The outward flux of the vector field $\mathbf{F} = 4x \hat{i} - 2y^2 \hat{j} + z^2 \hat{k}$ over the closed surface bounding the cylinder $x^2 + y^2 \le 4$ between $z = 0$ and $z = 3$ is $k \pi$. The value of $k$ is ________ (answer in integer).
Answer Key:
84
Step-by-Step Solution & Explanation
**Method 1: Gauss's Divergence Theorem**
$\iint_S \mathbf{F} \cdot \hat{n} \, dS = \iiint_V (\nabla \cdot \mathbf{F}) \, dV$
Compute the divergence:
$\nabla \cdot \mathbf{F} = \frac{\partial}{\partial x}(4x) + \frac{\partial}{\partial y}(-2y^2) + \frac{\partial}{\partial z}(z^2) = 4 - 4y + 2z$
Transform to cylindrical coordinates $(r, \theta, z)$:
$x = r \cos\theta, \quad y = r \sin\theta, \quad z = z$
where $0 \le r \le 2$, $0 \le \theta \le 2\pi$, and $0 \le z \le 3$, with $dV = r \, dr \, d\theta \, dz$.
Notice that $\int_0^{2\pi} y \, d\theta = \int_0^{2\pi} r \sin\theta \, d\theta = 0$ by symmetry.
Therefore:
$\iiint_V (4 + 2z) \, dV = \left( \int_0^{2\pi} d\theta \right) \left( \int_0^2 r \, dr \right) \left( \int_0^3 (4 + 2z) \, dz \right)$
$= (2\pi) \cdot \left[ \frac{r^2}{2} \right]_0^2 \cdot \left[ 4z + z^2 \right]_0^3$
$= (2\pi) \cdot (2) \cdot (12 + 9) = 4\pi \times 21 = 84\pi$
Hence $k = 84$.