Let's solve the given line integral step by step: ### **Given:** - **Vector Field:** \( \vec{F} = \langle x - y,\, y - z,\, z - x \rangle \) - **Curve:** \( \vec{r}(t) = \langle 2\cos t,\, t,\, 2\sin t \rangle \), where \( t \) ranges from \( 0 \) to \( 2\pi \) - **Line Integral to Compute:** \[ \int_C \vec{F}(\vec{r}) \cdot d\vec{r} \] --- ## **Step 1: Find \( \vec{F}(\vec{r}(t)) \)** Plug \( x = 2\cos t \), \( y = t \), \( z = 2\sin t \) into \( \vec{F} \): - \( x - y = 2\cos t - t \) - \( y - z = t - 2\sin t \) - \( z - x = 2\sin t - 2\cos t \) So, \[ \vec{F}(\vec{r}(t)) = \langle 2\cos t - t,\, t - 2\sin t,\, 2\sin t - 2\cos t \rangle \] --- ## **Step 2: Compute \( \frac{d\vec{r}}{dt} \)** Differentiate \( \vec{r}(t) \): - \( \frac{dx}{dt} = -2\sin t \) - \( \frac{dy}{dt} = 1 \) - \( \frac{dz}{dt} = 2\cos t \) So, \[ \frac{d\vec{r}}{dt} = \langle -2\sin t,\, 1,\, 2\cos t \rangle \] --- ## **Step 3: Dot Product \( \vec{F}(\vec{r}(t)) \cdot \frac{d\vec{r}}{dt} \)** \[ \begin{align*} &= (2\cos t - t)(-2\sin t) + (t - 2\sin t)(1) + (2\sin t - 2\cos t)(2\cos t) \\ \\ &= -4\cos t \sin t + 2t\sin t + t - 2\sin t + 4\sin t \cos t - 4\cos^2 t \\ \end{align*} \] Combine like terms: - \( -4\cos t \sin t + 4\sin t \cos t = 0 \) - \( 2t\sin t \) remains - \( t \) remains - \( -2\sin t \) remains - \( -4\cos^2 t \) remains So, \[ \vec{F}(\vec{r}(t)) \cdot \frac{d\vec{r}}{dt} = 2t\sin t + t - 2\sin t - 4\cos^2 t \] --- ## **Step 4: Set Up the Integral** \[ \int_0^{2\pi} \left(2t\sin t + t - 2\sin t - 4\cos^2 t \right) dt \] --- ## **Step 5: Integrate Each Term** ### 1. \( \int_0^{2\pi} 2t \sin t \, dt \) Use integration by parts: Let \( u = t \), \( dv = \sin t dt \) \( du = dt \), \( v = -\cos t \) \[ \int t\sin t dt = -t\cos t + \int \cos t dt = -t\cos t + \sin t \] So, \[ \int_0^{2\pi} 2t\sin t dt = 2\left[ -t\cos t + \sin t \right]_0^{2\pi} \] Plug in: - At \( t = 2\pi \): \( -2\pi \cos(2\pi) + \sin(2\pi) = -2\pi(1) + 0 = -2\pi \) - At \( t = 0 \): \( 0 + 0 = 0 \) So, \[ = 2(-2\pi - 0) = -4\pi \] ### 2. \( \int_0^{2\pi} t dt = \left[ \frac{t^2}{2} \right]_0^{2\pi} = \frac{(2\pi)^2}{2} = 2\pi^2 \) ### 3. \( \int_0^{2\pi} -2\sin t dt = -2 \left[ -\cos t \right]_0^{2\pi} = -2 ( -1 + 1 ) = 0 \) ### 4. \( \int_0^{2\pi} -4\cos^2 t dt \) Recall: \[ \cos^2 t = \frac{1 + \cos 2t}{2} \] So, \[ -4 \int_0^{2\pi} \cos^2 t dt = -4 \int_0^{2\pi} \frac{1 + \cos 2t}{2} dt = -2 \int_0^{2\pi} (1 + \cos 2t) dt \] \[ = -2 \left[ t + \frac{1}{2}\sin 2t \right]_0^{2\pi} = -2 \left( 2\pi + 0 - 0 \right) = -4\pi \] --- ## **Step 6: Add Results** \[ \begin{align*} &(-4\pi) + (2\pi^2) + (0) + (-4\pi) \\ &= 2\pi^2 - 8\pi \end{align*} \] --- ## **Final Answer** \[ \boxed{2\pi^2 - 8\pi} \] This is the value of the line integral: \[ \int_C \vec{F}(\vec{r}) \cdot d\vec{r} = 2\pi^2 - 8\pi \]

