Use Green's Theorem to evaluate

C
F · dr.
(Check the orientation of the curve before applying the theorem.)
F(x, y) =
y2 cos x, x2 + 2y sin x

C is the triangle from (0, 0) to (1, 3) to (1, 0) to (0, 0)

To evaluate the line integral using Green's Theorem, we need to find the double integral of the curl of F over the region enclosed by the curve C.

1. Orientation of the curve C:
The curve C is a triangle formed by the points (0, 0), (1, 3), (1, 0), and it is a closed curve since it starts and ends at the same point. The orientation of the curve C can be determined by tracing along the curve in a counterclockwise direction. Verify that the orientation is counterclockwise before applying Green's Theorem.

2. Calculating the curl of F:
Let's find the curl of F by taking the partial derivatives with respect to x and y:

∂F/∂x = ∂/∂x (y^2 cos x) = -y^2 sin x
∂F/∂y = ∂/∂y (x^2 + 2y sin x) = 2y sin x

The curl of F is given by ∇ × F = (∂F/∂y, -∂F/∂x):

∇ × F = (2y sin x, y^2 sin x)

3. Applying Green's Theorem:
Green's Theorem states that for a vector field F = (P, Q) and a simple closed curve C with orientation given by a counterclockwise direction, the line integral of F · dr along C is equal to the double integral of the curl of F over the region D enclosed by C:

∮C F · dr = ∬D ∇ × F dA

In this case, the region D is the triangle formed by the points (0, 0), (1, 3), and (1, 0).

4. Evaluating the double integral:
We need to find the double integral of the curl of F over the region D. The orientation of the curve C is counterclockwise, so we can directly apply Green's Theorem.

∬D ∇ × F dA = ∬D (2y sin x, y^2 sin x) dA

Since the region D is a triangle, we can rewrite the double integral as an iterated integral:

∬D ∇ × F dA = ∫[x=a..b] ∫[y=g(x)..h(x)] (2y sin x, y^2 sin x) dy dx

In this case, since D is a triangle, the range of integration for y is determined by the lines forming the sides of the triangle.

For the line segment from (0, 0) to (1, 3), the range of y is from 0 to 3x.
For the line segment from (0, 0) to (1, 0), the range of y is from 0 to 0.

Thus, the double integral becomes:

∬D (2y sin x, y^2 sin x) dA = ∫[x=0..1] ∫[y=0..3x] (2y sin x, y^2 sin x) dy dx + ∫[x=1..1] ∫[y=0..0] (2y sin x, y^2 sin x) dy dx

Simplifying the integrals:

∫[x=0..1] ∫[y=0..3x] (2y sin x, y^2 sin x) dy dx + ∫[x=1..1] ∫[y=0..0] (2y sin x, y^2 sin x) dy dx
= ∫[x=0..1] [(y^2 sin x/2)∣∣y=0..3x, (y^3 sin x/3)∣∣y=0..3x] dx + 0
= ∫[x=0..1] [9x^3 sin x/3 - 0] dx
= ∫[x=0..1] (3x^3 sin x) dx

Now, you can use any appropriate method to evaluate the integral, such as integration by parts or a numerical method if needed.

Once you find the value of the integral, that will be the result of evaluating the line integral using Green's Theorem for the given vector field F over the curve C.