Answer:
Explanation:
a. Use a parameterization of the curve:
The curve is the unit circle centered at the origin (x² + y² = 1). We can parameterize this curve using polar coordinates:
Let x = cos(t) and y = sin(t), where t goes from 0 to 2π (to cover the whole unit circle).
Now, the vector field F = x i + xy j becomes:
F = cos(t) i + (cos(t) * sin(t)) j
The line integral along the closed curve can be calculated as follows:
∮(F · dr) = ∮(F(x(t), y(t)) · (dx/dt i + dy/dt j) dt
∮(F · dr) = ∮((cos(t) i + (cos(t) * sin(t)) j) · ((-sin(t) i + cos(t) j) dt
∮(F · dr) = ∮(cos(t) * (-sin(t)) + cos(t) * sin(t)) dt
∮(F · dr) = ∮(0) dt
Since the integrand is 0, the line integral along the closed curve is zero.
b. Use Green's theorem:
Green's theorem relates the line integral of a vector field over a closed curve to the double integral of the curl of the vector field over the region enclosed by the curve.
For the vector field F = x i + xy j, the curl is:
curl(F) = (d/dx(xy)) - (d/dy(x)) = (y - 0) i - (1 - 0) j = y i - j
The curve is the unit circle centered at the origin (x² + y² = 1). The region enclosed by the curve is the disk with radius 1 centered at the origin.
Now, Green's theorem states:
∮(F · dr) = ∬(curl(F) · dA)
Since the curl of F is y i - j, the double integral of the curl over the disk is:
∬(curl(F) · dA) = ∬((y i - j) · (dxdy))
To evaluate this double integral, we use polar coordinates:
∬((y i - j) · (dxdy)) = ∫(from 0 to 2π) ∫(from 0 to 1) (r sinθ i - r j) dr dθ
∬((y i - j) · (dxdy)) = ∫(from 0 to 2π) ∫(from 0 to 1) r sinθ - r dr dθ
∬((y i - j) · (dxdy)) = ∫(from 0 to 2π) [(-1/2)r² cosθ] (from 0 to 1) dθ
∬((y i - j) · (dxdy)) = ∫(from 0 to 2π) (-1/2) cosθ dθ
∬((y i - j) · (dxdy)) = [-1/2 sinθ] (from 0 to 2π)
∬((y i - j) · (dxdy)) = 0
Since the double integral is zero, the line integral along the closed curve is also zero.
c. Use Stokes's theorem:
Stokes's theorem relates the line integral of a vector field around a closed curve to the surface integral of the curl of the vector field over a surface whose boundary is the closed curve.
Since we already computed the curl of F as y i - j, we can use it directly.
Now, for Stokes's theorem, we need to find a surface that has the unit circle as its boundary. A natural choice is the surface of the disk with radius 1 centered at the origin.
The normal vector of the surface points outward and is given by n = k (in the positive z-direction).
The surface integral of the curl of F over the disk is:
∬(curl(F) · dS) = ∬((y i - j) · n) dS
∬(curl(F) · dS) = ∬((y i - j) · k) dS
∬(curl(F) · dS) = ∬(0) dS
Since the integrand is 0, the surface integral is zero.
Therefore, according to Stokes's theorem, the line integral of F around the closed curve is also zero