212658013-Bai-17-Tich-Phan-Mat-Loai-2.ppt
-
Upload
chang-trai-lang-man -
Category
Documents
-
view
8 -
download
8
Transcript of 212658013-Bai-17-Tich-Phan-Mat-Loai-2.ppt
-
TCH PHN MT LOI 2
-
PHP TUYN CA MT CONG.Cho mt cong S: F(x, y, z) = 0, M(x0,y0,z0) SL l ng cong trong S i qua M. Tip tuyn ca L ti M gi l tip tuyn ca S ti M.Cc tip tuyn ny cng thuc 1 mt phng gi l mt tip din ca S ti M.Php tuyn ca mt tip din ti M gi l php tuyn ca S ti M.
-
PHP TUYN MT CONGGi s L S c pt: x = x(t), y = y(t), z = z(t)M = (x(t0), y(t0), z(t0)) LVt ch phng ca tip tuyn ti M l :M S: F(x,y,z) = 0, ta c:
-
l php vector ca S ti MMt k hiu khc:(gradient ca F ti M)(ng vi mi ng cong trong S v qua M)v cc vector t l
-
Mt s v d tm php vectora/ Mt cu(v cc vector t l)
-
a/ Mt trMt s v d tm php vector(v cc vector t l)Ma/ Mt tr
-
Mt s v d tm php vectora/ Mt nn
-
MT NH HNGS c gi l mt nh hng (mt 2 pha) nu cho php vector ti MS di chuyn dc theo 1 ng cong kn khng ct bin, khi quay v im xut pht vn khng i chiu.Ngc li, php vector o chiu, th S c gi l mt khng nh hng (mt 1 pha ).Pha ca S l pha m ng trn , php vector hng t chn ln u.(Chng trnh ch xt mt 2 pha)
-
Mt mt pha
-
Mt hai pha
-
V d tm PVT tng ng vi pha mt conga/ Mt cuphp VT ngoiphp VT trong
-
b/ Mt trMPVT ngoiPVT trong
-
PVT ngoiPVT trongc/ Mt nn
-
Php vector n vxzy
-
NH NGHA TCH PHN MT LOI 2Cho cc hm P, Q, R lin tc trn mt nh hng S.Gi php vector n v ca S lTch phn mt loi 2 ca P, Q, R trn S nh ngha bi
-
V D1/ Cho S l pha ngoi ca na mt cutnhTi M (x, y, z) trn S, php vector n v l
-
2/ Cho S l ca phn mpb chn bi cc mt ta , ly pha trc nhn t pha dng trc Oz, tnh
-
S: z = 1 x y ,
-
CCH TNH TP MT LOI 2V php vector n v thng thng rt phc tp nn ta c th dng cch tnh sau thay th:
-
Vit pt S dng: z = z(x,y) (bt buc)Tm hnh chiu Dxy ca S ln mp z = 0 (Oxy) ( bt buc)Tnh
-
Du + () nu php vt hp vi chiu dng Oz mt gc nhn ( t )Nu = /2 (S//Oz hoc S cha Oz) I3 = 0(p dng vi I3)Lu
-
Tng t:I2 :Pt ca S: y = y(x, z)Dzx = hc ca S ln OzxGc ca PVT so vi Oy+I1:Pt ca S: x = x(y, z)Dyz = hc ca S ln OyzGc ca PVT so vi Ox+S // Ox (hoc cha Ox) I1 = 0S // Oy (hoc cha Oy) I2 = 0S // Oz (hoc cha Oz) I3 = 0
-
V D1/ Cho S l pha ngoi ca na mt cutnh
-
2/ Cho S l pha ngoi ca na mt cutnhI = I2S = S1 S2 :Lu : S1 v S2 i xng qua mp x = 0
-
3/ Cho S l pha ngoi ca mt cutnhLu : S1 v S2 i xng qua mp z = 0S = S1 S2 :
-
Lu v tnh i xngS gm S1 v S2 i xng qua mp z = 0 R(x, y, z) chn theo z : I3 = 0 R(x, y, z) l theo z: Tng t cho I1(xt P v mp x=0), I2(xt Q v mp y=0)
-
4/ Cho S l pha trn ca phn mt tr z = y2 b chn bi mt tr x2 + y2 = 1, tnhI = I1 + I2 + I3 S cha Ox I1 = 0 S i xng qua mp y = 0, Q = 2z cosy chn theo y I2 = 0
-
NH L GAUSS - OSTROGRATXKICho l min ng v b chn trong R3, S l pha ngoi mt bin ca (S l mt cong kn). P, Q, R l cc hm lin tc trn .Tch phn mt loi 2Tch phn bi ba
-
V D1/ Cho S l pha ngoi mt bao khi : x2 + y2 z 1. Tnh
-
: x2 + y2 z 1
-
2/ Cho S l pha ngoi phn mt paraboloid z = x2 + y2 b chn bi mp z = 1. TnhS l mt h.
-
Thm S1 vo to thnh mt knS1 l pha trn phn mp z = 1 b chn trong paraboloid.Gi l vt th c bao bi S S1.
-
(xem v d trc)p dng cng thc G-O:
-
S1: z = 1, trong tr x2+y2 =1
-
3/ Cho S l pha trong mt bao khi gii hn bi:z = 4 y2 , x = 0, x = 4, z = 0. Tnh:
-
CNG THC STOKESCho ng cong C l bin ca mt nh hng S. C c gi l nh hng dng theo S nu khi ng trn S(php tuyn hng t chn ln u) s nhn thy C i ngc chiu kim ng h.
-
CNG THC STOKESCho P(x,y,z), Q(x,y,z), R(x,y,z) v cc o hm ring lin tc trn S, C l bin nh hng dng ca S. Khi :
-
V D1/ Cho C l giao tuyn ca tr x2 + y2 = 1 v tr z = y2 ly ngc chiu kim ng h nhn t pha dng Oz. Tnh:
-
Chn S l pha trn mt tr z = y2P = x + yQ = 2x2 z R = xy2
-
z = y2 b chn trong tr x2+y2=1
-
2/ Cho C l giao tuyn ca tr x2 + y2 = 1 v mt phng x + z = 1 ly ngc chiu kim ng h nhn t gc ta . Tnh:
-
Chn S l pha di phn mt phng x + z = 1, b chn bn trong tr.
-
Chuyn sang tp mt loi 1S: x + z = 1,
-
S: z = 1 x , b chn trong tr x2+y2=1