The problem asks for the value of the line integral of a vector field **F** along a piecewise path (O → (2,1) → (3,3)). However, the vector field **F** is **not** given in the image. Usually, the field is specified in the question or a prior part. ### What We Can Do: - **Assume** the missing information or guide you on how to solve it **if** the field is provided. Let's proceed with a **general approach** for computing line integrals along a piecewise linear path. --- ## **Step 1: Break the Path into Segments** The path is: 1. From \( (0,0) \) to \( (2,1) \) 2. From \( (2,1) \) to \( (3,3) \) ## **Step 2: Parameterize Each Segment** ### **Segment 1:** \( (0,0) \) to \( (2,1) \) Let \( \vec{r}_1(t) = (2t, t) \), \( t \in [0,1] \) ### **Segment 2:** \( (2,1) \) to \( (3,3) \) Let \( \vec{r}_2(s) = (2 + s, 1 + 2s) \), \( s \in [0,1] \) --- ## **Step 3: Set Up the Line Integrals** For a vector field \( \vec{F}(x, y) = (P(x, y), Q(x, y)) \), the line integral is: \[ \int_C \vec{F} \cdot d\vec{r} = \int_C P(x, y) dx + Q(x, y) dy \] Or, for a parameterization \( \vec{r}(t) = (x(t), y(t)) \): \[ \int_a^b \vec{F}(\vec{r}(t)) \cdot \vec{r}\,'(t) dt \] --- ## **Step 4: Substitute the Field (Once Given)** If, for example, \( \vec{F}(x, y) = (P, Q) \) is provided, substitute into the parameterizations and compute each segment's integral. --- ## **Step 5: Add The Integrals** Sum the values from both segments for the total line integral. --- ### **Final Step: Plug in the Actual Field** - **If you provide the vector field \( \vec{F} \), I can complete the calculations and give the final answer.** - **If you're expected to use a potential function (conservative field), you can simply use:** \[ \int_C \vec{F} \cdot d\vec{r} = f(\text{end}) - f(\text{start}) \] if \( \vec{F} \) is conservative and \( f \) is a potential function. --- ## **Summary** **I need the vector field \( \vec{F} \) to compute the value. Please provide it, or check if it was given earlier in your materials!** If you give me \( \vec{F} \), I'll complete the calculation step-by-step.

Let's carefully evaluate the line integral of the vector field: \[ \mathbf{F}(x,y,z) = \frac{1}{y^2 z^2} ( y z, - z x, - x y + 4 y^2 z^2 ) \] along the piecewise path from \( (0,0) \) to \( (2,1) \) and then from \( (2,1) \) to \( (3,3) \). --- ### **Step 1: Understand the Path** Given the points: - Segment 1: \( (0,0) \to (2,1) \) - Segment 2: \( (2,1) \to (3,3) \) **Note:** The vector field depends on \( x, y, z \). **But**, the path is only specified in the \( xy \)-plane, i.e., no \( z \)-coordinate information is provided. **Assumption:** Since the path is given in 2D and the field is in 3D, **it's common in such problems to assume \( z \) is constant along the path**—probably \( z=1 \), or similar. **Alternatively**, if the problem is purely in 2D, perhaps \( z=1 \), and the field reduces accordingly. --- ### **Step 2: Simplify the Field for \( z \) along the Path** Assuming \( z=1 \) along the path: \[ \mathbf{F}(x,y) = \frac{1}{y^2 \cdot 1^2} ( y \cdot 1, -1 \cdot x, - x y + 4 y^2 \cdot 1^2 ) = \frac{1}{ y^2 } ( y, - x, - x y + 4 y^2 ) \] Simplify each component: \[ \mathbf{F}(x,y) = \left( \frac{y}{ y^2 }, \frac{ -x }{ y^2 }, \frac{ - x y + 4 y^2 }{ y^2 } \right ) = \left( \frac{1}{ y }, - \frac{x}{ y^2 }, - \frac{x y}{ y^2 } + 4 \right) \] which simplifies to: \[ \boxed{ \mathbf{F}(x,y) = \left( \frac{1}{ y }, - \frac{x}{ y^2 }, - \frac{x}{ y } + 4 \right) } \] --- ### **Step 3: Parameterize the Path** Since the path is in the \( xy \)-plane: **Segment 1:** \[ \mathbf{r}_1(t) = (x(t), y(t)) = (2t, t), \quad t \in [0,1] \] which goes from \( (0,0) \) to \( (2,1) \). Calculate derivatives: \[ \frac{d\mathbf{r}_1}{dt} = (2, 1) \] --- ### **Step 4: Compute the Line Integral over Segment 1** The line integral: \[ I_1 = \int_0^1 \mathbf{F}(x(t), y(t)) \cdot \frac{d\mathbf{r}_1}{dt} dt \] where: \[ \mathbf{F}(x(t), y(t)) = \left( \frac{1}{ y(t) }, - \frac{x(t)}{ y(t)^2 }, - \frac{x(t)}{ y(t) } + 4 \right) \] and \[ \frac{d\mathbf{r}_1}{dt} = (2, 1) \] **Note:** the field is 3D, but our path is 2D. To evaluate the dot product, we need the third component of the field and the \( z \)-component of \( d\mathbf{r} \). Since the path is in 2D, we can assume \( z \) is constant, say \( z=1 \). **Alternatively**, if the integral is over the \( xy \)-plane, and the field is 3D, the line integral is: \[ \int_C \mathbf{F} \cdot d\mathbf{r} = \int_C P dx + Q dy + R dz \] but \( dz=0 \), as the path is in \( xy \)-plane. So, the line integral reduces to: \[ I_1 = \int_0^1 \left[ P \frac{dx}{dt} + Q \frac{dy}{dt} \right] dt \] where: \[ P = \frac{1}{ y }, \quad Q= - \frac{x}{ y^2 } \] and \[ \frac{dx}{dt} = 2, \quad \frac{dy}{dt} = 1 \] Compute: \[ I_1 = \int_0^1 \left[ \frac{1}{ y(t) } \cdot 2 + \left( - \frac{x(t)}{ y(t)^2 } \right) \cdot 1 \right] dt \] with \( x(t)=2t \), \( y(t)=t \). Substitute: \[ I_1 = \int_0^1 \left[ \frac{2}{ t } - \frac{ 2t }{ t^2 } \right] dt = \int_0^1 \left[ \frac{2}{ t } - \frac{ 2t }{ t^2 } \right] dt \] Simplify: \[ I_1 = \int_0^1 \left[ \frac{2}{ t } - 2 \cdot \frac{ t }{ t^2 } \right] dt = \int_0^1 \left[ \frac{2}{ t } - 2 \cdot \frac{1}{ t } \right] dt \] \[ I_1 = \int_0^1 \left( \frac{2}{ t } - \frac{2}{ t } \right) dt = \int_0^1 0 \, dt = 0 \] --- ### **Step 5: Compute the Integral over Segment 2** Parameterize from \( (2,1) \to (3,3) \): \[ \mathbf{r}_2(s) = (x(s), y(s)) = (2 + s, 1 + 2s), \quad s \in [0,1] \] Derivatives: \[ \frac{dx}{ds} = 1, \quad \frac{dy}{ds} = 2 \] Evaluate \( \mathbf{F} \) at \( (x(s), y(s)) \): \[ x(s) = 2 + s, \quad y(s) = 1 + 2s \] Recall: \[ \mathbf{F}(x,y) = \left( \frac{1}{ y }, - \frac{x}{ y^2 }, - \frac{x}{ y } + 4 \right) \] Compute: \[ I_2 = \int_0^1 \left[ P \frac{dx}{ds} + Q \frac{dy}{ds} \right] ds \] where: \[ P = \frac{1}{ y(s) }, \quad Q= - \frac{x(s)}{ y(s)^2 } \] Substitute: \[ I_2 = \int_0^1 \left[ \frac{1}{ 1 + 2s } \cdot 1 + \left( - \frac{ 2 + s }{ (1 + 2s)^2 } \right) \cdot 2 \right] ds \] Simplify: \[ I_2 = \int_0^1 \left[ \frac{1}{ 1 + 2s } - 2 \cdot \frac{ 2 + s }{ (1 + 2s)^2 } \right] ds \] --- ### **Step 6: Simplify the Integral \( I_2 \)** Let \( u = 1 + 2s \). Then: \[ du = 2 ds \Rightarrow ds = \frac{du}{2} \] When \( s=0 \), \( u=1 \) When \( s=1 \), \( u=3 \) Rewriting the integral: \[ I_2 = \int_{u=1}^{3} \left[ \frac{1}{u} - 2 \cdot \frac{ 2 + \frac{u-1}{2} }{ u^2 } \right] \cdot \frac{du}{2} \] Express \( 2 + \frac{u-1}{2} \): \[ 2 + \frac{u-1}{2} = \frac{4 + u - 1}{2} = \frac{u + 3}{2} \] Substituting back: \[ I_2 = \frac{1}{2} \int_1^{3} \left[ \frac{1}{u} - 2 \cdot \frac{ (u+3)/2 }{ u^2 } \right] du \] Simplify: \[ I_2 = \frac{1}{2} \int_1^{3} \left[ \frac{1}{u} - \frac{ u + 3 }{ u^2 } \right] du \] Break into two integrals: \[ I_2 = \frac{1}{2} \left[ \int_1^{3} \frac{1}{u} du - \int_1^{3} \frac{ u + 3 }{ u^2 } du \right] \] Compute each: 1. \( \int_1^{3} \frac{1}{u} du = \left[ \ln|u| \right]_1^{3} = \ln 3 \) 2. \( \int_1^{3} \frac{ u + 3 }{ u^2 } du \) Rewrite numerator: \[ \frac{ u + 3 }{ u^2 } = \frac{ u }{ u^2 } + \frac{ 3 }{ u^2 } = \frac{1}{u} + 3 u^{-2} \] Integrate: \[ \int_1^{3} \left( \frac{1}{u} + 3 u^{-2} \right) du = \int_1^{3} \frac{1}{u} du + 3 \int_1^{3} u^{-2} du \] Compute: \[ \int_1^{3} \frac{1}{u} du = \ln 3 \] \[ \int_1^{3} u^{-2} du = \left[ - u^{-1} \right]_1^{3} = - \frac{1}{3} + 1 = \frac{2}{3} \] Thus, \[ \int_1^{3} \frac{ u + 3 }{ u^2 } du = \ln 3 + 3 \times \frac{2}{3} = \ln 3 + 2 \] Now, assemble the integral: \[ I_2 = \frac{1}{2} \left[ \ln 3 - (\ln 3 + 2) \right] = \frac{1}{2} ( \ln 3 - \ln 3 - 2 ) = \frac{1}{2} (-2) = -1 \] --- ### **### **Final Answer:** \[ \boxed{ \text{Total line integral} = I_1 + I_2 = 0 + (-1) = -1 } \] --- ## **Final Result:** \[ \boxed{ \boxed{ \text{The value of the line integral} = -1 } } \] --- **Note:** The calculation assumes \( z=1 \) along the path, based on the form of the field and typical problem context. If \( z \) varies or other information is provided, adjustments might be needed.