Answer:
52 units
Explanation:
We require ∫꜀ F.Tds along the line segment given
We obtain the equation of the line in parameterized vector form
r(t) = (1 - t) (1,0) + t (3,3) (where 0≤t≤1)
r(t) = (1 - t, 0) + (3t, 3t)
r(t) = (1 + 2t, 3t) = (1 + 2t)î + 3tj
x = (1 + 2t) and y = 3t
∫꜀ F.Tds = ∫¹₀ F(r(t)).r'(t) dt
r'(t) = 2î + 3j
F(x,y) = (x+6)î +(6y+3)j
x = (1 + 2t) and y = 3t
F(r(t)) = (1 + 2t + 6)î + (6(3t) + 3)j
F(r(t)) = (2t + 7)î + (18t + 3)j
∫꜀ F.Tds = ∫¹₀ F(r(t)).r'(t) dt = ∫¹₀ [(2t + 7)î + (18t + 3)j] . [2î + 3j] dt = ∫¹₀ [(4t + 14) + (54t + 9)] dt
∫꜀ F.ds = ∫¹₀ (58t + 23) dt = [29t² + 23t]¹₀ = (29 + 23) = 52 units.