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:

To calculate the divergence of a vector field, take the partial derivative of each component with respect to its respective variable and sum them up.

Answer 1

The Correct Option is C

The given vector field is: \[ \vec{F} = x(x^2 + y^2 + z^2) \hat{i} + 2y(x^2 + y^2 + z^2) \hat{j} + 3z(x^2 + y^2 + z^2) \hat{k} \] The divergence of a vector field is given by: \[ \text{div} \, \vec{F} = \frac{\partial F_x}{\partial x} + \frac{\partial F_y}{\partial y} + \frac{\partial F_z}{\partial z} \] where $F_x = x(x^2 + y^2 + z^2)$, $F_y = 2y(x^2 + y^2 + z^2)$, and $F_z = 3z(x^2 + y^2 + z^2)$. 
Now, calculate the partial derivatives: 
1. For $F_x = x(x^2 + y^2 + z^2)$: \[ \frac{\partial F_x}{\partial x} = \frac{\partial}{\partial x} \left( x(x^2 + y^2 + z^2) \right) = (x^2 + y^2 + z^2) + x(2x) = 3x^2 + y^2 + z^2 \] 2. For $F_y = 2y(x^2 + y^2 + z^2)$: \[ \frac{\partial F_y}{\partial y} = \frac{\partial}{\partial y} \left( 2y(x^2 + y^2 + z^2) \right) = 2(x^2 + y^2 + z^2) + 2y(2y) = 2(x^2 + y^2 + z^2) + 4y^2 \] 3. For $F_z = 3z(x^2 + y^2 + z^2)$: \[ \frac{\partial F_z}{\partial z} = \frac{\partial}{\partial z} \left( 3z(x^2 + y^2 + z^2) \right) = 3(x^2 + y^2 + z^2) + 3z(2z) = 3(x^2 + y^2 + z^2) + 6z^2 \] Now, evaluate the divergence at the point $(1, 1, 1)$: \[ \text{div} \, \vec{F} = \left( 3x^2 + y^2 + z^2 \right) + \left( 2(x^2 + y^2 + z^2) + 4y^2 \right) + \left( 3(x^2 + y^2 + z^2) + 6z^2 \right) \] Substituting $x = 1$, $y = 1$, and $z = 1$: \[ \text{div} \, \vec{F} = \left( 3(1)^2 + (1)^2 + (1)^2 \right) + \left( 2((1)^2 + (1)^2 + (1)^2) + 4(1)^2 \right) + \left( 3((1)^2 + (1)^2 + (1)^2) + 6(1)^2 \right) \] \[ \text{div} \, \vec{F} = (3 + 1 + 1) + (2(3) + 4) + (3(3) + 6) \] \[ \text{div} \, \vec{F} = 5 + 10 + 15 = 30 \] Therefore, the correct answer is 30.