The work done by the force field F along path C, parametrized by y = 4x² from (0,0) to (1,4), is 5 Joules.
To evaluate the line integral ∫c F.dr along the path y = 4x² from (0,0) to (1,4), we substitute y with 4x² and dy with 8x dx.
The vector field F = (5xy,7y²) becomes (20x³,112x´) after substitution.
Consequently:
∫c F.dr = ∫0¹ (20x³ dx, 112x´ dx).⋅dr
The work done along path C is a scalar, so we calculate the dot product of F with dr (dx, 0):
W = ∫0¹ 20x³ dx = 20 ∫0¹ x³ dx = 20 [x´/4] from 0 to 1 = 20 [1´/4 - 0´/4] = 5 Joules.