vektornyy_analiz_23 (Векторный анализ (Кузнецов Л.А.))

8 _ 01_ 23u = x3 + y 2 + z 2l = j − k,M (1, -3, 4 )∂U ∂U∂U∂U=cos α +cos β +cos γ∂I∂x∂y∂zI = 1 + 1 = 2 ⇒ cos α = 0;cos β = 1/ 2;cos γ = −1/ 2∂U= 3x 2∂x1∂U=⋅ 2y =∂y 2 y 2 + z 2∂U1=⋅ 2z =∂z 2 y 2 + z 2yy2 + z2zy2 + z24∂U∂U−3 ∂U| = 3;| =;| = ⇒∂x M∂y M5 ∂z M 5∂U ∂U∂U∂U−3 1 4 1−7cos α +cos β +⇒=cos γ =⋅−=∂I∂x∂y∂z52 5 2 5 28 _ 02 _ 23V =∂V∂x∂V∂y∂V∂z∂U∂x∂U∂y∂U∂z⎛ 13 2x + 3 y 2 − 2 z 2 , U = x 2 yz 3 , M ⎜⎜ 2, ,2⎝ 33⎞⎟.2 ⎟⎠= 3x 2= 6y= −4 z= 2 xyz 3= x2 z3= 3 x 2 yz 2grad V = {∂V ∂V ∂V∂U ∂U ∂U;; };grad U = {;;}∂x ∂y ∂z∂x ∂y ∂zgrad V ( M ) = {6; 2; −2 6}; grad V ( M ) = 8grad U ( M ) = { 6,3 6, 6}; grad U ( M ) = 4 6grad V ( M ) ⋅ grad U ( M )6 2 + 6 6 − 12 6==0⇒grad V ( M ) ⋅ grad U ( M )8⋅4 6⇒ α = arccos(0) = π / 2cos α =8 _ 04 _ 23 _1GGGGa = ( x + z )i + ( y + z ) j + ( z − x − y )kS : x 2 + y 2 + z 2 = 4; P : z = 0( z ≥ 0)П = Пн + Пбок ⇒ Пбок = П − − ПнG GG∂a y ∂az ⎞⎛ ∂aП = ∫∫ (a, n)dS = ∫∫∫ div a dx dy dz = ∫∫∫ ⎜ x ++⎟ dx dy dz =∂x∂y∂z ⎠SVV ⎝1 41 4= ∫∫∫ (1 + 1 + 1) dx dy dz = 3∫∫∫ dx dy dz = 3 ⋅ ⋅ π R 3 = 3 ⋅ ⋅ π ⋅ 23 = 16π2 32 3VVGG GG Gn н = (0;0; −1); a ⋅ n н = x + y − z;(a ⋅ n н ) |z =0 = x − y2G Gx = r cos ϕ 2πПн = ∫∫ (a ⋅ n н )dS = ∫∫ ( x − y )dS == ∫ dϕ ∫ r 2 (cos ϕ − sin ϕ ) dr =y = r sin ϕSS00=2π2002π 22222∫ (cos ϕ − sin ϕ )dϕ ∫ r dr = ( sin ϕ + cos ϕ ) | ⋅ ∫ r dr = 0 ⋅ ∫ r dr = 0Пбок = П − Пн = 16π − 0 = 16π0008_04_23_2Проекция на плоскость OXY8 _ 05 _ 23 _1GGGGa = xi + 2 y j + 5 zkP : x + 2y + z / 2 = 1GGП = ∫∫ (an)dS = ∫∫ ( ax cos α + a y cos β + cz cos γ ) dSSSGGx + 2 y + z / 2 = 1 ⇒ 2 x + 4 y + z = 2 ⇒ n = {2; 4;1} ⇒ n = 21 ⇒⇒ cos α = 2 / 21;cos β = 4 / 21;cos γ = 1/ 21dS = 1 + ( z x′ ) + ( z ′y ) dx dy = 1 + ( −2 ) + ( −4 ) dx dy = 21 dx dy2211/ 2 − x / 200П = 21∫ dx1= ∫ dx∫1∫ (10 − 8x − 12 y ) dy = ∫ dx (10 y − 8 xy − 6 y )20⎛ 5x7x ⎞ 1 4=⎜− 3x 2 +⎟| =2 ⎠0 3⎝ 63241⎛ 2⎞x+2y +5(2 − 2 x − 4 y ) ⎟ dy =⎜2121⎝ 21⎠1/ 2 − x / 2020(1− x ) / 2|01= ∫ ( 5 x 2 / 2 − 6 x + 7 / 2 ) dx =08_05_23_2Проекция на OXY1.00.80.60.40.20.00.00.20.40.60.81.08 _ 08 _ 23GGGGa = xi − 2 y j + 3 zk⎧ x2 + y 2 = z 2S :⎨⎩z = 2xТ .