计算∫Γx2dx+y2dy+z2dz,其中Γ是从点(1,1,1)到点(2,3,4)的直线段.

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问题 计算∫Γx2dx+y2dy+z2dz,其中Γ是从点(1,1,1)到点(2,3,4)的直线段.

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答案设Γ的参数方程为x=t+1,y=2t+1,z=3t+1,t:0→1.则 ∫Γx2dx+y2dy+z2dz=∫01(t+1)2d(t+1)+(2t+1)2d(2t+1)+(3t+1)2d(3t+1) =∫01[t2+2t+1+2(4t2+4t+1)+3(9t2+6t+1)]dt =∫01(36t2+28t+6)dt =(12t3+14t2+6t)|01 =32

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