Evaluate R
C 3xy dx − xz dy + 2 dz where C is the curve formed by
traveling along the unit circle from (0, 1, 1) to (1, 0, 1) and then traveling along
the straight line from (1, 0, 1) to (2, 1, 1).

To evaluate the line integral of the vector field \(\mathbf{F} = \langle 3xy, -xz, 2 \rangle\) along the curve \(C\) formed by traveling along the unit circle from \((0, 1, 1)\) to \((1, 0, 1)\) and then traveling along the straight line from \((1, 0, 1)\) to \((2, 1, 1)\), we can split the problem into two parts corresponding to the two segments of the curve and then sum the results.

### 1. Traveling along the unit circle from \((0, 1, 1)\) to \((1, 0, 1)\)

Parametrize the unit circle segment in the \(xy\)-plane while \(z = 1\):
\[ \mathbf{r}_1(t) = (\cos t, \sin t, 1), \quad t \in \left[ \frac{\pi}{2}, 0 \right] \]

The differential \(d\mathbf{r}_1 = \frac{d\mathbf{r}_1}{dt} dt = (-\sin t, \cos t, 0) dt\).

Evaluate the integral:
\[ \int_{C_1} 3xy \, dx - xz \, dy + 2 \, dz = \int_{\frac{\pi}{2}}^{0} \left( 3 (\cos t)(\sin t)(-\sin t) - (\cos t)(1)(\cos t)(\cos t) + 2 \cdot 0 \right) dt \]

Simplify the integrand:
\[ = \int_{\frac{\pi}{2}}^{0} \left( -3\cos t \sin^2 t - \cos^3 t \right) dt \]

\[ = \int_{\frac{\pi}{2}}^{0} \left( -3\cos t (1 - \cos^2 t) - \cos^3 t \right) dt \]

\[ = \int_{\frac{\pi}{2}}^{0} \left( -3\cos t + 3\cos^3 t - \cos^3 t \right) dt \]

\[ = \int_{\frac{\pi}{2}}^{0} \left( -3\cos t + 2\cos^3 t \right) dt \]

Evaluate this integral by parts or appropriate substitutions:
\[ \int_{\frac{\pi}{2}}^{0} -3\cos t \, dt + \int_{\frac{\pi}{2}}^{0} 2\cos^3 t \, dt \]

The first integral is straightforward:
\[ -3 \int_{\frac{\pi}{2}}^{0} \cos t \, dt = -3 \sin t

Learn how to evaluate the line integral of the vector field F = <3xy, -xz, 2> along the curve C formed by traveling along the unit circle and a straight line in 3D space. Detailed step-by-step solution using parametrization and integral calculus. Suitable for advanced undergraduate students and beyond.

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