gayss theory
1
Gauss’ theorem
Chapter 14 Gauss’ theorem
We now present the third great theorem of integral vector calculus. It is interesting that
Green’s theorem is again the basic starting point. In Chapter 13 we saw how Green’s theorem directly translates to the case of surfaces in R3 and produces Stokes’ theorem. Now we are going to see how a reinterpretation of Green’s theorem leads to Gauss’ theorem for R2 , and then we shall learn from that how to use the proof of Green’s theorem to extend it to Rn ; the result is called Gauss’ theorem for Rn .
A. Green’s theorem reinterpreted
We begin with the situation obtained in Section 12C for a region R in R2 . With the positive orientation for bdR, we have
∂f
dxdy =
∂x
R
f dy, bdR ∂g dxdy = −
∂y
R
gdx.
bdR
We now consider a vector field F = (F1 , F2 ) on R, and we obtain from Green’s theorem
∂F1 ∂F2
+
∂x
∂y
R
dxdy =
F1 dy − F2 dx. bdR (Notice if we were thinking in terms of Stokes’ theorem, we would put the minus sign on the left side instead of the right.)
We now work on the line integral, first writing it symbolically in the form
F • (dy, −dx). bdR We now express what this actually means. Of course, bdR may come in several disjoint pieces, so we focus on just one such piece, say the closed curve γ = γ(t) = (γ1 (t), γ2 (t)), a ≤ t ≤ b.
Then the part of the line integral corresponding to this piece is actually b a
F (γ(t)) • (γ2 (t), −γ1 (t))dt.
R
2
Chapter 14
This in turn equals
b
F (γ(t)) • a (γ2 (t), −γ1 (t)) γ (t) dt. γ (t)
The unit vector in the dot product has a beautiful geometric interpretation. Namely, it is orthogonal to the unit tangent vector
(γ1 (t), γ2 (t))
,
γ (t) and it is oriented 90◦ clockwise from the tangent vector. (You can use the matrix J of
Section 8D to see this, as γ1 −γ2
.
J
=
γ1 γ2 You can also think in complex arithmetic: −i(γ1 + iγ2 ) = γ2 − iγ1 , and multiplication by −i rotates 90◦
Gauss’ theorem
Chapter 14 Gauss’ theorem
We now present the third great theorem of integral vector calculus. It is interesting that
Green’s theorem is again the basic starting point. In Chapter 13 we saw how Green’s theorem directly translates to the case of surfaces in R3 and produces Stokes’ theorem. Now we are going to see how a reinterpretation of Green’s theorem leads to Gauss’ theorem for R2 , and then we shall learn from that how to use the proof of Green’s theorem to extend it to Rn ; the result is called Gauss’ theorem for Rn .
A. Green’s theorem reinterpreted
We begin with the situation obtained in Section 12C for a region R in R2 . With the positive orientation for bdR, we have
∂f
dxdy =
∂x
R
f dy, bdR ∂g dxdy = −
∂y
R
gdx.
bdR
We now consider a vector field F = (F1 , F2 ) on R, and we obtain from Green’s theorem
∂F1 ∂F2
+
∂x
∂y
R
dxdy =
F1 dy − F2 dx. bdR (Notice if we were thinking in terms of Stokes’ theorem, we would put the minus sign on the left side instead of the right.)
We now work on the line integral, first writing it symbolically in the form
F • (dy, −dx). bdR We now express what this actually means. Of course, bdR may come in several disjoint pieces, so we focus on just one such piece, say the closed curve γ = γ(t) = (γ1 (t), γ2 (t)), a ≤ t ≤ b.
Then the part of the line integral corresponding to this piece is actually b a
F (γ(t)) • (γ2 (t), −γ1 (t))dt.
R
2
Chapter 14
This in turn equals
b
F (γ(t)) • a (γ2 (t), −γ1 (t)) γ (t) dt. γ (t)
The unit vector in the dot product has a beautiful geometric interpretation. Namely, it is orthogonal to the unit tangent vector
(γ1 (t), γ2 (t))
,
γ (t) and it is oriented 90◦ clockwise from the tangent vector. (You can use the matrix J of
Section 8D to see this, as γ1 −γ2
.
J
=
γ1 γ2 You can also think in complex arithmetic: −i(γ1 + iγ2 ) = γ2 − iγ1 , and multiplication by −i rotates 90◦