Answer to Question #159836 in Physics for Haji

Question

A Velocity Vector V is given by, V= 5x3i +2y2 j +9z2 k, . Calculate curl of V (∇*V ) at (3, 0, 1).

Answer 1

Let the vector be:

$$\mathbf{V} = (5x^3, 2y^2,9z^2)$$

By definition, the curl is:

$$\nabla \times \mathbf{V} = \left(\frac{\partial V_z}{\partial y} - \frac{\partial V_y}{\partial z}\right) \mathbf e_x+ \left(\frac{\partial V_x}{\partial z} - \frac{\partial V_z}{\partial x}\right) \mathbf e_y+ \left(\frac{\partial V_y}{\partial x} - \frac{\partial V_x}{\partial y}\right) \mathbf e_z$$

Subsituting the coordinates of the vector, obtain:

$$\nabla \times \mathbf{V} = \left(\frac{\partial (9z^2)}{\partial y} - \frac{\partial (2y^2)}{\partial z}\right) \mathbf e_x+ \left(\frac{\partial (5x^3)}{\partial z} - \frac{\partial (9z^2)}{\partial x}\right) \mathbf e_y+ \left(\frac{\partial (2y^2)}{\partial x} - \frac{\partial (5x^3)}{\partial y}\right) \mathbf e_z$$

It is clear, that each derivative is equal to zero. Thus:

$$\nabla \times \mathbf{V} = 0$$

Answer. 0.