Answer:
75 - 25π.
Explanation:
To find the work done by a force field on a particle moving along a curve, we use the line integral of the force field over that curve.
In this case, the curve is a circle of radius 5 centered at (0, 0, 3) lying on the plane z = 3. We can parameterize this curve using polar coordinates as:
r(t) = (5cos(t), 5sin(t), 3), where t goes from 0 to 2π.
The differential of the curve, dr(t), is given by:
dr(t) = (-5sin(t), 5cos(t), 0) dt
Now we need to calculate the work done by the force field F along this curve. The line integral of F over the curve is given by:
W = ∫ F · dr = ∫ (2x +y²Z)dx + (2x-4y+Z)dy + (x-2y-Z²)dz
Substituting x = 5cos(t), y = 5sin(t), and z = 3, we get:
W = ∫ (10cos(t) + 25sin²(t)·3) (-5sin(t))dt
∫ (10cos(t) - 20sin(t) + 3) (5cos(t))dt
∫ (5cos(t) - 10sin(t) - 9) (0)dt
Simplifying, we get:
W = -75∫sin(t)cos(t)dt + 50∫cos²(t)dt + 0
Using the trigonometric identities sin(2t) = 2sin(t)cos(t) and cos²(t) = (1 + cos(2t))/2, we can simplify this further:
W = -75∫(1/2)sin(2t)dt + 25∫(1 + cos(2t))dt
= -75·(1/2)·(-cos(2t))∣₀^(2π) + 25·(t + (1/2)sin(2t))∣₀^(2π)
= 75 - 25π
Therefore, the work done by the force field F on the particle moving along the circle of radius 5 centered at (0, 0, 3) lying on the plane z = 3 is 75 - 25π.