一道封闭空间求积分题目5ycos(4x)dx+3xydy 这个求积分,从(0,0)到(4,2)再到(0,2)再到(0,0
2个回答
(1)从(0,0)到(4,2)：x:0→4 y=x/2 dx=dx dy=1/2dx
∫[L1]5ycos(4x)dx+3xydy =∫[0→4][5x/2cos(4x)+3x^2/2*1/2]dx
=[5/8xsin(4x)+5/32cos(4x)+1/4x^3][0→4]
=5/2sin(16)+5/32cos(16)+16-5/32
(2)从(4,2)到(0,2):y=2 dy=0 x:4→0 dx=dx
∫[L2]5ycos(4x)dx+3xydy =∫[4→0][10cos(4x)dx=5/2sin(4x)[4→0]=-5/2sin(16)
(3)从(0,2)到(0,0)：x=0 dx=0 y:2→0 dy=dy
∫[L3]5ycos(4x)dx+3xydy =∫[2→0][0+0dy=0
∴∫[L]5ycos(4x)dx+3xydy =5/2sin(16)+5/32cos(16)+16-5/32]-5/2sin(16)
=5/32cos(16)+16-5/32]
=5/32[cos(16)-1]+16
=16-5(sin8)^2/16
