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Use Green's theorem to evaluate ∫CF⋅dr. (Check the orientation of the curve before applying the theorem.) F(x,y)=⟨x2+4,tan−1(x)⟩,

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

User Hanser
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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:

User Takuya
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