�� 33. Gauss Theorem Theorem 33.1 (Gauss’ Theorem). Let M ⊂ R3 be a smooth 3-manifold with boundary, and let F�: M −→ R3 be a smooth vector field with compaac support. Then ��� �� div F�dx dy dz = F�d�S, · M ∂M where ∂M is given the outward orientation. Example 33.2. Three point charges are located at the points P1, P2 and P3. There is an electric field E�: R3 \{P1,P2,P3} −→ R3 , which satisfies div E�=0. Suppose there are four closed surfaces S1, S2, S3 and S4. Each Si divides R3 into two pieces, which we will informally call the inside and the outside. S1 and S2 and S3 are completely contained in the inside of S4. The inside of S1 contains the point P1 but neither P2 nor P3, the inside of S2 contains the point P2 but neither P1 nor P3, and the inside of S3 contains the point P3 but neither P1 nor P2. The inside of S4, together with S4, minus the inside of S1, S2 and S3 is a smooth 3-manifold with boundary. We have ∂M = S1�� S2�� S3�� S4. Recall that primes denote the reverse orientation. (33.1) implies that �� �� �� �� E�dS�− E�dS�− E�dS�− E�dS�···· S4 S1 S2 S3�� �� �� �� = E�dS�+ E�dS�+ E�dS�+ E�dS�···· S4 S1�S2�S3�= E�dS�· �� ∂M = div E�dx dy dz M =0. In other words, we have �� �� �� �� �d��d��d��d�ES = ES + ES + E S. · ··· S4 S1 S2 S3 Proof of (33.1). The proof (as usual) is divided into three steps. 1 �� �� �� �� Step 1: We first suppose that M = H3, upper half space. Suppose that we are given a vector field G�: H3 −→ R3, which is zero outside some box K =[−a/2,a/2] × [−b/2,b/2] × [0,c/2]. We calculate: ��� � c � b � a �� div G�du dv dw = ∂G1 + ∂G2 + ∂G1 du dv dw H3 0 ∂u ∂v ∂w � c �−bb −a =(G1(a,v,w) − G1(−a,v,w)) dv dw �0 �−b ca +(G2(u,b,w) − G2(u, −b,w)) du dw � 0 b �−aa +(G3(u,v,c) − G3(u,v, 0)) du dw −� bb −� aa = − G3(u,v, 0)du dw. −b −a On the other hand, let’s parametrise the boundary ∂H3, by �g : R2 −→ ∂H3 , where �g(u,v)=(u,v, 0). In this case ∂�g ∂�g ˆ∂u × ∂v =ˆı × jˆ= k. It follows that G�dS�= G�kˆdu dv·· (∂H2)� R2 � b � a = G3(u,v, 0)du dv. −b −a Therefore G�dS�= G�dS�∂H2 · (∂H2)� · � b � a = − G3(u,v, 0)du dv. −b −a Putting all of this together, we have��� �� div G�du dv dw = �d�G S. H3 ∂H2 · This completes step 1. 2 � �� �� Step 2: We suppose that there is a compact subset K ⊂ M and a parametrisation �g : H3 ∩ U −→ M ∩ W, such that (1) F�(�x)= �0 for any �x ∈ M \ K. (2) K ⊂ M ∩ W .We may write�g(u,v,w)=(x(u,v,w),y(u,v,w),z(u,v,w)). Define G�: H3 −→ R3 , by ∂(y,z) ∂(x,z) ∂(x,y)G1 = F1 + F3∂(v,w) − F2 ∂(v,w) ∂(v,w) ∂(y,z) ∂(x,z) ∂(x,y)G2 = −F1 ∂(u,w)+ F2 ∂(u,w) − F3 ∂(u,w) ∂(y,z) ∂(x,z) ∂(x,y)G3 = F1 + F3 ,∂(u,v) − F2 ∂(u,v) ∂(u,v)for any (u,v,w) ∈ V and otherwise zero. Put differently, G�(u,v,w)= F�· A if (u,v,w) ∈ U �0 otherwise, where A is the matrix of cofactors of the derivative D�g. One can check (that is, there is a somewhat long and involved calculaation similar, but much worse, than ones that appear in the proof of Green’s Theorem or Stokes’ Theorem) that div G�= div F�det D�g = div F�∂(x,y,z) . ∂(u,v,w) We have ��� ��� div F�dx dy dz = div G�dx dy dz M H3 �d�= G S, ∂H2 · d�= F�S, · ∂M 3 � � � � � where the last equality needs to be checked (this is relatively straightforwward) This completes step 2. Step 3: We finish off in the standard way. We may find a partition of unity k1= ρi, i=1 where ρi is a smooth function which is zero outside a compact subset Ki such that F�i = ρiF�is a smooth vector field, which satisfies the hypothesis of step 2, for each 1 ≤ i ≤ k. We have kF�= F�i. i=1 and so �� k�� curl F�dS�= curl F�i · dS�· SSi=1 k� = F�i · d�s i=1 ∂M = F�d�s. �· ∂M 4 MIT OpenCourseWarehttp://ocw.mit.edu 18.022 Calculus of Several Variables Fall 2010 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.