Question

Evaluate the line integral H C F · dr for the vector field F(x, y, z) = −y i + x j − z k, where the closed curve C is the boundary of the triangle with vertices (0, 0, 5), (2, 0, 1), and (0, 3, 2) traced in this order.

Answer 1

The line integral of the vector field F(x, y, z) over the closed curve C can be evaluated by parameterizing each side of the triangle and computing the integral over these paths.

Step-by-step explanation:

The problem asks to evaluate the line integral of the vector field F(x, y, z) = −y i + x j − z k over the closed curve C, which is the boundary of the triangle with vertices (0, 0, 5), (2, 0, 1), and (0, 3, 2) traced in the given order. To solve this, we need to parameterize each side of the triangle and then compute the line integral over each parameterized path.

The triangle has three sides that can be represented as vectors:

From (0, 0, 5) to (2, 0, 1): The vector representing this side is r₁(t) = (2t, 0, 5 - 4t) where t ranges from 0 to 1.

From (2, 0, 1) to (0, 3, 2): The vector representing this side is r₂(t) = (2 - 2t, 3t, 1 + t) where t also ranges from 0 to 1.

From (0, 3, 2) to (0, 0, 5): The vector representing this side is r₃(t) = (0, 3 - 3t, 2 + 3t) where t ranges from 0 to 1.

To find the value of the line integral, we will plug in these parameterized paths into the integral of F · dr. The line integral over the closed path is the sum of the line integrals over each piece of the path. We will have to calculate the line integral over each of these paths and then add them together to get the final result.

Answer 2

The value of the line integral is 9.5.

Break the curve C into three line segments:

C1: from (0, 0, 5) to (2, 0, 1)

C2: from (2, 0, 1) to (0, 3, 2)

C3: from (0, 3, 2) back to (0, 0, 5)

Parameterize each line segment:

C1: r(t) = (2t, 0, 5 - 4t), 0 ≤ t ≤ 1

C2: r(t) = (2 - 2t, 3t, 1 + t), 0 ≤ t ≤ 1

C3: r(t) = (0, 3 - 3t, 2 - t), 0 ≤ t ≤ 1

Evaluate the line integral along each segment:

C1:

F(r(t)) = (0, 2t, -5 + 4t)

dr/dt = (2, 0, -4)

F · dr = 0 + 0 + (-5 + 4t)(-4) = 20 - 16t

∫_C1 F · dr = ∫_0^1 (20 - 16t) dt = 4

C2:

F(r(t)) = (-3t, 2 - 2t, -1 - t)

dr/dt = (-2, 3, 1)

F · dr = (-3t)(-2) + (2 - 2t)(3) + (-1 - t)(1) = 11 - 7t

∫_C2 F · dr = ∫_0^1 (11 - 7t) dt = 4

C3:

F(r(t)) = (-3 + 3t, 0, -2 + t)

dr/dt = (0, -3, -1)

F · dr = (-3 + 3t)(0) + 0 + (-2 + t)(-1) = 2 - t

∫_C3 F · dr = ∫_0^1 (2 - t) dt = 1.5

Add the line integrals along each segment:

∫_C F · dr = ∫_C1 F · dr + ∫_C2 F · dr + ∫_C3 F · dr = 4 + 4 + 1.5 = 9.5