The following problem appears as 533 in Putnam and Beyond (2007 edition). I have some doubts about the solution given at the end of the book. I would appreciate it if someone could answer my questions (to appear at the end).
Problem: Let $\vec{F} = F_1 \vec{i} + F_2 \vec{j}$ be a vector field, and let $G(x, y, t)$ be a smooth function. Assume for any rectangle $D$ with boundary $C$ we have $\frac{d}{dt}\iint_D G(x, y, t) dxdy = -\oint_C \vec{F} \cdot d\vec{R}$. Prove that $\frac{\partial G}{\partial t} + \frac{\partial F_2}{\partial x} + \frac{\partial F_1}{\partial y} = 0$.
The official solution: Pick an arbitrary box $[a_1, b_1] \times [a_2, b_2] \times [a, b]$ and let $D = [a_1, b_1] \times [a_2, b_2]$. Using the assumption, one can derive
\begin{align*} 0 & = \int_{D \times \{b\}} G(x, y, t) \vec{k} \cdot d\vec{n} + \int_{D \times \{a\}} G(x, y, t) \vec{k} \cdot d\vec{n}\\ & + \int_a^b \int_{a_1}^{b_1} F_1(x, a_2) dx - \int_a^b \int_{b_1}^{a_1} F_1(x, b_2) dx\\ & + \int_a^b \int_{a_2}^{b_2} F_2(b_1, y) dy - \int_a^b \int_{b_2}^{a_2} F_2(a_1, y) dy. \end{align*}
This can be written as $\iint_{\partial V} \vec{H} \cdot \vec{n} dS = 0$, where $\vec{H} = F_2 \vec{i} + F_1 \vec{j} + G\vec{k}$. (The rest follows from Gauss' theorem.)
My questions:
Should $-\int_a^b \int_{b_1}^{a_1} F_1(x, b_2) dx$ have been $-\int_a^b \int_{a_1}^{b_1} F_1(x, b_2) dx$ instead?
Is the definition of $H$ correct? I feel like it should be $F_2 \vec{i} - F_1 \vec{j} + G \vec{k}$. For, consider the side $[a_1, b_1] \times \{a_2\} \times [a, b]$. The outward normal is $-\vec{j}$, so on this side, $\iint_S \vec{H} \cdot \vec{n} dS$ would be $-\int_a^b \int_{a_1}^{b_1} F_1(x, a_2) dx$ if $\vec{H}$ were defined as given in the solution.
Is the desired conclusion of the problem correct? If my previous question makes sense, then I think the conclusion should be changed to $\frac{\partial G}{\partial t} + \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} = 0$.