Let's carefully solve the line integral step-by-step. --- ### **Given Data:** - **Force field:** \(\vec{F} = \langle x - y, \ y - z, \ z - x \rangle \) - **Curve:** Parameterized line segment from \((x_0, y_0, z_0)\) to \((x_1, y_1, z_1)\): \[ x(t) = (1 - t)x_0 + t x_1 \] \[ y(t) = (1 - t)y_0 + t y_1 \] \[ z(t) = (1 - t)z_0 + t z_1 \] for \( t \in [0, 1] \). --- ### **Step 1: Write the parameterization** The parametric form of the curve: \[ \vec{r}(t) = \langle x(t), y(t), z(t) \rangle \] The derivatives: \[ \frac{d\vec{r}}{dt} = \langle x_1 - x_0, \ y_1 - y_0, \ z_1 - z_0 \rangle \] ### **Step 2: Express \(\vec{F}(\vec{r}(t))\)** \[ \vec{F}(\vec{r}(t)) = \langle x(t) - y(t), \ y(t) - z(t), \ z(t) - x(t) \rangle \] --- ### **Step 3: Compute the dot product \(\vec{F}(\vec{r}(t)) \cdot \frac{d\vec{r}}{dt}\)** \[ \text{Line integral} = \int_0^1 \vec{F}(\vec{r}(t)) \cdot \frac{d\vec{r}}{dt} \, dt \] which simplifies to: \[ \int_0^1 [ (x(t) - y(t))(x_1 - x_0) + (y(t) - z(t))(y_1 - y_0) + (z(t) - x(t))(z_1 - z_0) ] dt \] --- ### **Step 4: Expand the integrand** Define: \[ \Delta x = x_1 - x_0, \quad \Delta y = y_1 - y_0, \quad \Delta z = z_1 - z_0 \] then: \[ \text{Integrand} = (x(t) - y(t)) \Delta x + (y(t) - z(t)) \Delta y + (z(t) - x(t)) \Delta z \] --- ### **Step 5: Write \(x(t), y(t), z(t)\)** \[ x(t) = x_0 + t \Delta x \] \[ y(t) = y_0 + t \Delta y \] \[ z(t) = z_0 + t \Delta z \] So: \[ x(t) - y(t) = (x_0 - y_0) + t(\Delta x - \Delta y) \] \[ y(t) - z(t) = (y_0 - z_0) + t(\Delta y - \Delta z) \] \[ z(t) - x(t) = (z_0 - x_0) + t(\Delta z - \Delta x) \] --- ### **Step 6: Write the integral explicitly** \[ \int_0^1 \left[ [(x_0 - y_0) + t(\Delta x - \Delta y)] \Delta x + [(y_0 - z_0) + t(\Delta y - \Delta z)] \Delta y + [(z_0 - x_0) + t(\Delta z - \Delta x)] \Delta z \right] dt \] Split into parts: \[ \int_0^1 \left[ (x_0 - y_0) \Delta x + (y_0 - z_0) \Delta y + (z_0 - x_0) \Delta z \right] dt + \int_0^1 t \left[ (\Delta x - \Delta y) \Delta x + (\Delta y - \Delta z) \Delta y + (\Delta z - \Delta x) \Delta z \right] dt \] The first integral is just a constant multiplied by 1 (since integrating over 0 to 1): \[ \left[ (x_0 - y_0) \Delta x + (y_0 - z_0) \Delta y + (z_0 - x_0) \Delta z \right] \] The second integral: \[ \frac{1}{2} \left[ (\Delta x - \Delta y) \Delta x + (\Delta y - \Delta z) \Delta y + (\Delta z - \Delta x) \Delta z \right] \] --- ### **Final formula:** \[ \boxed{ \text{Line integral} = (x_0 - y_0) \Delta x + (y_0 - z_0) \Delta y + (z_0 - x_0) \Delta z + \frac{1}{2} \left[ (\Delta x - \Delta y) \Delta x + (\Delta y - \Delta z) \Delta y + (\Delta z - \Delta x) \Delta z \right] } \] --- ### **Summary:** - Determine the differences \(\Delta x, \Delta y, \Delta z\). - Plug into the above formula for the line integral. --- **Note:** If you have specific points \((x_0, y_0, z_0)\) and \((x_1, y_1, z_1)\), just substitute these values into the formula to get the numerical result.