Question

Find the divergence of the vector field \(\mathbf{F}\) at the given point. $$ \begin{array}{lll} \text { Vector Field } & & \text { Point } \\ \hline \mathbf{F}(x, y, z)=x y z \mathbf{i}+y \mathbf{j}+z \mathbf{k} & & (1,2,1) \\ \end{array} $$

Short answer

The divergence of the vector field \(\mathbf{F}\) at the point (1,2,1) is 4.

Step-by-step solution

Find the Partial Derivatives

First, find the partial derivatives of \(P = xyz\), \(Q = y\), and \(R = z\) with respect to x, y, z: \(\frac{\partial P}{\partial x} = yz\), \(\frac{\partial Q}{\partial y} = 1\), and \(\frac{\partial R}{\partial z} = 1\).

Sum the partial derivatives

Next, sum the partial derivatives to find the divergence of \(\mathbf{F}\): \(\nabla \cdot \mathbf{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} = yz + 1 + 1.\)

Substitute the given point into the divergence

Then, substitute the given point (1,2,1) into the divergence: \(\nabla \cdot \mathbf{F} = 2*1 + 1 + 1 = 4.\)