18 02A Topic 45 More surface integrals divergence theorem Author Jeremy Orloff Read SN V10 Closed Surfaces with outward normals Divergence F hM N P i divF Mx Ny Pz Divergence Theorem Also called Gauss Theorem This is like the flux form of Green s Theorem Take S a closed surface with interior D and F a vector field continuously differentiable on all of D Then ZZ ZZZ F n dS divF dV S D where n is the outward normal Notation Sometimes we write dS for n dS Example Verify the theorem for F xi yj zk and S sphere of radius a RR F n a RRRF n dS flux a area of S a 4 a2 4 a3 3 divF 3 D divF dV 3 vol 4 a QED Example Let S1 part of the paraboloid z 1 x2 y 2 above the xy plane and S2 the unit disk in the xy plane Take F hyz xz xyi and find the flux of F upward through S1 using the divergence theorem Write F M i N j P k where M yz N xz P xy divF Mx Ny Pz 0 ZZ ZZZ ZZZ The divergence theorem says flux F n dS divF dV 0 dV 0 S1 S2 D D ZZ ZZ ZZ ZZ F n dS F n dS 0 F n dS F n dS S1 S2 S1 S2 Therefore to find what we want we only need to compute the flux through S2 But S2 is in the xy plane so dS dx dy n k F n dS xy dx dy on S2 ZZ ZZ ZZ Since S2 is the unit disk symmetry gives xy dx dy 0 F n dS S2 S1 1 S2 F n dS 0 18 02A topic 45 2 Proof of the divergence theorem First we look at a small box Then just like for Green s Theorem we handle the general volume by adding together a lot of boxes Let F hM N P i P0 x0 y0 z0 S the box as shown The box has limits x x0 to x0 x y y0 to y0 y z z0 to z0 z Look at the flux through the top and bottom Z x0 x Z y0 y P x y z0 z dy dx Top n k F n dS P dS P dx dy flux x0 y0 x0 x Z y0 y Z Bottom n k F n dS P dS P dx dy flux P x y z0 dy dx x0 Z y0 x0 x Z y0 y P x y z0 z P x y z0 dy dx Z x0 x Z y0 y Z z0 z P x y z dz dy dx z x0 z0 RRR y0 P dV In brief net flux through top and bottom RRR D Z Likewise net flux through front and back Mx dV RRR D Likewise net flux through left and right D Ny dV ZZZ net flux out of the box Mx Ny Pz dV QED net flux through top and bottom x0 y0 D Example Let S be the part of the plane shown Compute the upward flux of F y j two ways i directly ii using the divergence theorem i The plane has equation x y 2 z 3 1 N h1 1 2 1 3i n dS h3 3 2 1i dx dy We know this because we know the k component in the vector must be 1 3 Thus F n dS y dx dy 2 Z 1 Z 2 2x ZZ 3 3 y dx dy y dy dx 1 flux 2 2 0 0 R ii To use the divergence theorem we need a closed surface S1 We use the tetrahedron with faces S S1 S2 S3 Computing flux S1 n j F n y 0 on S1 flux 0 1 S2 n i F n 0 on S2 flux 0 x S3 n k F n 0 on S3 flux 0 Now using all these zeros and the divergence theorem we have ZZ ZZ ZZZ F n dS F n dS divF dV S S S1 S2 S3 ZZZ z 3 S2 S 2 S3 y 2 D 1 1 dV volume base height 1 3 D R 1 End of topic 45 notes x y
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