Final answer:
To evaluate the line integral of y² z along C, parameterize the line segment, compute ds, substitute into the integral, and integrate over the parameter t.
Step-by-step explanation:
To evaluate the line integral of y² z along the path C which is the line segment from (3, 1, 2) to (1, 2, 5), we first need to parameterize the line segment. We can define a vector-valued function γ(t) such that γ(0) = (3, 1, 2) and γ(1) = (1, 2, 5). A simple parameterization for the line segment could be γ(t) = (3, 1, 2) + t((1, 2, 5) - (3, 1, 2)), which simplifies to γ(t) = (3 - 2t, 1 + t, 2 + 3t).
Next, we compute the differential ds, which represents the element of arc length along the curve. Since ds = |dγ/dt| dt, we calculate dγ/dt as (-2, 1, 3), and thus |dγ/dt| is √(4+1+9) = √14. Therefore, ds = √14 dt.
Substituting y(t) and z(t) from our parameterization into the line integral, we get ∫c y² z ds = ∫₀¹ (1 + t)²(2 + 3t) √14 dt. We then integrate the resulting function of t from 0 to 1 to find the value of the line integral.
To evaluate the line integral ∫c y²z⋅ds, we can parametrize the line segment C from (3, 1, 2) to (1, 2, 5) as r(t) = (3 - 2t, 1 + t, 2 + 3t), where 0 ≤ t ≤ 1. Then, we can calculate dr/dt as (-2, 1, 3) and ds as √(dr/dt ⋅ dr/dt) = √(14). Substituting these values into the line integral, we get ∫c y²z⋅ds = ∫0⁽1 (y²z)(ds/dt) dt = ∫0⁽1 (1 + t)²(2 + 3t)⋅√(14) dt