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1 Vector lculus 16.7 tokes Theorem Nme: toke's Theorem is higher dimensionl nlogue to Green's Theorem nd the Fundmentl Theorem of clculus. Why, you sk? Well, let us revisit these theorems. Fundmentl Theorem of lculus: If f :[, b] is function differentible on the interior of (,b) nd continous on [,b], then b df f () b f () o let us describe this in the lnguge of vector clculus. Given the properties on f nd n open smooth intervl, then the integrl of the derivtive of f on the interior of the region is the difference of the vlues of f on the boundry of the intervl. Now let us look t Green's Theorem. As you cn see, Green's Theorem sys tht if you tke the double integrl of the projection of the curl of vector field onto the norml vector the plnr region over which you re integrting, then it is equilvent to tking the integrl of the vlues of the tngentil component of the vector field on the boundry of the plnr region. Note tht this theorem lso reltes the double integrl of the curl (derivtive) of f on the interior of region with the single integrl of the vlues of f on the boundry. The Del Opertor For toke's Theorem, we re going to hve to define the curl of vector field in generl. In order to do this, we will hve to define the del opertor, : ( ) ( ), by If f is rel-vlued multivrible function, then (,, ) x y z ( f, f f, f ) x y z

2 Vector lculus 16.7 tokes Theorem Nme: nd if F (x, y, z) = ( f1( x, yz, ), f2( xyz,, ), f3( xy,, z) ) is vector field. i j k F x y z f f f f3 f2 f1 f3 f2 f1,, y z z x x y curl F The third component function is the second version of Green's Theorem nd ws clled the k-th component of the curl. The vector field F is the generlized curl of vector field, F. o now we cn define curl F F Now, if F is vector field with potentil function f, then it is the grdient of function. This mens tht 3 f : hs continuous second-order prtil derivtives, then

3 Vector lculus 16.7 tokes Theorem Nme: crl u ( f) ( f) i j k x y z f f f x y z Therefore, the curl of conservtive vector field is f f f f f f (,, ) y z z y z x x z x y y x 0 curl ( f ) 0 The converse of this sttement is not true in generl nd we won't prove it here becuse we hve enough 3 to prove in this section. If F is vector field defined on ll whose component functions hve continuous prtil derivtives nd curl(f) = 0, then F is conservtive vector field. toke's Theorem Remember tht Green's Theorem relted double integrl over plne region D to line integrl round its plne boundry curve. toke's Theorem reltes surfce integrl over surfce to line integrl round the boundry curve of, which is spce curve. (tewrt's lculus p. 1121) The orienttion on the surfce induces positive orienttion on the boundry curve.

4 Vector lculus 16.7 tokes Theorem Nme: Recll tht nd b F dr ( f1, f2, f3) dr ( F) n d curl F n d We will prove this theorem by proving three identities: f3 f2 f1 f3 f2 f1,, nd y z z x x y (1.1) (1.2) f f f1 0) dr (0, n d z y 1 1 (,0,, ) f f ( f2 dr (, 0 z x 2 2 0,,0), ) n d f3 f3 (1.3) (0,0, f3) dr (,,0) n d y z Then the sum of these will be equivlent to proving toke's Theorem. We will prove the first theorem, the other two re proven similrly. Let : :[0,1] be simple prmeteriztion of with positive orienttion given by n, nd be simple prmeteriztion of the counterclockwise oriented boundry of the region which defines the prmeteriztion of. Then r is the curve whose grph is the boundry of.

5 Vector lculus 16.7 tokes Theorem Nme: 2 3 Before we strt the computtion, I wnt to remind you tht : is function from two vribles on region R onto the surfce. o we will define the prmeteriztion of the surfce, by ( u, v) ( 1( u, v), 2( u, v), 3( u, v) ) nd Fxyz (,, ) ( f1( xyz,, ), f2( x, yz, ), f3( xyz,, )) is defined on. Therefore, we cn write F s composition functions, F( 1( u, v), 2( u, v), 3( u, v)). Hence b d( ) ( f1,0,0) dr ( 1,0,0) f b d = ( f1,0,0) ( 1( 1(), t 2()), t 2( 1(), t 2()), t 3( 1(), t 2())) t by definition of r(t) b 1 d1() t d1 d2( t) f1( ( ())( t ( ()) t ( () t ) ) by chin rule u dv 1 1 ( f1) du ( f1) dv since ( t) ( u( t), v( t)) u v 1 1 ( f1) ( f1) v u d by Green's Theorem u v 1 If the lst step is difficult, then note tht if we define M ( f1 ) 1 nd N ( f1 ) u v then Green s Theorem is stisfied. By product rule, the lst eqution cn be written s: 2 2 ( f1) 1 1 ( f1) 1 1 ( f1) ( f1 ) d u v uv v u vu ince the mixed prtils re equl, we cn simplify this eqution to * ( f1) 1 ( f1) 1 u v v u d

6 Vector lculus 16.7 tokes Theorem Nme: Also, since 1 x, 2 y, nd 3 z, then the chin rule gives us f f f ( f ) u x u y u z u nd f1 1 f1 2 f1 3 ( f1 ) v x v y v z v Therefore, 1 f1 1 1 f1 2 1 f1 3 1 ( f1 ) u v x u v y u v z u v nd 1 f1 1 1 f1 2 1 f1 3 1 ( f1 ) v u x v u y v u z v u o, s you cn see, the integrl (*) cn ctully be written s f f f f f f d x u v y u v z u v x v u y v u z v u f1 2 1 f1 2 1 f1 3 1 f1 3 1 d y u v y v u z u v z v u You will see in the lst expression on the previous pge, we cn fctor out the prtils of the components of f with respect to y nd z. f f d z u v v u y v u u v Now, note tht the expression is the second component to the norml vector given u v v u by the prmeteriztion ρ nd is the third component to the norml vector given by v u u v the prmeteriztion ρ. You cn see this by looking t the curl of ρ. i j k u v u u u v v v

7 Vector lculus 16.7 tokes Theorem Nme: Therefore, f f d z u v v u y v u u v f z f N d y 1 1 0,, The other identities re your tke home quiz due TOMORROW. MUA HA HA HA HA (EVIL laugh) Just Kidding