[微積] 向量微積分
Erwin Kreyszig Advanced Engineering Mathematics 的 Ch9.9 521頁
10.
F = (y,z/2,3y/2)，C為圓形 x^2 + y^2 + z^2 = 6z, z = x+3
▽ × F = (1,0,-1)；N = (-r/√2,0,-r/√2)
∮(y)dx+(z/2)dy+(3y/2)dz = ∫∫(▽ × F)．N drdψ
C R
2π 3 2π 3
= ∫∫(1,0,-1)．(-r/√2,0,-r/√2)drdψ = ∫∫-√2rdrdψ = -9√2π
0 0 0 0
可是解答卻是-18π
13.
F = (y^3,0,x^3)，C為三角形邊界其頂點為(1,0,0),(0,1,0),(0,0,1)
▽ × F = (0,-3x^2,-3y^2)；N = (1,1,1)
∮(y^3)dx+(0)dy+(x^3)dz = ∫∫(▽ × F)．N dxdy
C R
1 1-y 1 1-y
= ∫∫(0,-3x^2,-3y^2)．(1,1,1)dxdy = ∫∫(-3x^2-3y^2)dxdy = -1/2
0 0 0 0
可是解答卻是-√3/10
Ch9 Review 522 頁
38.
F = (Sin[x],z,y)，S：y^2 +z^2 = 4，-1/2≦x≦1/2，y≧0，z≧0
為什麼第38題不能用底下的右式而要用以下的左式去計算？
^
∫∫F．n dA = ∫∫∫▽．FdV
S T
用左式算的結果是4，用右式算的結果是2πSin[1/2]，其解答是4
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