Question

If $\vec{F} = x(x^2 + y^2 + z^2)\hat{i} + 2y(x^2 + y^2 + z^2)\hat{j} + 3z(x^2 + y^2 + z^2)\hat{k}$, then $\text{div} \, \vec{F}$ at $(1,1,1)$ is equal to ______

• A. 12
• B. 21
• C. 30
• D. 33

Hint:

Use the product rule when differentiating expressions like $x f$, $y f$, and $z f$, especially when $f$ is a function of multiple variables.

Answer 1

The Correct Option is C

We are given the vector field:
$\vec{F} = x(x^2 + y^2 + z^2)\hat{i} + 2y(x^2 + y^2 + z^2)\hat{j} + 3z(x^2 + y^2 + z^2)\hat{k}$
Let $f = x^2 + y^2 + z^2$
Then, $\vec{F} = x f \hat{i} + 2y f \hat{j} + 3z f \hat{k}$
The divergence is:
$\nabla \cdot \vec{F} = \frac{\partial}{\partial x}(x f) + \frac{\partial}{\partial y}(2y f) + \frac{\partial}{\partial z}(3z f)$
Using the product rule:
$\frac{\partial}{\partial x}(x f) = f + x \frac{\partial f}{\partial x} = f + x(2x) = f + 2x^2$
$\frac{\partial}{\partial y}(2y f) = 2f + 2y \frac{\partial f}{\partial y} = 2f + 2y(2y) = 2f + 4y^2$
$\frac{\partial}{\partial z}(3z f) = 3f + 3z \frac{\partial f}{\partial z} = 3f + 3z(2z) = 3f + 6z^2$
Add all components:
$\nabla \cdot \vec{F} = f + 2x^2 + 2f + 4y^2 + 3f + 6z^2 = (f + 2f + 3f) + (2x^2 + 4y^2 + 6z^2)$
$= 6f + 2x^2 + 4y^2 + 6z^2$
Now substitute $(1,1,1)$:
$f = 1^2 + 1^2 + 1^2 = 3$
So, $\nabla \cdot \vec{F} = 6(3) + 2(1^2) + 4(1^2) + 6(1^2) = 18 + 2 + 4 + 6 = 30$