Answer:
To evaluate the line integral ∮_C F · dr using Green's theorem, we first need to compute the curl of the vector field F = ⟨ycos(x) - xysin(x), xy + xcos(x)⟩:
1. Calculate ∇ × F (the curl):
∇ × F = (∂Q/∂x - ∂P/∂y) i + (∂P/∂x + ∂Q/∂y) j
Where P = ycos(x) - xysin(x) and Q = xy + xcos(x).
∂P/∂y = -xsin(x) - xsin(x) - ycos(x)
∂Q/∂x = y + cos(x) - xsin(x) + xcos(x)
Now, calculate the curl:
∇ × F = (-2xsin(x) - 2ysin(x) - 2xcos(x) + y + cos(x)) i + (2x - 2ycos(x)) j
2. Green's theorem states that for a simply connected region D, the line integral ∮_C F · dr is equal to the double integral of the curl of F over D:
∮_C F · dr = ∬_D (∇ × F) · dA
Now, let's apply Green's theorem to evaluate the line integral over the triangle C, which consists of three line segments from (0,0) to (0,6), from (0,6) to (3,0), and from (3,0) to (0,0).
3. Orient the curve C counterclockwise. Since you didn't specify the orientation, we will assume a counterclockwise orientation.
4. Parameterize the three line segments:
- Line segment from (0,0) to (0,6): r(t) = ⟨0, t⟩ for 0 ≤ t ≤ 6.
- Line segment from (0,6) to (3,0): r(t) = ⟨3t/6, 6 - t⟩ for 0 ≤ t ≤ 6.
- Line segment from (3,0) to (0,0): r(t) = ⟨3 - 3t/3, 0⟩ for 0 ≤ t ≤ 3.
5. Calculate the line integrals over each segment:
a. Line integral over the first segment:
∫_(C1) F · dr = ∫_(t=0 to 6) F(r(t)) · r'(t) dt
= ∫_(t=0 to 6) ⟨tcos(0) - 0tsin(0), t*0 + tcos(0)⟩ · ⟨0, 1⟩ dt
= ∫_(t=0 to 6) ⟨0, t⟩ · ⟨0, 1⟩ dt
= ∫_(t=0 to 6) 0 dt
= 0
b. Line integral over the second segment:
∫_(C2) F · dr = ∫_(t=0 to 6) F(r(t)) · r'(t) dt
= ∫_(t=0 to 6) ⟨(6-t)cos(t) - 0t*sin(t), (6-t)*t + (6-t)cos(t)⟩ · ⟨3/6, -1⟩ dt
= ∫_(t=0 to 6) ⟨(6-t)cos(t), (6-t)t⟩ · ⟨1/2, -1⟩ dt
= (1/2)∫_(t=0 to 6) ((6-t)cos(t) - (6-t)t) dt
This integral is a bit more complicated to solve analytically, but it can be numerically approximated.
c. Line integral over the third segment:
∫_(C3) F · dr = ∫_(t=0 to 3) F(r(t)) · r'(t) dt
= ∫_(t=0 to 3) ⟨(3-t)cos(t) - 0t*sin(t), (3-t)*t + (3-t)cos(t)⟩ · ⟨-1/3, 0⟩ dt
= ∫_(t=0 to 3) ⟨-(3-t)cos(t), 0⟩ · ⟨-1/3, 0⟩ dt
= (1/3)∫_(t=0 to 3) (3-t)cos(t) dt
Again, this integral is a bit more complicated to solve analytically, but it can be numerically approximated.
6. Now, calculate the double integral of the curl over the region D enclosed by C:
∬_D (∇ × F) · dA = ∬_D ((-2xsin(x) - 2ysin(x) - 2xcos(x) + y + cos(x)) i + (2x - 2ycos(x)) j) · dA
This integral can also be evaluated numerically.
In summary, you'll need to approximate the line integrals over the second and third segments and the double integral of the curl over the region D using numerical methods or appropriate software.
Explanation: