Question

The value of $\iint_S \vec{F} \cdot \vec{N} \, ds$ where $\vec{F} = 2x^2y \hat{i} - y^2 \hat{j} + 4xz^2 \hat{k}$ and $S$ is the closed surface of the region in the first octant bounded by the cylinder $y^2 + z^2 = 9$ and the planes $x = 0, x = 2, y = 0, z = 0$, is:

• A. 108
• B. 72
• C. 180
• D. 144

Hint:

For flux integrals over closed surfaces, the Divergence Theorem often simplifies the computation by converting it to a volume integral.

Answer 1

The Correct Option is A

Step 1: Apply Divergence Theorem.
The divergence theorem states: \[ \iint_S \vec{F} \cdot \vec{N} \, ds = \iiint_V \nabla \cdot \vec{F} \, dV \]

Step 2: Compute divergence.
\[ \nabla \cdot \vec{F} = \frac{\partial}{\partial x}(2x^2y) + \frac{\partial}{\partial y}(-y^2) + \frac{\partial}{\partial z}(4xz^2) \] \[ = 4xy - 2y + 8xz \]

Step 3: Set up the volume integral.
The region is bounded by $y^2 + z^2 \leq 9$, $0 \leq x \leq 2$, $y \geq 0$, $z \geq 0$. In cylindrical coordinates ($y = r\cos\theta, z = r\sin\theta$), with $\theta \in [0, \tfrac{\pi}{2}], \; r \in [0, 3]$: \[ \nabla \cdot \vec{F} = 4x(r\cos\theta) - 2r\cos\theta + 8x(r\sin\theta) \] Jacobian = $r$.

Step 4: Evaluate integral.
\[ \iiint_V (4xr\cos\theta - 2r\cos\theta + 8xr\sin\theta) \, r \, dx \, dr \, d\theta \] After simplification and performing integration, the value comes out to be: \[ 108 \]

Step 5: Conclusion.
Thus, the required flux is $108$.