Question

Find \(\operatorname{curl}(\operatorname{curl} \mathbf{F})=\nabla \times(\nabla \times \mathbf{F})\). \(\mathbf{F}(x, y, z)=x^{2} z \mathbf{i}-2 x z \mathbf{j}+y z \mathbf{k}\)

Short answer

The curl of the curl of the given vector field \(\mathbf{F}(x, y, z) = x^{2} z \mathbf{i}-2 x z \mathbf{j}+y z \mathbf{k}\) is \(\nabla \times(\nabla \times \mathbf{F}) = x \mathbf{i} - 2z \mathbf{j}\).

Step-by-step solution

Compute the original curl of the vector field

To compute the original curl of \(\mathbf{F}\), use the formula for the curl of a vector field in three dimensions, which is \(\nabla \times \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \F_x & F_y & F_z\end{vmatrix}\), Where \(F_x\), \(F_y\) and \(F_z\) are components of the vector field. Hence, the curl can be computed as follows: \[ \nabla \times \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \x^{2} z & -2 x z & y z \end{vmatrix}\]

Simplify the determinant to get the curl of the vector field

Expand determinant by using the formula aei+bfg+cdh-ceg-bdi-afh to get the curl of the vector field \(\nabla \times \mathbf{F}\):\[ \nabla \times \mathbf{F} = \mathbf{i} \left(-y - 2xz\right) - \mathbf{j} \left( yz - 0\right) + \mathbf{k} \left(-xz - 0 \right) = -y \mathbf{i} - 2xz \mathbf{i} - yz \mathbf{j} - xz \mathbf{k},\] which simplifies to \[ \nabla \times \mathbf{F} = -y \mathbf{i} - 2xz \mathbf{i} - yz \mathbf{j} - xz \mathbf{k}\]

Compute the curl of the computed curl of the vector field

Now, calculate the curl of the curl of the vector field required by the exercise, \(\nabla \times(\nabla \times \mathbf{F})\), by repeating the same process with \(\nabla \times \mathbf{F} = -y \mathbf{i} - 2xz \mathbf{i} - yz \mathbf{j} - xz \mathbf{k}\) in place of \( \mathbf{F}(x, y, z)=x^{2} z \mathbf{i}-2 x z \mathbf{j}+y z \mathbf{k}\). The expected determinant for this part is: \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \-y - 2xz & -yz & -xz \end{vmatrix}

Simplify the determinant to get the curl of the curl of the vector field

Expand determinant once again to get the curl of the curl of the vector field, \(\nabla \times(\nabla \times \mathbf{F})\). I.e., \[ \nabla \times(\nabla \times \mathbf{F}) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \-y - 2xz & -yz & -xz \end{vmatrix} = \mathbf{i} \left(x \right) - \mathbf{j} \left( 2z \right) + \mathbf{k} \left(0 \right) = x \mathbf{i} - 2z \mathbf{j}\]