Answer:
the work done by the force field = 24 π
Step-by-step explanation:
From the information given:
r(t) = 3 cos (t)i + 3 sin (t) j + 2 tk
= xi + yj + zk
∴
x = 3 cos (t)
y = 3 sin (t)
z = 2t
dr = (-3 sin (t)i + 3 cos (t) j + 2 k ) dt
Also F(x,y,z) = 6xi + 6yj + 6k
∴ F(t) = 18 cos (t) i + 18 sin (t) j +6 k
Workdone = 0 to 2π ∫ F(t) dr
[tex]\mathbf{= \int \limits ^{2 \pi} _{0} (18 cos (t) i + 18 sin (t) j +6k)(-3 sin (t)i+3cos (t) j +2k)\ dt}[/tex]
[tex]\mathbf{= \int \limits ^{2 \pi} _{0} (-54 \ cos (t).sin(t) + 54 \ sin (t).cos (t) + 12 ) \ dt}[/tex]
[tex]\mathbf{= \int \limits ^{2 \pi} _{0} 12 \ dt}[/tex]
[tex]\mathbf{= 12 \times 2 \pi}[/tex]
= 24 π
