Lecture: Section 20.2 The Divergence Theorem The divergence theorem relates the integral of the flux density over a solid region to the flux integral through the boundary of that region. The Boundary of a Solid Region W = ball (volume) ; S = sphere (surface) W = solid cylinder ; S = Tube and 2 disks W = solid cube ; S = 6 square faces A surface that encloses a solid region is called a closed surface. T. King: MA3160 Page 1 Lecture 20.2 Calculating Flux from Flux Density Consider a solid region W in 3‐D space whose boundary is the closed surface S. There are two ways of calculating the total flux of a vector field out of W. 1. Calculate Flux of through S. This is done using a flux (surface) integral as we learned in Chapter 19. .
·
2. Use div (flux density) Flux from W = Flux density ∙ Volume One way to do this would be to divide W into small regions of size ∆V. Then: ·∆ In the limit as ∆V → 0: .
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Divergence Theorem The divergence theorem is simply the equation relating these two approaches. .
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Please note that this theorem is very limited in scope! However, it is often much easier to compute the volume integral rather than the surface integral for the case of a closed surface. Example Compute the flux of through S, a sphere with R = 2 centered at the origin and oriented outward by (1) a surface integral and (2) the divergence theorem. T. King: MA3160 Page 2 Lecture 20.2 F = zk
3
2
1
0
-1
-2
-3
2
1
2
1
0
0
-1
-1
-2 -2
Y
X
We will start with the divergence theorem since it will be easier. .
1 .
4
3
32
3
2
Now check using a surface integral via the approach from section 19.2 .
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·
, ,
, therefore: · sin cos
̂
̂
.
2
̂
̂
2
.
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T. King: MA3160 · sin cos
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Page 3 Lecture 20.2 8
3
8
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0
16
3
2
16
·2
3
32
3
So our answer checks. Example: Exercise 20.2.6 Calculate the flux through a sphere of R = 5 centered at origin for ̂
2
̂ 10 using the divergence theorem F = x 2i + (y-2xy)j + 10zk
6
4
2
0
10
-2
5
-4
0
-6
-5
10
5
0
X
-10
-5
-10
Y
.
2
1
.
11
11
2
10
4
3
5
11 5500
3
This would be a bit more involved using a surface integral. T. King: MA3160 Page 4 Lecture 20.2 Example: Exercise 20.2.15 Find the flux due to origin. ̂
̂
out of a sphere of r = 1 centered at the .
Break into 3 integrals and convert to spherical coordinates. x = ρ sinφcosθ dV = ρ2 sinφ dρdφdθ .
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1
4
1
4
4
2
0
0 The other integrals are 0 by a similar calculation. One could also use a symmetry argument since x, y, z each take equal positive and negative values on each side of the sphere. T. King: MA3160 Page 5