Question

Find the curl of the vector field \(F\). \(\mathbf{F}(x, y, z)=2 z \mathbf{i}-4 x^{2} \mathbf{j}+\arctan x \mathbf{k}\)

Short answer

The curl of the vector field \( \mathbf{F}(x, y, z)=2 z \mathbf{i}-4 x^{2} \mathbf{j}+\arctan x \mathbf{k} \) is given by the vector field \( \nabla \times \mathbf{F} \).

Step-by-step solution

Define the Vector Field

The vector field \( \mathbf{F}(x, y, z)=2 z \mathbf{i}-4 x^{2} \mathbf{j}+\arctan x \mathbf{k} \) is defined as a field with three component functions \( F_x = 2z, F_y = -4x^2, F_z = \arctan x \).

Apply the Curl Operation

The curl operation is denoted by \( \nabla \times \mathbf{F} \) and it's calculated as the determinant of the following 3x3 matrix: \[ \begin{bmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ F_x & F_y & F_z \end{bmatrix} \]

Calculate the Determinant

Calculating the determinant gives: \[ \nabla \times \mathbf{F} = \Bigg( \frac{\partial (\arctan x)}{\partial y} - \frac{\partial (-4x^2)}{\partial z} \Bigg) \mathbf{i} - \Bigg( \frac{\partial (\arctan x)}{\partial x} - \frac{\partial (2z)}{\partial z} \Bigg) \mathbf{j} + \Bigg( \frac{\partial (-4x^2)}{\partial x} - \frac{\partial (2z)}{\partial y} \Bigg) \mathbf{k} \]