Multivariable Calculus Question of the Day

1. $0$ (correct answer)
2. $8\pi$
3. $24\pi$
4. $-24\pi$

Explanation: Let $E$ be the solid cone bounded by $S_1$ and $S_2$. The boundary of $E$ with outward orientation is $S_{out} = S_{2,up} \cup S_{1,down}$. By the Divergence Theorem, the total outward flux is $\iiint_E \text{div}(\mathbf{F}) \, dV$. Here, $\text{div}(\mathbf{F}) = 3+3+3=9$. The volume of the cone is $V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (2^2)(2) = 8\pi/3$. The total flux is $9V = 9(8\pi/3) = 24\pi$. So, $\iint_{S_{2,up}} \mathbf{F} \cdot d\mathbf{S} + \iint_{S_{1,down}} \mathbf{F} \cdot d\mathbf{S} = 24\pi$. Let's calculate the flux through the top disk $S_2$. The normal is $\mathbf{n}=\langle 0,0,1 \rangle$. On $S_2$, $z=2$, so $\mathbf{F} = \langle 3x-y, 3y+2, 6-x \rangle$. The flux is $\iint_{S_2} (6-x) \, dA = 6 \cdot \text{Area}(S_2) - \iint_{S_2} x \, dA = 6(\pi 2^2) - 0 = 24\pi$. Substituting this into the Divergence Theorem equation: $24\pi + \iint_{S_{1,down}} \mathbf{F} \cdot d\mathbf{S} = 24\pi$, which implies $\iint_{S_{1,down}} \mathbf{F} \cdot d\mathbf{S} = 0$. The question asks for the flux through $S_1$ oriented upward, which is the opposite orientation: $\iint_{S_{1,up}} \mathbf{F} \cdot d\mathbf{S} = -\iint_{S_{1,down}} \mathbf{F} \cdot d\mathbf{S} = -0 = 0$.