к поверхность замкнутая, то воспользуемся формулойОстроградского − ГауссаG GG⎛ ∂ax ∂a y ∂az ⎞⋅⋅==++П=wandSadivdxdydz⎜⎟ dx dy dz =∫∫S∫∫∫∫∫∫∂x∂y∂z ⎠VV ⎝x = r cos ϕ= ∫∫∫ (1 − 2 + 3) dx dy dz = 2 ∫∫∫ dx dy dz = y = r sin ϕ =Vπ /2=2∫πdϕ− /2π /2=2z=zV∫−π / 2π /22 cos ϕ∫0dϕ2 cos ϕ∫2 r cos ϕr ⋅ dr∫π /2dz = 2∫πdϕ− /2r2( 2r 2 cos ϕ − r 3 ) ⋅ dr = 202 cos ϕ∫ r ( 2r cos ϕ − r ) ⋅ dr =20π /2⎛ 2r 3 cos ϕ r 4 ⎞ 2 cos ϕdϕ− ⎟ | =⎜∫34⎠ 0⎝−π / 2π /2π /24 cos 4 ϕ 82= 2 ∫ dϕ= ∫ cos 4 ϕ ⋅ dϕ =(1 + cos 2ϕ ) 2 ⋅ d ϕ =∫33 −π / 23 −π / 2−π / 2π /2π /2=221 + cos 4ϕ ⎞⎛(1 + 2 cos 2ϕ + cos 2 2ϕ ) ⋅ d ϕ =⎜1 + 2 cos 2ϕ +⎟ ⋅ dϕ =∫∫3 −π / 23 −π / 2 ⎝2⎠=2 ⎛ 3ϕsin 4ϕ ⎞ π / 2+ sin 2ϕ +⎜⎟ | =π3⎝ 28 ⎠ −π / 28 _10 _ 23JGGGF = ( y 2 − y )i + (2 xy + x) jL : x 2 + y 2 = 3( y ≥ 0)M (3;0); N (−3;0)⎧ x = 3cos t⎧dx = −3sin t dtA = ∫ ( Fx dx + Fy dy ) = ⎨⇒⎨=⎩ y = 3sin t⎩dy = 3cos t dtLπ()= ∫ ( 9sin 2 t − 3sin t ) (−3sin t dt) + (2 ⋅ 3cos t ⋅ 3sin t + 3cos t )(3cos t dt) =0π= 3∫ ( −9sin 3 t + 3sin 2 t + 18cos 2 t sin t + 3cos 2 t ) dt =0πππ⎛ π1 − cos 2t1 + cos 2t ⎞= 3 ⎜ −9 ∫ sin 2 t ⋅ sin t dt + 3∫dt − 18∫ cos 2 t ⋅ (− sin t dt) + 3∫⎟=22000⎝ 0⎠π⎛cos3 t π ⎛ t sin 2t ⎞ π ⎞⎛ t sin 2t ⎞ π= 3 ⎜ 9 ∫ (1 − cos 2 t ) ⋅ (− sin t dt) + 3 ⎜ −|−18⋅| + 3⎜ +⎟⎟|⎟=4 ⎠03 0 ⎝24 ⎠0⎠⎝2⎝ 0ππ⎛ π⎛⎞3π3π ⎞cos3 t π= 3 ⎜ 9 ∫ − sin t dt − 9∫ cos 2 t ⋅ (− sin t dt) ++ 12 +| + 3π + 12 ⎟⎟ = 3 ⎜ 9 cos t | − 922 ⎠3 00⎝⎠0⎝ 0= 3 ( −18 + 6 + 3π + 12 ) = 9π8 _11_ 23 _1GG GGa = 7 zi − x j + yzk⎧ x = 6 cos t⎪Г : ⎨ y = 6sin t⎪ z = 1/ 3⎩dx = −6sin t dt;dy = 6 cos t dt;dz = 0Ц = ∫ ( ax dx + a y dy + az dz ) =Г2π⎛02π=⎞1∫ ⎜⎝ 7 ⋅ 3 ⋅ (−6sin t dt) − 6 cos t ⋅ 6 cos t dt ⎟⎠ =2π2∫ ( 7 ⋅ (−2sin t dt) − 36 cos t dt ) = − ∫ (14sin t + 18 (1 + cos 2t ) ) dt =00⎛⎛ sin 2t ⎞ ⎞= − ⎜ −14 cos t + 18 ⎜ t +⎟ | = −36π2 ⎠ ⎟⎠ 0⎝⎝2π50.60.40.20.00 50 558_11_23_2Проекция на плоскость OXY642642224646.