91Ó°ÊÓ

The vector field is \( \mathbf{F} = (x^2 + y) \mathbf{i} + (y^2 + x) \mathbf{j} + z e^z \mathbf{k} \) and the line segment is along the \( z \)-axis where \( x = 1 \), \( y = 0 \) and \( z \) varies from 0 to 1. The path can be parametrized as \( \mathbf{r}(z) = \mathbf{i} + z \mathbf{k} \) with \( z \in [0,1] \). Then, \( d\mathbf{r} = \mathbf{k} \, dz \).On this path, \( F \cdot d\mathbf{r} = (xe^z) \mathbf{k} \cdot \mathbf{k} \, dz = z e^z \, dz \).Integrate from 0 to 1:\[ W = \int_{0}^{1} z e^z \, dz \]This integration can be solved using integration by parts, letting \( u = z \) and \( dv = e^z \, dz \). Thus, we get \( du = dz \) and \( v = e^z \). Applying integration by parts:\[ W = \left[ z e^z \right]_0^1 - \int_{0}^{1} e^z \, dz \]\[ = (1 \cdot e^1 - 0 \cdot e^0) - (\left[e^z\right]_0^1) \]\[ = e - (e - 1) = 1 \]

For the helix, \( \mathbf{r}(t) = \cos(t) \mathbf{i} + \sin(t) \mathbf{j} + \frac{t}{2\pi} \mathbf{k} \) for \( 0 \leq t \leq 2\pi \).Thus, \( d\mathbf{r} = (-\sin(t) \mathbf{i} + \cos(t) \mathbf{j} + \frac{1}{2\pi} \mathbf{k}) \, dt \).Substitute into the line integral of the vector field along this path:\[ F \cdot d\mathbf{r} = ((\cos(t))^2 + \sin(t))(-\sin(t)) + ((\sin(t))^2 + \cos(t)) \cos(t) + \frac{t}{2\pi} e^{\frac{t}{2\pi}} \frac{1}{2\pi} \, dt \]= \[(\cos^2(t) + \sin(t))( -\sin(t)) + (\sin^2(t) + \cos(t)) \cos(t) + \frac{t}{2\pi} e^{\frac{t}{2\pi}} \frac{1}{2\pi} \, dt \]Simplify each term before solving the integral. Due to complexity, use numeric techniques as necessary or break down terms using trigonometric identities if closed-form solutions are feasible. Solving this expression typically requires multiple steps or computational assistance for a numeric answer. The anticipated answer should verify as little to no work is done on a closed loop, due to conservative nature of field properties.

The first segment of the path is the \( x \)-axis from \( (1,0,0) \) to \( (0,0,0) \). Along this path, \( y = 0 \), \( z = 0 \), and \( \mathbf{F} = (x^2 + y) \mathbf{i} \), and \( d\mathbf{r} = dx \mathbf{i} \).Calculate work along the first part:\[ W_1 = \int_{1}^{0} (x^2) dx = -\int_{0}^{1} x^2 \, dx = -\left[ \frac{x^3}{3} \right]_0^1 = -\frac{1}{3} \]The second segment is the parabola from \((0,0,0)\) to \((1,0,1)\). Parametrize as \( \mathbf{r}(x) = x \mathbf{i} + 0 \mathbf{j} + x^2 \mathbf{k} \) ranging from \( x = 0 \) to \( x = 1 \), then \( d\mathbf{r} = (\mathbf{i} + 2x \mathbf{k}) \, dx \).Calculate work along the parabola:\[ W_2 = \int_0^1 ((x^2) \mathbf{i} + e^{x^2} \mathbf{k}) \cdot (\mathbf{i} + 2x \mathbf{k}) \, dx \]\[ = \int_0^1 (x^2 + 2x e^{x^2}) \, dx \]Break this integral by equivalent small manageable components or estimation:For example, simplification or numeric solutions of specific components may facilitate stitched portions achievable through composite analysis.Finally, total work is \( W = W_1 + W_2 \) converging results preamble.