Question

Determine the work done by the force F =(xy +3z)i + (2y2 - x2)+(z - 2yk in
taking a particle from x = 0 to x = 1 along a curve defined by the equations
x² = 2y 2x3 = 32​

Answer 1

Here, we have to find work done by force F= (xy + 3z)i + (2y - x²)j + (z - 2y)k in taking particle from x=0 to x=1 along the curve defined by x²=2y , 2x³=32 .

We know that work done is  W=∫F. dl           where F and dl are vectors

Since the particle is to be taken in x direction only so dl=dx i

$F.dl =( (xy + 3z)i + (2y - x²)j + (z-2y)k ). (dx i)$

       $= xy + 3z$

given,

$x^{2} =2y\\ y= \frac{x^{2} }{2}$

$2x^{3} = 3z \\ z= \frac{2x^{3} }{3}$

F.dl= x³/2 + 2x³

      = 5x³/2

$\int\ {F} \, dl = \int\limits^1_0 {\frac{5x^{3} }{2} } \, dx$

hence, W = ∫F.dl = 5/8