91Ó°ÊÓ

The given vector field is defined as \( \mathbf{F}(x, y, z) = \sin yz \mathbf{i} + \sin zx \mathbf{j} + \sin xy \mathbf{k} \). Identify the components: \( F_1 = \sin yz \), \( F_2 = \sin zx \), and \( F_3 = \sin xy \).

The curl of a vector field \( \mathbf{F} = (F_1, F_2, F_3) \) is given by: \[ abla \times \mathbf{F} = \left( \frac{\partial F_3}{\partial y} - \frac{\partial F_2}{\partial z} \right) \mathbf{i} + \left( \frac{\partial F_1}{\partial z} - \frac{\partial F_3}{\partial x} \right) \mathbf{j} + \left( \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} \right) \mathbf{k} \] Calculate each component:- \( \mathbf{i} \) component: \( \frac{\partial F_3}{\partial y} = x \cos xy \), \( \frac{\partial F_2}{\partial z} = x \cos zx \) Result: \( x \cos xy - x \cos zx \)- \( \mathbf{j} \) component: \( \frac{\partial F_1}{\partial z} = y \cos yz \), \( \frac{\partial F_3}{\partial x} = y \cos xy \) Result: \( y \cos yz - y \cos xy \)- \( \mathbf{k} \) component: \( \frac{\partial F_2}{\partial x} = z \cos zx \), \( \frac{\partial F_1}{\partial y} = z \cos yz \) Result: \( z \cos zx - z \cos yz \)The curl is: \[ abla \times \mathbf{F} = (x \cos xy - x \cos zx) \mathbf{i} + (y \cos yz - y \cos xy) \mathbf{j} + (z \cos zx - z \cos yz) \mathbf{k} \]

The divergence of a vector field \( \mathbf{F} = (F_1, F_2, F_3) \) is given by: \[ abla \cdot \mathbf{F} = \frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z} \] Calculate each term:- \( \frac{\partial F_1}{\partial x} = 0 \), since \( \sin yz \) does not depend on \( x \).- \( \frac{\partial F_2}{\partial y} = 0 \), since \( \sin zx \) does not depend on \( y \).- \( \frac{\partial F_3}{\partial z} = 0 \), since \( \sin xy \) does not depend on \( z \).Thus, the divergence is:\[ abla \cdot \mathbf{F} = 0 + 0 + 0 = 0 \]