A particle starts at the point
(−3, 0),
moves along the x-axis to (3, 0), and then along the semicircle
y = sqrt(9 − x^2)
to the starting point. Use Green's Theorem to find the work done on this particle by the force field
F(x, y) = <2x, x^3 + 3xy^2>
.
To find the work done on the particle by the force field using Green's Theorem, we need to calculate the line integral of the force field around the closed path formed by the x-axis and the semicircle.
Step 1: Parameterize the curve
Let's first parameterize the x-axis and the semicircle separately.
For the x-axis segment, we can use the parameterization:
r(t) = (x(t), y(t)) = (t, 0)
where -3 ≤ t ≤ 3
For the semicircle segment, we can use the parameterization:
r(t) = (x(t), y(t)) = (cos(t), sqrt(9 - cos^2(t)))
where 0 ≤ t ≤ π
Step 2: Calculate the line integrals
Now, we need to calculate the line integral along both segments of the closed curve using the Green's Theorem formula:
∬R (∂Q/∂x - ∂P/∂y) dA = ∮C P dx + Q dy
where R is the region enclosed by the closed curve, C is the curve itself, P and Q are the components of the force field F(x, y), and dA is the differential area element.
For the x-axis segment, we have:
∫C1 P dx + Q dy = ∫C1 (2x) dx + (x^3 + 3xy^2) dy
Substituting the parameterization, we get:
∫C1 (2t) dt + (t^3 + 3t(0)^2) (0) = ∫C1 2t dt
Evaluating this integral from -3 to 3:
∫C1 2t dt = [t^2] from -3 to 3 = 3^2 - (-3)^2 = 18
For the semicircle segment, we have:
∫C2 P dx + Q dy = ∫C2 (2x) dx + (x^3 + 3xy^2) dy
Substituting the parameterization, we get:
∫C2 (2cos(t)) (-sin(t)) + (cos^3(t) + 3cos(t)(sqrt(9 - cos^2(t)))^2) (sqrt(9 - cos^2(t))') dt
Simplifying and using the trigonometric identity cos^2(t) + sin^2(t) = 1, we get:
∫C2 -2cos(t) sin(t) + (cos^3(t) + 3cos(t)(9 - cos^2(t))) (-sin(t)) dt
Expanding and simplifying further:
∫C2 -2sin(t)cos(t) - 3cos(t)sin(t) + 27cos(t)sin(t) - cos^4(t)sin(t) - 3cos^2(t)sin(t) dt
∫C2 (26cos(t)sin(t) - cos^4(t)sin(t) + 24cos(t)sin(t)) dt
Combining like terms:
∫C2 (50cos(t)sin(t) - cos^4(t)sin(t)) dt
Since cos(t) and sin(t) are periodic with period 2π, the integral over a semicircle centered at the origin will be zero:
∫C2 (50cos(t)sin(t) - cos^4(t)sin(t)) dt = 0
Step 3: Calculate the total work
The total work done on the particle is the sum of the line integrals over both segments:
Work = ∫C P dx + Q dy = ∫C1 P dx + Q dy + ∫C2 P dx + Q dy = 18 + 0 = 18
Therefore, the work done on the particle by the force field F(x, y) = <2x, x^3 + 3xy^2> is 18 units of work.