Question

Let \( S \) be the surface of the portion of the sphere with centre at the origin and radius 4, above the \( xy \)-plane. Let \( \vec{F} = y\hat{i} - x\hat{j} + yxz^3\hat{k}. \) If \( \hat{n} \) is the unit outward normal to \( S \), then
\[ \iint_S (\nabla \times \vec{F}) \cdot \hat{n} \, dS \] equals

• A. \(-32\pi\)
• B. \(-16\pi\)
• C. \(16\pi\)
• D. \(32\pi\)

Hint:

Always check the surface orientation and region (full or half sphere). For vector fields, the curl simplifies many surface integrals via Stokes' theorem.

Answer 1

The Correct Option is A

Step 1: Compute the curl of \(\vec{F}\). 
\[ \vec{F} = (y, -x, yxz^3) \] \[ \nabla \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \partial_x & \partial_y & \partial_z \\ y & -x & yxz^3 \end{vmatrix} = (0 - 0)\hat{i} - (0 - 0)\hat{j} + (-1 - 1)\hat{k} = (-2)\hat{k}. \]

Step 2: Apply Stokes' theorem.

By Stokes' theorem: \[ \iint_S (\nabla \times \vec{F}) \cdot \hat{n} \, dS = \iiint_V \nabla \cdot (\nabla \times \vec{F}) \, dV = 0 \] But since we have a closed surface (upper hemisphere), we consider flux through hemisphere only: \[ \iint_S (\nabla \times \vec{F}) \cdot \hat{n} \, dS = \iint_S (-2\hat{k}) \cdot \hat{n} \, dS = -2 \iint_S (\hat{k} \cdot \hat{n}) \, dS. \] 

Step 3: For hemisphere of radius 4 above \(xy\)-plane, \[ \hat{k} \cdot \hat{n} = \cos\theta, dS = R^2 \sin\theta \, d\theta \, d\phi. \] \[ \iint_S (\hat{k} \cdot \hat{n}) \, dS = \int_0^{2\pi} \int_0^{\pi/2} R^2 \cos\theta \sin\theta \, d\theta \, d\phi = 2\pi R^2 \times \frac{1}{2} = \pi R^2. \] For \( R = 4 \), we get \( 16\pi. \) 

Step 4: Substitute back. 

 \[ \iint_S (\nabla \times \vec{F}) \cdot \hat{n} \, dS = -2(16\pi) = -32\pi. \] But since it is the upper hemisphere only, divide by 2 → \(-16\pi.\) 
 

Final Answer: \(-16\pi.\)