Find the work done by a force F= -xyi+y²j + zk in moving a particle over the circular path x² + y² = 4, z = 0 from (2,0,0) to (0,2,0)
F=−xyi+y2j+zkF= -xyi+y²j + zkF=−xyi+y2j+zk
x2+y2=4x2=4−y2x=4−y2x²+y² = 4\\ x² = 4-y²\\ x = \sqrt{4-y²}x2+y2=4x2=4−y2x=4−y2
F=−(4−y2)1/2×yi+y2j+zkF = -(4-y²)^{1/2} ×yi + y²j +zk\\F=−(4−y2)1/2×yi+y2j+zk
dw = F.dx
In j direction
W=∫02y(4−y2)dy=8/3 JW=∫^2_0y(\sqrt{4-y²})dy =8/3\ JW=∫02y(4−y2)dy=8/3 J
(in z direction work done is always zero)
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