91Ó°ÊÓ

To determine if \( \mathbf{F} \) is conservative, check if \( abla \times \mathbf{F} = \mathbf{0} \). Calculate the curl: \[ abla \times \mathbf{F} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \ y^2 \cos z & 2xy \cos z & -xy^2 \sin z \end{vmatrix} \] Calculating the determinant, we find all components to be zero, hence \( abla \times \mathbf{F} = \mathbf{0} \). Thus, \( \mathbf{F} \) is conservative.

Since \( \mathbf{F} \) is conservative, there exists a function \( f \) such that \( abla f = \mathbf{F} \). Integrate each component of \( \mathbf{F} \) to find \( f(x, y, z) \).1. Integrate \( y^2 \cos z \) with respect to \( x \): \[ f(x, y, z) = xy^2 \cos z + g(y, z) \] 2. Differentiate with respect to \( y \), \[ \frac{\partial f}{\partial y} = 2xy \cos z + \frac{\partial g}{\partial y} = 2xy \cos z \] which satisfies the second component. Thus, \( g(y, z) \) is constant in \( y \).3. Integrate \( \frac{\partial g}{\partial y} = 0 \) to adjust constant: \[ g(y, z) = h(z) \]4. Differentiate with respect to \( z \) using the \( f(x, y, z) = xy^2 \cos z + h(z) \), \( -xy^2 \sin z + \frac{dh}{dz} = -xy^2 \sin z \), giving \( h(z) \) is constant. Thus, \( f(x, y, z) = xy^2 \cos z \).

Since \( \mathbf{F} \) is conservative, use the fundamental theorem for line integrals: \( \int_{C} \mathbf{F} \cdot d\mathbf{r} = f(x(\pi), y(\pi), z(\pi)) - f(x(0), y(0), z(0)) \). Substitute \( \mathbf{r}(t) = t^2 \mathbf{i} + \sin t \mathbf{j} + t \mathbf{k} \):1. \( x(\pi) = \pi^2, y(\pi) = 0, z(\pi) = \pi \) gives \( f(\pi^2, 0, \pi) = 0 \).2. \( x(0) = 0, y(0) = 0, z(0) = 0 \) gives \( f(0, 0, 0) = 0 \).Thus, the integral evaluates to 0: \( 0 - 0 = 0 \).