Question

Let \( \vec{F}, \vec{G} \) be two vector point functions such that \( \vec{F} = \nabla \times \vec{G} \). Let \( S \) be a closed surface enclosing region \( E \). Then \[ \iint_S \vec{F} \cdot \vec{n} \, ds = ? \]

• A. \( \iiint_E \nabla \times \vec{F} \cdot \vec{n} \, ds \)
• B. \( \iiint_E \vec{F} \cdot \vec{G} \, dx\,dy\,dz \)
• C. \( 0 \)
• D. \( \iiint_V \vec{G} \, dv \)

Hint:

Divergence of a curl is always zero: \( \nabla \cdot (\nabla \times \vec{G}) = 0 \).

Answer 1

The Correct Option is C

To solve this problem, we apply the Divergence Theorem, also known as Gauss's theorem. The Divergence Theorem relates the flux of a vector field through a closed surface to the divergence of the field inside the volume bounded by the surface.

We are given that \( \vec{F} = \nabla \times \vec{G} \). Therefore, \(\vec{F}\) is a curl of some vector field \(\vec{G}\).

The Divergence Theorem states:

\[\iint_S \vec{F} \cdot \vec{n} \, ds = \iiint_E \nabla \cdot \vec{F} \, dV\]

Where \( S \) is the boundary of the volume \( E \) and \(\vec{n}\) is the outward normal to \( S \). However, in our case, since \( \vec{F} = \nabla \times \vec{G} \), we use the identity:

\[\nabla \cdot (\nabla \times \vec{G}) = 0\]

The divergence of a curl is always zero. Thus, we have:

\[\iiint_E \nabla \cdot (\nabla \times \vec{G}) \, dV = 0\]

Therefore, by the Divergence Theorem:

\[\iint_S \vec{F} \cdot \vec{n} \, ds = \iiint_E \nabla \cdot \vec{F} \, dV = \iiint_E 0 \, dV = 0\]

Hence, the integral over the closed surface \( S \) is zero.