Calculus- Early Transcendentals, 2021a

552 Vector Calculus
16.5 The Divergence Theorem
The third version of Green’s Theorem (Equation 16.3)wesawwas:
∫
∫∫
f · Nds = ∇ · fdA.
With minor changes this turns into another equation, the Divergence Theorem:
Theorem 16.15: Divergence Theorem
∂D
Under suitable conditions, if E is a region of three dimensional space and D is its boundary surface,
oriented outward, then
∫∫
∫∫∫
f · NdS = ∇ · fdV .
D
D
E
Proof. Again this theorem is too difficult to prove here, but a special case is easier. In the proof of a
special case of Green’s Theorem, we needed to know that we could describe the region of integration in
both possible orders, so that we could set up one double integral using dxdy and another using dydx.
Similarly here, we need to be able to describe the three-dimensional region E in different ways.
We start by rewriting the triple integral:
∫∫∫
E
∫∫∫
∇ · fdV =
E
The double integral may be rewritten:
∫∫
∫∫
∫∫
f · NdS = ( f 1 i + f 2 j + f 3 k) · NdS =
D
D
( ∂ f 1
∂x + ∂ f 2
∂y + ∂ f 3
∂z
∫∫∫E
)dV = ∂ f 1
∂x
∫∫∫E
dV + ∂ f 2
∂y
∫∫∫E
dV + ∂ f 3
∂z dV .
D
∫∫
∫∫
f 1 i · NdS+ f 2 j · NdS+ f 3 k · NdS.
D
D
To prove that these give the same value it is sufficient to prove that
∫∫
∫∫∫
∂ f 1
f 1 i · NdS = dV , (16.4)
D
E ∂x
∫∫
∫∫∫
∂ f 2
f 2 j · NdS = dV , and (16.5)
D
E ∂y
∫∫
∫∫∫
∂ f 3
f 3 k · NdS = dV . (16.6)
D
E ∂z
Not surprisingly, these are all pretty much the same; we’ll do the first one.
We set the triple integral up with dx innermost:
∫∫∫
E
∫
∂ f g2 1
∂x
∫∫B
dV = (y,z) ∂ f 1
g 1 (y,z) ∂x
∫∫B
dxdA = f 1 (g 2 (y,z),y,z) − f 1 (g 1 (y,z),y,z)dA,
where B is the region in the yz-plane over which we integrate. The boundary surface of E consists of a
“top” x = g 2 (y,z), a “bottom” x = g 1 (y,z), and a “wrap-around side” that is vertical to the yz-plane. To