Answer to Question #159838 in Physics for Haji

Question

A Force Vector F is given by, F= 3x3 i +2y2 j +4z2k, . Calculate divF (∇.F) at (0, 1, 1).

Answer 1

By definition, the divergence of any vector field $\mathbf{F} = (F_x,F_y,F_z)$ is:

$$\nabla \cdot \mathbf{F} = \dfrac{\partial F_x}{\partial x} +\dfrac{\partial F_y}{\partial y} + \dfrac{\partial F_z}{\partial z}$$

In our case $\mathbf{F} = (3x^3,2y^2,4z^2)$, and the divergence is:

$$\nabla \cdot \mathbf{F}(x,y,z) = \dfrac{\partial (3x^3)}{\partial x} +\dfrac{\partial (2y^2)}{\partial y} + \dfrac{\partial (4z^2)}{\partial z} = 9x^2 + 4y + 8z$$

At the point $(0,1,1)$ this divergence is:

$$\nabla \cdot \mathbf{F}(0,1,1) = 9\cdot 0^2 + 4\cdot 1 + 8\cdot 1 = 12$$

Answer. 12.