Question

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

Short answer

The curl of the vector field \( \mathbf{F}(x, y, z)=x^{2} \mathbf{i}+y^{2} \mathbf{j}+x^{2} \mathbf{k} \) is \( \nabla \times \mathbf{F} = 0 \mathbf{i} - 2x \mathbf{j} + 2x \mathbf{k} \).

Step-by-step solution

Representation of the Vector Field

The vector field \(F\) is represented as \( \mathbf{F}(x, y, z)=x^{2} \mathbf{i}+y^{2} \mathbf{j}+x^{2} \mathbf{k} \).

Derivation of Curl

The curl of a three-dimensional vector field is defined as \( \nabla \times \mathbf{F} \), the cross product of the del operator and \(\mathbf F\). Using del in component form, the curl is given by the determinant of the following matrix: \[ \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \ F_x & F_y & F_z \end{vmatrix} \] where \(F_x\), \(F_y\), and \(F_z\) are the components of the vector \(\mathbf F\).

Calculation of Curl

Substitute \(F_x = x^2\), \(F_y = y^2\), and \(F_z = x^2\) to the formula and calculate the determinant. This gives the curl as: \[ \nabla \times \mathbf{F} = \left( \frac{\partial (x^2)}{\partial y} - \frac{\partial (y^2)}{\partial z} \right) \mathbf{i} - \left( \frac{\partial (x^2)}{\partial x} - \frac{\partial (x^2)}{\partial z} \right) \mathbf{j} + \left( \frac{\partial (x^2)}{\partial x} - \frac{\partial (x^2)}{\partial y} \right) \mathbf{k} \] Calculate each partial derivative separately: \(\frac{\partial (x^2)}{\partial y} = 0\), \(\frac{\partial (y^2)}{\partial z} = 0\), \(\frac{\partial (x^2)}{\partial x} = 2x\), \(\frac{\partial (x^2)}{\partial z} = 0\), \(\frac{\partial (x^2)}{\partial x} = 2x\), and \(\frac{\partial (x^2)}{\partial y} = 0\). Substituting each partial derivative into the curl formula yields: \[ \nabla \times \mathbf{F} = 0 \mathbf{i} - 2x \mathbf{j} + 2x \mathbf{k} \]