Use Green's Theorem to evaluate ∫ C
​
F⋅dr. (Check the orientation of the curve before applying the theorem.) F(x,y)=⟨ycos(x)−xysin(x),xy+xcos(x)⟩,C is the triangle from (0,0) to (0,10) to (2,0) to (0,0)

The value of the line integral is ∫ C F⋅dr = 88/3 + 4cos(2) + 8/3sin(2) - 10sin(2)

To use Green's Theorem, we first need to calculate the curl of the vector field F(x, y). The curl of a vector field F = ⟨P, Q⟩ is given by the following formula:

curl(F) = ∂Q/∂x - ∂P/∂y

Let's calculate the curl of F(x, y):

P = ycos(x) - xysin(x)

Q = xy + xcos(x)

∂Q/∂x = y + cos(x) - xsin(x) - xsin(x) - xcos(x) = y - 2xsin(x) - xcos(x)

∂P/∂y = cos(x)

curl(F) = ∂Q/∂x - ∂P/∂y = (y - 2xsin(x) - xcos(x)) - cos(x)

        = y - 2xsin(x) - xcos(x) - cos(x)

Now, we can apply Green's Theorem. Green's Theorem states that for a vector field F = ⟨P, Q⟩ and a curve C oriented counterclockwise,

∫ C F⋅dr = ∬ R curl(F) dA

Here, R represents the region enclosed by the curve C. In our case, the curve C is the triangle from (0, 0) to (0, 10) to (2, 0) to (0, 0).

To apply Green's Theorem, we need to determine the region R enclosed by the curve C. In this case, R is the entire triangular region.

Since the curve C is a triangle, we can express the region R as follows:

R = {(x, y) | 0 ≤ x ≤ 2, 0 ≤ y ≤ (10 - x/2)}

Now, we can evaluate the double integral:

∫ C F⋅dr = ∬ R curl(F) dA

        = ∫[0,2]∫[0,10 - x/2] (y - 2xsin(x) - xcos(x) - cos(x)) dy dx

Evaluating this double integral will give us the desired result.

∫[0,2]∫[0,10 - x/2] (y - 2xsin(x) - xcos(x) - cos(x)) dy dx

Let's integrate with respect to y first and then with respect to x:

∫[0,2]∫[0,10 - x/2] (y - 2xsin(x) - xcos(x) - cos(x)) dy dx

= ∫[0,2] [(1/2)[tex]y^2[/tex] - 2xsin(x)y - xcos(x)y - ycos(x)] [0,10 - x/2] dx

= ∫[0,2] [(1/2)[tex](10 - x/2)^2[/tex]- 2xsin(x)(10 - x/2) - xcos(x)(10 - x/2) - (10 - x/2)cos(x)] dx

Now, let's simplify and evaluate this integral:

= ∫[0,2] [(1/2)(100 - 20x + x^2/4) - (20x - [tex]x^2[/tex]sin(x)/2) - (10x -[tex]x^2[/tex]cos(x)/2) - (10 - x/2)cos(x)] dx

= ∫[0,2] [50 - 10x + [tex]x^2/8[/tex] - 20x + [tex]x^2[/tex]sin(x)/2 - 10x +[tex]x^2[/tex]cos(x)/2 - 10cos(x) + xcos(x)/2] dx

Now, we can integrate term by term:

= [50x - 5[tex]x^2/2[/tex] + [tex]x^3/24[/tex]- [tex]10x^2[/tex] + [tex]x^2cos(x)[/tex]- [tex]5x^2 + x^3sin(x)/3 - 10sin(x) + xsin(x)/2[/tex]] evaluated from 0 to 2

= [100 - 20 + 8/24 - 40 + 4cos(2) - 20 + 8/3sin(2) - 10sin(2) + sin(2)] - [0]

Simplifying further:

= 88/3 + 4cos(2) + 8/3sin(2) - 10sin(2)

Therefore, the value of the given line integral using Green's Theorem is:

∫ C F⋅dr = 88/3 + 4cos(2) + 8/3sin(2) - 10sin(2)

Learn more about green's theorem

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