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Both sides of the divergence theorem for F(x,y,z)=(x2,y2,z2) on the closed unit box

Example

Let B=[0,1]3 with the six-patch presentation of The closed unit box, with its six faces, is an elementary solid region and let F(x,y,z)=(x2,y2,z2) on R3. Then both sides of the divergence theorem equal 3: the volume integral of divF=2(x+y+z) over B is 3, and the six face fluxes are 1 for each of the faces x=1, y=1 and z=1 and 0 for each of the faces x=0, y=0 and z=0, so the boundary flux is 3 as well.

Facts & Assumptions

Given: The box B=[0,1]3 with the six patches φz+,φz,φx+,φx,φy+,φy on S=[0,1]2 of The closed unit box, with its six faces, is an elementary solid region, and the field F(x,y,z)=(x2,y2,z2).

[F1]

The divergence of a C1 field is divG=i<niGi (Divergence and curl of a C1 vector field), and a map is Ck when each component is (Ck Euclidean maps and diffeomorphisms).

[F2]

For a compatible finite patch presentation the oriented flux is the sum of the patch values, each being S(Gφj)(φj,u×φj,v) (Finitely patched regular surfaces, their area, scalar integrals, and flux, Unit normal fields, orientations, and flux through a regular surface patch).

[L1]

The six faces above, with those parametrizations, make B an elementary solid region, and the oriented area vectors are the constants ez,ez,ex,ex,ey,ey respectively (The closed unit box, with its six faces, is an elementary solid region).

[L2]

For an elementary solid region E with presentation Σ and a C1 field G on an open set containing E, EdivG=EG,n (The divergence theorem on an elementary solid region).

[L3]

For a bounded Jordan set ERp+q and integrable g whose sections are integrable outside a content-zero set, Eg=h(x)dx with h(x)=Exgx (Fubini over a bounded Jordan set when all but a content-zero family of sections are integrable).

[L4]

If G is differentiable at every point of [a,b] with a<b and G is integrable there, then abG=G(b)G(a) (The second fundamental theorem: if G is differentiable on [a,b] with G=f and f is integrable, then abf=G(b)G(a)).

[L6]

Every continuous real function on a compact Jordan measurable set is Riemann integrable over it (A continuous real function on a compact Jordan measurable set is Riemann integrable over that set).

Verification

technique · direct
1.1

The three components of F are x2, y2 and z2, so by [L5] the partial derivatives xFx=2x, yFy=2y and zFz=2z exist and are continuous on R3, as are the six off-diagonal ones, which vanish. Hence F is C1 on R3 by [F1] and divF=2x+2y+2z=2(x+y+z).

givenF1L5
1.2

By [L1] the box B with those six patches is an elementary solid region, and R3 is an open set containing it, so [L2] applies to F on B.

givenL1L2
2.1

The function 2(x+y+z) is continuous on the compact Jordan set B, hence integrable by [L6]. Applying [L3] to split off the z coordinate and then again to split off y, and evaluating each inner integral by [L4] and [L5]: 012(x+y+z)dz=2(x+y)+1, then 01(2(x+y)+1)dy=2x+2, then 01(2x+2)dx=3. So BdivF=3 by step 1.1.

step 1.1L3L4L5L6
2.2

By [F2] and [F3] the six face fluxes are the integrals over S of F(φj),φj,u×φj,v, which by [L1] is ± one coordinate of F along the patch. Face by face: on φz+ it is Fz(u,v,1)=1; on φz it is Fz(v,u,0)=0; on φx+ it is Fx(1,u,v)=1; on φx it is Fx(0,v,u)=0; on φy+ it is Fy(v,1,u)=1; and on φy it is Fy(u,0,v)=0. Since S=[0,1]2, [L3], [L4], and [L5] give S1=01011dudv=1, while S0=0; so the six patch fluxes are 1,0,1,0,1,0 and the boundary flux is their sum, 3.

step 1.2F2F3L1L3L4L5
3.1

Steps 2.1 and 2.2 give 3 for the volume integral and 3 for the boundary flux, which is what [L2] asserts of them.

step 2.1step 2.2L2

Remarks

  • The three vanishing faces vanish for a reason worth naming. On the face x=0 the flux integrand is Fx evaluated there, and Fx=x2 is zero on that face; the same happens for y and z. It is the choice of integrand, not any symmetry of the box, that makes half the faces contribute nothing, and each of the six was evaluated rather than inferred.

Depends on

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