f x y z ds f P S (,, ) lim ( ) ds f P S S S.

Transcription

1 4.7 rface Integral We diide srface into patches patch, mltiply by the area with area, and form the sm n n i j f P ( ). We ealate f at a point P, in each Then we take the limit as the patch size approaches 0, and define the srface integral of f oer the srface as n n (,, ) lim ( ) P 0 i j f x y z d f P Property : n n n n lim ( ) area of P 0 i j i j d f P. rface integral formla: If a smooth parametric srface is gien by the eqation r(, ) x(, ) i y(, ) j z(, ) k, (, ) then the srface integral of f oer is () f ( x, y, z) d f ( r(, )) r r da x y z x y z where r i j k, r i j k pecial case: If the srface is gien by then srface integral of f oer is z g( x, y), ( x, y), () Example : Ealate g g f ( x, y, z) d f ( x, y, z) da x y yd, where is the srface z x y, 0 x, 0 y. z z oltion: ince and y, x y

2 z z yd y da y y da y 4y dydx x y Example : Ealate zd, where is the srface whose sides are gien by the cylinder x y, whose bottom is the disk x of the plane z x that lies aboe. Ths, the srface integral oer is y in the plane z 0, and whose top is the part oltion: For, we se and z as parameters and write its parametric eqations as x cos, y sin, z z, where 0 and 0 z x cos. Therefore, r r sin cos 0 cos i sin j z i j k r r (cos ) (sin ) cos zd z r r da zdzd ince lies in the plane z 0, we hae zd 0d 0 The top srface lies aboe the nit disk : x y and is part of the plane z x. o, taking g( x, y) x in Formla (), and conerting to polar coordinates, we hae z z zd ( x) da ( r cos ) 0rdrd x y zd zd + zd + zd = 0

3 Oriented rfaces: From now on, we consider only orientable srfaces. We start with a srface that has a tangent plane at eery point ( x, y, z ) on. There are two nit normal ectors n and n n at ( x, y, z ). If it is possible to choose a nit normal ector n at eery sch point ( x, y, z) so that n aries continosly oer, then is called oriented srface and the gien choice of n proides with an orientation. There are two possible orientations for any orientable srface. For a srface z g( x, y), the indced orientation is gien by the nit normal ector ince the k( z axis) is positie, this gies the pward orientation of the srface. If is a smooth orientable srface gien in parametric form by a ector fnction r(, ), then it is atomatically spplied with the orientation of the nit normal ector n r r r r and the opposite orientation is gien by n.

4 For a closed srface, that is, a srface that is the bondary of a solid region E, the conention is that the positie orientation is the one for which the normal ectors point otward from E, and inwardpointing normal gie the negatie orientation. Positie orientation negatie orientation rface Integrals of Vector Fields: efinition: If F Pi Qj Rk is a ector field defined on an oriented srface with nit normal ector n, then the srface integral of F oer is F d F nd This integral is also called the flx of F across. Formlas:. For a srface z g( x, y) with pward nit normal ector, we hae g g F d P Q R da x y For a downward nit normal ector, we mltiply by -.. For a parametric srface is gien by a ector fnction r(, ) with nit normal ector is gien n r r by r r, then we hae the following F d F r r da

5 Example : Ealate F d, where F yi xj zk and is the bondary of the solid region E enclosed by the paraboloid z x y and 0 z. oltion: consists of a parabolic top srface and a circlar bottom srface. ince is a closed srface, we se the conention of positie (otward) orientation. This means that is oriented pward, is the projection of on xy plane, namely, the disk x y. ince P y Q x R z x y,,, g x g y g( x, y) x y, x, y, F d P Q R da y( x) x( y) x y da = 4 xy x y da ( 4r cos sin r ) rdrd g g x y The disk is oriented downward, so its nit normal ector is n k and we hae F d F ( k) d zd 0d 0 Ths, zd zd + zd = Example 4: Find the flx of the ector field F( x, y, z) zi yj xk across the nit sphere x y z. oltion: Using the parametric representation We hae and, we hae r(, ) sin cos i + sin sin j+cos k, (, ) [0, ] [0, ] F( r(, )) cos i sin sin j sin cos k

6 i j k i j k x y z r r cos cos cos sin sin sin ( sin ) sin cos 0 x y z = sin cos isin sin jsin cos k Ths, F( r(, )) r r cos sin cos sin sin sin cos cos The flx is (cos sin cos sin sin sin cos cos ) F d F r r da dd 4 ( cos sin cos sin sin ) dd 0 0