Question

Let $\phi(x,y,z) = (x^2+y^2+z^2)^{n/2}$ and $f(x,y,z)=(xyz)^{-n/2}$. Then div Curl grad$\phi$ + (Curl grad $f$) $\cdot$ grad$\phi =$

• A. $\frac{n}{2}(x^2+y^2+z^2)^{\frac{n+1}{2}} - \frac{n}{2}(xyz)^{\frac{n-1}{2}}$
• B. $\frac{n(n+1)}{2}(x^2+y^2+z^2)(xyz)$
• C. 0
• D. $x+y+z$

Hint:

• Key vector calculus identities:
  • Curl of a gradient of any scalar field $\psi$ is zero: $\nabla \times (\nabla \psi) = \mathbf{0}$.
  • Divergence of a curl of any vector field $\vec{G}$ is zero: $\nabla \cdot (\nabla \times \vec{G}) = 0$.
• Apply these identities:
  • Curl grad $\phi = \mathbf{0}$. Then div($\mathbf{0}$) = 0. (First term is 0).
  • Or, grad$\phi$ is a vector field. Let $\vec{V} = \text{grad}\phi$. Then $\text{div}(\text{curl } \vec{V})=0$.
  • Curl grad $f = \mathbf{0}$. Then $(\mathbf{0}) \cdot \text{grad}\phi = 0$. (Second term is 0).
• The sum is $0+0=0$. The specific functions for $\phi$ and $f$ do not matter as long as they are smooth enough.

Answer 1

The Correct Option is C

Given:

\[ \phi(x,y,z) = \left( x^2 + y^2 + z^2 \right)^{\frac{n}{2}} \quad \text{and} \quad f(x,y,z) = (xyz)^{-\frac{n}{2}} \]

We need to evaluate:

\[ \operatorname{div} \left( \operatorname{Curl}(\nabla \phi) \right) + \left( \operatorname{Curl}(\nabla f) \right) \cdot \nabla \phi \]

Step 1: Use vector calculus identities

• The curl of a gradient is always zero: \[ \operatorname{Curl}(\nabla \phi) = \mathbf{0} \] and \[ \operatorname{Curl}(\nabla f) = \mathbf{0} \] for any scalar functions \(\phi\) and \(f\).
• Therefore, the divergence of the curl of a gradient is also zero: \[ \operatorname{div}(\operatorname{Curl}(\nabla \phi)) = 0 \] and similarly: \[ (\operatorname{Curl}(\nabla f)) \cdot \nabla \phi = \mathbf{0} \cdot \nabla \phi = 0 \]

Step 2: Summing both terms

\[ \operatorname{div}(\operatorname{Curl}(\nabla \phi)) + (\operatorname{Curl}(\nabla f)) \cdot \nabla \phi = 0 + 0 = 0 \]

Answer: The value of the given expression is 0.

This corresponds to option 3.