※ 引述《KOREALee (韓國最高)》之銘言：
: Evaluate the following integral
: I = ∫ 2xyz^2 dx + (x^2z^2 + zcos(yz)) dy + (2x^2yz + ycos(yz)) dz
: L
: along the line segment L from P:(0,0,1) to Q:(2,pi/4,2)
There are two ways for this problem.
Do line integral
But is complex.
Stoke Theorem
∮F dot dr = ∮Curl F da
if Curl F = 0 -> ∮ F dot dr = 0 -> path independent.
You can do by yourself that here Curl F = 0 .
And if Curl F = 0 we can find g that satisfy " ▽g = f "
▽g = (gx,gy,gz) = (fx,fy,fz)
gi means partial to i
fi means i-component
gx = 2xyz^2 -> g = x^2yz^2 + h(y,z)
partial respect to y
gy = x^2z^2 + hy(y,z) = x^2z^2 + zcos(yz)
so we know h (y,z) = sin(yz) + k(z)
g = x^2yz^2 + sin(yz) +k(z)
gz = 2x^2yz + ycos(yz) + k'(z) = 2x^2yz + ycos(yz)
k'(z) = 0 -> k(z) = constant
so we get g = x^2yz^2 + sin(yz) + c
dg
∫F dot dr = ∫▽g dot dr = ∫--- dr = ∫dg = g +c
dr
so we can get the answer by put the last position and initial position
into g ,
B(x,y,z) |(2,1/4π,2)
∫f dot dr = g | = 4π + 1
A(x,y,z) |(0,0,1)
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