Question

Consider a vector field $\vec{V} = x^{3}\,\hat{\imath} + 2y^{2}x\,\hat{\jmath} + 0.5\,z\,\hat{k}$, where $\hat{\imath}$, $\hat{\jmath}$, and $\hat{k}$ are the unit vectors in $x$, $y$, and $z$ directions, respectively. The divergence of $\vec{V}$ at the point $(1,2,1)$ is ............. (rounded to one decimal place).

Hint:

For polynomial fields, compute divergence by differentiating each component with respect to its own coordinate and then substitute the point—very quick arithmetic.

Answer 1

Step 1: Write divergence formula.
$\nabla\!\cdot\!\vec{V}=\dfrac{\partial V_x}{\partial x}+\dfrac{\partial V_y}{\partial y}+\dfrac{\partial V_z}{\partial z}$, where $V_x=x^3,\; V_y=2y^2x,\; V_z=0.5\,z$.
[2pt] Step 2: Differentiate components.
$\dfrac{\partial}{\partial x}(x^3)=3x^2,\quad \dfrac{\partial}{\partial y}(2y^2x)=4yx,\quad \dfrac{\partial}{\partial z}(0.5\,z)=0.5.$
[2pt] Step 3: Evaluate at $(x,y,z)=(1,2,1)$.
$\nabla\!\cdot\!\vec{V}=3(1)^2+4(2)(1)+0.5=3+8+0.5=11.5\Rightarrow 11.5.$