The line integral ∫y²z ds over the line segment from (3,1,2) to (1,2,5) is equal to -19/3.
The line integral ∫y²z ds is a path integral along the given curve C, which is a line segment from the point (3,1,2) to the point (1,2,5). In this context, ds represents the differential arc length along the curve.
To compute the line integral, we parameterize the curve with vector-valued functions. Let's denote the position vector as r(t) = [x(t), y(t), z(t)]. The differential arc length ds is given by ds = ||r'(t)|| dt, where r'(t) is the derivative of the position vector with respect to t.
For the given curve, we can parameterize it with a linear interpolation between the initial and final points. Let r(t) = [3 - 2t, 1 + t, 2 + 3t], where t ranges from 0 to 1.
Now, compute the integral ∫y²z ds by substituting the parameterization and evaluating the integral along the curve. The resulting value is -19/3, representing the line integral over the specified line segment from (3,1,2) to (1,2,5). This value quantifies the contribution of the function y²z along the given curve.