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(1 point) For each of the following equations, find θ∈[0,2π) correct to at least 3 decimal places. Use commas to separate your answers if there are multiple solutions. Enter "none" if there are no solutions. 0.135sin(θ)−0.826=0⇒θ= 0.135−0.826sin(θ)=0⇒θ= 0.135cos(θ)+0.826=0⇒θ= 0.135+0.826cos(θ)=0⇒θ= 0.135cos(θ)−0.826sin(θ)=0⇒θ= 0.135sin(θ)+0.826cos(θ)=0⇒θ=

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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

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