※ 引述《hwujialuen (松原拓)》之銘言：
: 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 (奇函數消掉)
-1 √(1-x^2)
let x=sint
pi/2 pi/2
= ∫ (-2) (sint)^2 dt (因為 ∫ (sint)^2 +(cost)^2 dt
-pi/2 -pi/2
pi/2
1 = ∫ 1 dt = pi )
= (-2)(---)(pi) -pi/2
2
= -pi
最後因為有上下兩面(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
這題用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
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※ 編輯: beatitude 來自: 218.168.242.213 (07/17 13:17)