18 02A Topic 46 Applications and interpretations of the divergence theorem Author Jeremy Orloff Read SN V15 sec 1 for div in notation V15 sec 2 to middle p 3 Del Notation i j k x y z x y z f f f f f f grad f f i j k x y z x y z M N P hM N P i divF F x y z x y z curlF F more on this later Aside on the Laplacian Laplacian 2 f f divgradf 2f 2f 2f x2 y 2 z 2 Interpretation divF source rate of F at every point Reason Take a small volume around x0 y0 z0 Call the RRRvolume D and its boundary S The divergence theorem says the flux of F through S D divF dV For small D the triple integral divF x0 y0 z0 volume D flux of F through S amount out from x0 y0 z8 divF source rate Vol D unit volume unit time Extended Divergence Theorem just like Green s Theorem If a solid volume D has a boundary consisting of several closed surfaces then the divergence theorem still holds if for each surface we use the normals that point away from the volume ZZ ZZZ ZZZ F dV divF dV S1 S2 D D Example Let F be the gravitational field of a mass m1 For a closed surface S show ZZ 4 Gm1 if S surrounds m1 F n dS 0 otherwise S hx y zi We set our coordinate system so m1 is at the origin F x y z G m1 3 x To compute divF use the chain rule and the formula x 1 x2 1 y2 1 z2 divF G m1 3 5 G m1 3 5 G m1 3 5 3 3 3 2 2 2 3 x y z G m1 3 3 5 0 Because F is not defined at the origin we have to be careful in applying the divergence theorem 1 18 02A topic 46 In preparation for considering an arbitrary closed surface we compute the flux directly when S is a sphere of radius a centered at the origin with outward pointing normals On S F and n are parallel in opposite directions F n F G m1 a2 ZZ RR G m1 G m1 flux S F n dS 2 dS 2 area G m1 4 a a S In general there are two cases i S does not surround 0 and ii S surrounds 0 In case i see picture F is continuously differentiable on the entire volume inside S RR RRR RRR flux S F n dS D divF dV D 0 dV 0 as claimed In case ii see picture let S1 be a small sphere centered at 0 that is entirely inside S Using the extended divergence theorem we have note the signs RR RRR divF dV 0 S S1 F n dS RR RR D F n dS S1 F n dS 0 RRS RR S F n dS S1 F n dS 4 Gm1 again as claimed Example Let F be the gravitational field from a mass distribution Show that the flux through a closed surface S is 4 GM where M is the total mass enclosed by S We divide the mass into infinitesimal pieces dm and let dF be the field due to dm By the previous example if dm is inside S then the flux of dF through S is 4 G dm If dm is outside S then the flux is 0 4 G dm if it s inside S That is each dm contributes 0 if it s outside S Since F is the sum of all the dF the net flux is just a sum of 4 G dm over all dm inside S that is flux 4 G M End of topic 46 notes 2
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