※ 引述《KOREALee (韩国最高)》之铭言： : 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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