Question

Let \[ \vec{F}=x^2yz\,\hat{i}+y^2z\,\hat{j}+zx^3\,\hat{k} \] be a vector point function. Let \(S\) be the surface of the sphere \[ x^2+y^2+z^2=a^2. \] Then \[ \iint_{S}(\nabla\times\vec{F})\cdot\vec{N}\,dS \] is equal to (\(\vec{N}\) is any unit outward normal vector to \(S\)).

• A. \(2\pi a^2\)
• B. \(\dfrac{4}{3}\pi a^3\)
• C. \(4\pi\)
• D. \(0\)

Hint:

Remember that the flux of the curl of any vector field through a closed surface is always zero. This follows directly from Stokes' Theorem (or equivalently, the Divergence Theorem together with \(\nabla\cdot(\nabla\times\vec{F})=0\)).

Answer 1

The Correct Option is D

Concept: By Stokes' Theorem, \[ \iint_S (\nabla\times\vec{F})\cdot\vec{N}\,dS = \oint_C \vec{F}\cdot d\vec{r}, \] where \(C\) is the boundary of the surface \(S\). If the surface is a closed surface (such as a sphere), then its boundary is empty. Hence, the line integral is zero. Equivalently, \[ \nabla\cdot(\nabla\times\vec{F})=0, \] and applying the Divergence Theorem also gives zero flux of the curl through any closed surface.

Step 1: Identify the nature of the surface. The given surface \[ x^2+y^2+z^2=a^2 \] is a closed sphere. Hence, \[ \partial S=\varnothing. \]

Step 2: Apply Stokes' Theorem. Therefore, \[ \iint_S(\nabla\times\vec{F})\cdot\vec{N}\,dS = \oint_{\partial S}\vec{F}\cdot d\vec r. \] Since \[ \partial S=\varnothing, \] the line integral is \[ \oint_{\partial S}\vec{F}\cdot d\vec r=0. \] Thus, \[ \iint_S(\nabla\times\vec{F})\cdot\vec{N}\,dS=0. \] Hence, \[ \boxed{(D)\;0.} \]