \iint_R P(x,y,h) \dd{x} \dd{y} - \iint_R P(x,y,g) \dd{x} \dd{y}}\)

5 Why can you divide up a region in smaller domain for proofing divergence theorem?

5.1 Front

Why can you divide up a region in smaller domain for proofing divergence theorem?

5.2 Back

You can divide up DD into smaller domains DiD_i which are bounded by such surfaces SiS_i. Adding these up gives the divergence theorem for DD and SS, since the surface integrals over the new faces introduced by cutting up DD each occur twice, with opposite normal vectors \(\vu{n}\), so that they cancel out.

After addition, one ends up just with the surface integral over the original SS

Similar as Green’s Theorem for 2D