Question

Consider a vector function \(\vec{u}(\vec{r})\) and two scalar functions \(\psi(\vec{r})\) and \(\phi(\vec{r})\). The unit vector \(\hat{n}\) is normal to the elementary surface \(dS\), \(dV\) is an infinitesimal volume, \(d\vec{l}\) is an infinitesimal line element, and \(\partial / \partial n\) denotes the partial derivative along \(\hat{n}\). Which of the following identities is/are correct?

• A. \(\displaystyle \int_V \psi \, \nabla \cdot \vec{u} \, dV = \oint_S \psi \, \vec{u} \cdot \hat{n} \, dS\), where surface \(S\) bounds the volume \(V\).
• B. \(\displaystyle \int_V [\psi \nabla^2 \phi - \phi \nabla^2 \psi] dV = \oint_S \left[\psi \frac{\partial \phi}{\partial n} - \phi \frac{\partial \psi}{\partial n}\right] dS\), where \(S\) bounds \(V\).
• C. \(\displaystyle \int_V [\psi \nabla^2 \phi - \phi \nabla^2 \psi] dV = \oint_S \left[\psi \frac{\partial \phi}{\partial n} + \phi \frac{\partial \psi}{\partial n}\right] dS\), where \(S\) bounds \(V\).
• D. \(\displaystyle \oint_S \phi \, \vec{u} \cdot d\vec{l} = \iint_S (\nabla \times \vec{u}) \cdot \hat{n} \, dS\), where \(C\) is the boundary of \(S\).

Hint:

Green's identities are derived from the divergence theorem and are essential in vector calculus and electrostatics for relating surface and volume integrals.

Answer 1

The Correct Option is A, B, D

Analyzing each vector identity:

(A) $\int_V \vec{\nabla} \cdot \vec{u},dV = \oint_S \vec{u} \cdot \hat{n},dS$

This is the Divergence Theorem (Gauss's theorem): $$\int_V (\vec{\nabla} \cdot \vec{u}),dV = \oint_S \vec{u} \cdot \hat{n},dS$$

TRUE 

(B) $\int_V [\psi \nabla^2\phi - \phi \nabla^2\psi],dV = \oint_S \left[\psi\frac{\partial\phi}{\partial n} - \phi\frac{\partial\psi}{\partial n}\right]dS$

This is Green's second identity (also called Green's theorem):

Starting with the divergence theorem applied to $\vec{F} = \psi\vec{\nabla}\phi - \phi\vec{\nabla}\psi$:

$$\vec{\nabla} \cdot (\psi\vec{\nabla}\phi) = \psi\nabla^2\phi + \vec{\nabla}\psi \cdot \vec{\nabla}\phi$$

$$\vec{\nabla} \cdot (\phi\vec{\nabla}\psi) = \phi\nabla^2\psi + \vec{\nabla}\phi \cdot \vec{\nabla}\psi$$

Subtracting: $$\vec{\nabla} \cdot (\psi\vec{\nabla}\phi - \phi\vec{\nabla}\psi) = \psi\nabla^2\phi - \phi\nabla^2\psi$$

Applying divergence theorem: $$\int_V [\psi\nabla^2\phi - \phi\nabla^2\psi],dV = \oint_S (\psi\vec{\nabla}\phi - \phi\vec{\nabla}\psi) \cdot \hat{n},dS$$

Since $\frac{\partial\phi}{\partial n} = \vec{\nabla}\phi \cdot \hat{n}$:

$$= \oint_S \left[\psi\frac{\partial\phi}{\partial n} - \phi\frac{\partial\psi}{\partial n}\right]dS$$

TRUE 

(C) $\int_V [\psi \nabla^2\phi - \phi \nabla^2\psi],dV = \oint_S \left[\psi\frac{\partial\phi}{\partial n} + \phi\frac{\partial\psi}{\partial n}\right]dS$

From the analysis in (B), the correct form has a minus sign, not a plus sign.

FALSE 

(D) $\oint_C \vec{u} \cdot d\vec{l} = \iint_S (\vec{\nabla} \times \vec{u}) \cdot \hat{n},dS$

This is Stokes' theorem: $$\oint_C \vec{u} \cdot d\vec{l} = \iint_S (\vec{\nabla} \times \vec{u}) \cdot \hat{n},dS$$

where $C$ is the boundary of surface $S$.

TRUE 

Answer: (A), (B), and (D) are correct