Answer:
To use Green's theorem to evaluate ∮_C F ⋅ dr, where F(x, y) = ⟨x^2 + 4, tan(x)⟩, and C is the triangle from (0, 0) to (1, 1) to (0, 1) to (0, 0), we need to follow these steps:
1. Find the partial derivatives of F:
∂F/∂x = ⟨2x, sec^2(x)⟩
∂F/∂y = ⟨0, 0⟩
2. Apply Green's theorem, which states:
∮_C F ⋅ dr = ∬_D (∂F/∂y - ∂F/∂x) dA
3. Determine the region D enclosed by the curve C. In this case, it's the triangle with vertices (0, 0), (1, 1), and (0, 1).
4. Compute the double integral ∬_D (∂F/∂y - ∂F/∂x) dA.
Now, we need to calculate ∬_D (∂F/∂y - ∂F/∂x) dA. Since ∂F/∂x and ∂F/∂y are both constant vectors (not functions of x and y), the integral becomes:
∬_D (∂F/∂y - ∂F/∂x) dA = (∂F/∂y - ∂F/∂x) ∬_D dA
Now, let's evaluate this integral over the region D:
D is the triangle with vertices (0, 0), (1, 1), and (0, 1). We can set up the integral as follows:
∬_D dA = ∫[0 to 1] ∫[0 to x] dy dx
Now, we can compute the integral:
∫[0 to 1] ∫[0 to x] dy dx = ∫[0 to 1] [y]_[0 to x] dx
= ∫[0 to 1] (x - 0) dx
= ∫[0 to 1] x dx
= [1/2 * x^2]_[0 to 1]
= 1/2 * (1^2 - 0^2)
= 1/2
Now, multiply this result by (∂F/∂y - ∂F/∂x):
(∂F/∂y - ∂F/∂x) ∬_D dA = (0 - (2x - sec^2(x))) * (1/2)
= (-2x + sec^2(x)) * (1/2)
Now, we need to evaluate this expression over the region D. First, find the values at the bounds of the region:
At x = 1:
(-2(1) + sec^2(1)) * (1/2) = (-2 + sec^2(1)) * (1/2)
At x = 0:
(-2(0) + sec^2(0)) * (1/2) = (0 + 1) * (1/2) = 1/2
Now, we can calculate the difference:
(-2 + sec^2(1)) * (1/2) - (1/2) = (-2 + sec^2(1) - 1) * (1/2)
= (-2 + sec^2(1) - 1)/2
= (-1 + sec^2(1))/2
Now, we need to find sec^2(1). Recall that sec(x) = 1/cos(x). So, sec^2(1) = 1/cos^2(1).
Use a calculator to find the value of cos(1) and then take the reciprocal:
cos(1) ≈ 0.540302306
sec^2(1) ≈ 1/(0.540302306^2) ≈ 3.42551882
Now, plug this value back into the expression:
(-1 + sec^2(1))/2 ≈ (-1 + 3.42551882)/2 ≈ 1.21275941/2 ≈ 0.606379705
So, the value of ∮_C F ⋅ dr is approximately 0.606379705.
Explanation: