Answer to Question #229930 in Chemical Engineering for Lokika

Question

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)

Answer 1

$F= -xyi+y²j + zk$

$x²+y² = 4\\ x² = 4-y²\\ x = \sqrt{4-y²}$

$F = -(4-y²)^{1/2} ×yi + y²j +zk\\$

dw = F.dx

In j direction

$W=∫^2_0y(\sqrt{4-y²})dy =8/3\ J$

(in z direction work done is always zero)