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The third version of Green's Theorem (equation 16.5.2) we saw was: ∫∂DF⋅Nds=∫∫D∇⋅FdA. With minor changes this turns into another equation, the Divergence Theorem:

Theorem 16.9.1 (Divergence Theorem) Under suitable conditions, if E is a region of three dimensional space and D is its boundary surface, oriented outward, then ∫∫DF⋅NdS=∫∫∫E∇⋅FdV.

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(Px+Qy+Rz)dV=∫∫∫EPxdV+∫∫∫EQydV+∫∫∫ERzdV. The double integral may be rewritten: ∫∫DF⋅NdS=∫∫D(Pi+Qj+Rk)⋅NdS=∫∫DPi⋅NdS+∫∫DQj⋅NdS+∫∫DRk⋅NdS. To prove that these give the same value it is sufficient to prove that ∫∫DPi⋅NdS=∫∫∫EPxdV,∫∫DQj⋅NdS=∫∫∫EQydV,and∫∫DRk⋅NdS=∫∫∫ERzdV. Not surprisingly, these are all pretty much the same; we'll do the first one.

We set the triple integral up with dx innermost: ∫∫∫EPxdV=∫∫B∫g2(y,z)g1(y,z)PxdxdA=∫∫BP(g2(y,z),y,z)−P(g1(y,z),y,z)dA, where B is the region in the y-z plane over which we integrate. The boundary surface of E consists of a "top'' x=g2(y,z), a "bottom'' x=g1(y,z), and a "wrap-around side'' that is vertical to the y-z plane. To integrate over the entire boundary surface, we can integrate over each of these (top, bottom, side) and add the results. Over the side surface, the vector N is perpendicular to the vector i, so ∫∫sidePi⋅NdS=∫∫side0dS=0. Thus, we are left with just the surface integral over the top plus the surface integral over the bottom. For the top, we use the vector function r=⟨g2(y,z),y,z⟩ which gives ry×rz=⟨1,−g2y,−g2z⟩; the dot product of this with i=⟨1,0,0⟩ is 1. Then ∫∫topPi⋅NdS=∫∫BP(g2(y,z),y,z)dA. In almost identical fashion we get ∫∫bottomPi⋅NdS=−∫∫BP(g1(y,z),y,z)dA, where the negative sign is needed to make N point in the negative x direction. Now ∫∫DPi⋅NdS=∫∫BP(g2(y,z),y,z)dA−∫∫BP(g1(y,z),y,z)dA, which is the same as the value of the triple integral above.

Example 16.9.2 Let F=⟨2x,3y,z2⟩, and consider the three-dimensional volume inside the cube with faces parallel to the principal planes and opposite corners at (0,0,0) and (1,1,1). We compute the two integrals of the divergence theorem.

The triple integral is the easier of the two: ∫10∫10∫102+3+2zdxdydz=6. The surface integral must be separated into six parts, one for each face of the cube. One face is z=0 or r=⟨u,v,0⟩, 0≤u,v≤1. Then ru=⟨1,0,0⟩, rv=⟨0,1,0⟩, and ru×rv=⟨0,0,1⟩. We need this to be oriented downward (out of the cube), so we use ⟨0,0,−1⟩ and the corresponding integral is ∫10∫10−z2dudv=∫10∫100dudv=0.

Another face is y=1 or r=⟨u,1,v⟩. Then ru=⟨1,0,0⟩, rv=⟨0,0,1⟩, and ru×rv=⟨0,−1,0⟩. We need a normal in the positive y direction, so we convert this to ⟨0,1,0⟩, and the corresponding integral is ∫10∫103ydudv=∫10∫103dudv=3.

The remaining four integrals have values 0, 0, 2, and 1, and the sum of these is 6, in agreement with the triple integral. ◻

Example 16.9.3 Let F=⟨x3,y3,z2⟩, and consider the cylindrical volume x2+y2≤9, 0≤z≤2. The triple integral (using cylindrical coordinates) is ∫2π0∫30∫20(3r2+2z)rdzdrdθ=279π.

For the surface we need three integrals. The top of the cylinder can be represented by r=⟨vcosu,vsinu,2⟩; ru×rv=⟨0,0,−v⟩, which points down into the cylinder, so we convert it to ⟨0,0,v⟩. Then ∫2π0∫30⟨v3cos3u,v3sin3u,4⟩⋅⟨0,0,v⟩dvdu=∫2π0∫304vdvdu=36π. The bottom is r=⟨vcosu,vsinu,0⟩; ru×rv=⟨0,0,−v⟩ and ∫2π0∫30⟨v3cos3u,v3sin3u,0⟩⋅⟨0,0,−v⟩dvdu=∫2π0∫300dvdu=0. The side of the cylinder is r=⟨3cosu,3sinu,v⟩; ru×rv=⟨3cosu,3sinu,0⟩ which does point outward, so ∫2π0∫20⟨27cos3u,27sin3u,v2⟩⋅⟨3cosu,3sinu,0⟩dvdu=∫2π0∫2081cos4u+81sin4udvdu=243π. The total surface integral is thus 36π+0+243π=279π. ◻

Exercises 16.9

Here is a Sage cell if you'd like to use it.

Ex 16.9.1 Using F=⟨3x,y3,−2z2⟩ and the region bounded by x2+y2=9, z=0, and z=5, compute both integrals from the Divergence Theorem. (answer)

Ex 16.9.2 Let E be the volume described by 0≤x≤a, 0≤y≤b, 0≤z≤c, and F=⟨x2,y2,z2⟩. Compute ∫∫∂EF⋅NdS. (answer)

Ex 16.9.3 Let E be the volume described by 0≤x≤1, 0≤y≤1, 0≤z≤1, and F=⟨2xy,3xy,zex+y⟩. Compute ∫∫∂EF⋅NdS. (answer)

Ex 16.9.4 Let E be the volume described by 0≤x≤1, 0≤y≤x, 0≤z≤x+y, and F=⟨x,2y,3z⟩. Compute ∫∫∂EF⋅NdS. (answer)

Ex 16.9.5 Let E be the volume described by x2+y2+z2≤4, and F=⟨x3,y3,z3⟩. Compute ∫∫∂EF⋅NdS. (answer)

Ex 16.9.6 Let E be the hemisphere described by 0≤z≤√1−x2−y2, and F=⟨√x2+y2+z2,√x2+y2+z2,√x2+y2+z2⟩. Compute ∫∫∂EF⋅NdS. (answer)

Ex 16.9.7 Let E be the volume described by x2+y2≤1, 0≤z≤4, and F=⟨xy2,yz,x2z⟩. Compute ∫∫∂EF⋅NdS. (answer)

Ex 16.9.8 Let E be the solid cone above the x-y plane and inside z=1−√x2+y2, and F=⟨xcos2z,ysin2z,√x2+y2z⟩. Compute ∫∫∂EF⋅NdS. (answer)

Ex 16.9.9 Prove the other two equations in the display 16.9.1.

Ex 16.9.10 Suppose D is a closed surface, and that D and F are sufficiently nice. Show that ∫∫D(∇×F)⋅NdS=0 where N is the outward pointing unit normal.

Ex 16.9.11 Suppose D is a closed surface, D is sufficiently nice, and F=⟨a,b,c⟩ is a constant vector field. Show that ∫∫DF⋅NdS=0 where N is the outward pointing unit normal.

Ex 16.9.12 We know that the volume of a region E may often be computed as ∫∫∫Edxdydz. Show that this volume may also be computed as 13∫∫∂E⟨x,y,z⟩⋅NdS where N is the outward pointing unit normal to ∂E.