※ 引述《hwujialuen (松原拓)》之铭言： : Evaluate the surface integral : 1 ∫ ∫xy dS : S : S is the boundary of the region enclosed by the cylinder : x^2 +z^2=1 and the planes y=0 and x+y =2 x=x , y=y , z=√(1-x^2) 1 J = ---------- √(1-x^2) 1 2-x xy ∫ ∫xy dS = ∫ ∫ ---------- dydx S -1 0 √(1-x^2) 1 x(2-x)^2 = ∫ ------------ dx -1 2 √(1-x^2) 1 x^3 - 4x^2 + 4x = ∫ -------------------- dx -1 2 √(1-x^2) 1 (-2) x^2 = ∫ ------------- dx (奇函数消掉) -1 √(1-x^2) let x=sint pi/2 pi/2 = ∫ (-2) (sint)^2 dt (因为 ∫ (sint)^2 +(cost)^2 dt -pi/2 -pi/2 pi/2 1 = ∫ 1 dt = pi ) = (-2)(---)(pi) -pi/2 2 = -pi 最後因为有上下两面(z>0,z<0) 所以要乘2 ans:"可能是" -2pi : 2 Evaluate the surface integral ∫ ∫F dS : S : F(x,y,z)=yj -zk : S consists of the paraboloid y= x^2 +z^2 ,0<y<1 ,and the : disk x^2 +z^2 ≦1 ,y=1 这题用div ∫∫∫(1-1)dv = 0 V : 3 ∫ ∫(x^2 +y^2 +z^2)dS : S : S consists of the cylinder in Exercise 10 together with : its top and bottom disks : Exercise10: S is the part of the cylinder x^2 +y^2=9 : between the planes z=0 and z=2 S = S1 + S2 + S3 let S1:surface of the cylinder x^2 +y^2=9 S2:the planes z=0 between cylinder x^2 +y^2=9 S3:the planes z=2 between cylinder x^2 +y^2=9 ∫ ∫(x^2 +y^2 +z^2)dS S1 2pi(3) 2 = ∫ ∫ 9 + z^2 dzds 0 0 = (18+ 8/3)6pi ∫ ∫(x^2 +y^2 +z^2)dS S2 = ∫ ∫ x^2+y^2 dS S2 2pi 3 = ∫ ∫ r^2 rdr dt 0 0 = (81/4)2pi ∫ ∫(x^2 +y^2 +z^2)dS S3 = ∫ ∫ x^2+y^2 + 4 dS S3 2pi 3 = ∫ ∫ (r^2 + 4 )r drdt 0 0 = (81/4 + 18)2pi -- ※ 编辑: beatitude 来自: 218.168.242.213 (07/17 13:17)