How to prove Green's Theorem over a general region R in $\mathbb{R}^2$?

Question

I have the following question on my homework:

Consider a general region of the form $$ R = \{(x,y) \in \mathbb{R}^2 | a \leq x \leq b, \phi_1(x) \leq y \leq \phi_2(x)\}, $$ where $a<b$ and $\phi_1(x) \leq \phi_2(x) \in C^1(\mathbb{R}, \mathbb{R})$) and some function $Q \in C^1(\mathbb{R}^2, \mathbb{R})$. Show that Green's Theorem is true for the vector field $(0, Q)$ and this region. Ie, show that $$ \oint_{\partial R}Q(x,y)dy= \iint_R \frac{\partial (Q(x,y))}{\partial x}dxdy. $$

The question also writes a hint afterwards to first use the chain rule and fundamental theorem of calculus to calculate $\frac{d}{dx} \int_{\phi_1(x)}^{\phi_2(x)}Q(x,y)dy$, which I found to be equal to $Q(x,\phi_2(x))\phi_2'(x)-Q(x,\phi_1(x))\phi_1'(x)$.

Onto my attempt at solving:

To make things clearer for my own benefit notationally, I started by letting $\mathbf{g}(\mathbf{x}) = (0,Q(x,y))$ be the vector field in question. Then, I started working on $\oint_{\partial R} \mathbf{g}(\mathbf{x}) \cdot d\mathbf{x}$. If we start by letting $C_1$ be the part of the curve $y=\phi_1(x)$ on $x \in [a,b]$, $C_2$ be the vertical line $x=b$ for $y \in [\phi_1(b),\phi_2(b)]$, $C_3$ be the curve $y=\phi_2(x)$ from $(b,\phi_2(b))$ to $(a,\phi_2(a))$ and $C_4$ be the vertical line down along $x=a$ from $(a,\phi_2(a))$ to $(a,\phi_1(a))$. Then, $$ \begin{align*} \oint_{\partial R} \mathbf{g}(\mathbf{x})\cdot d\mathbf{x} &= \oint_{\partial R} (0, Q(x,y)) \cdot (dx,dy) \\ &= \oint_{\partial R}Q(x,y)dy \\ &= \int_{C_1} \mathbf{g}(\mathbf{x})\cdot d\mathbf{x} + \int_{C_2} \mathbf{g}(\mathbf{x})\cdot d\mathbf{x} + \int_{C_3} \mathbf{g}(\mathbf{x})\cdot d\mathbf{x} + \int_{C_4} \mathbf{g}(\mathbf{x})\cdot d\mathbf{x} \end{align*} $$ Now, $C_1, C_2, C_3$ and $C_4$ can be parametrised by the following: $$ \begin{align*} \mathbf{p}_1:[a,b] \to \mathbb{R}^2, \mathbf{p}_1(t) &= (t, \phi_1(t)) \\ \mathbf{p}_2:[\phi_1(b),\phi_2(b)] \to \mathbb{R}^2, \mathbf{p}_2(t)&=(b,t) \\ \mathbf{p}_3:[a,b] \to \mathbb{R}^2, \mathbf{p}_3(t)&=(a+b-t,\phi_2(a+b-t)) \\ \mathbf{p}_4:[\phi_1(a),\phi_2(a)] \to \mathbb{R}^2, \mathbf{p}_4(t)&=(a,\phi_1(a)+\phi_2(a)-t) \end{align*} $$ Then, $$ \begin{align*} \oint_{\partial R}Q(x,y)dy &= \int_{a}^{b}(0,Q(t,\phi_1(t)))\cdot (1,\phi_1'(t))dt+\int_{\phi_1(b)}^{\phi_2(b)}(0,Q(b,t))\cdot(0,1)dt+ \\ &\int_{a}^{b}(0,Q(a+b-t,\phi_2(a+b-t)))\cdot(-1,-\phi_2'(a+b-t))dt+\int_{\phi_1(a)}^{\phi_2(a)}(0,Q(a,\phi_1(a)+\phi_2(a)-t))\cdot(0,-1)dt \\ &=\int_{a}^{b}\phi_1'(t)Q(t,\phi_1(t))dt+\int_{\phi_1(b)}^{\phi_2(b)}Q(b,t)dt-\int_{a}^{b}\phi_2'(a+b-t)Q(a+b-t,\phi_2(a+b-t))dt-\int_{\phi_1(a)}^{\phi_2(a)}Q(a,\phi_1(a)+\phi_2(a)-t)dt \end{align*} $$ Using the King's Property of Definite Integrals on the third and fourth integrals, we have $$ \begin{align*} \oint_{\partial R} Q(x,y) \, dy &= \int_{a}^{b} \phi_1'(t) Q(t, \phi_1(t)) \, dt + \int_{\phi_1(b)}^{\phi_2(b)} Q(b, t) \, dt - \int_{a}^{b} \phi_2'(t) Q(t, \phi_2(t)) \, dt - \int_{\phi_1(a)}^{\phi_2(a)} Q(a, t) \, dt \\ &= \int_{a}^{b} \left( \phi_1'(t) Q(t, \phi_1(t)) - \phi_2'(t) Q(t, \phi_2(t)) \right) \, dt + \int_{\phi_1(b)}^{\phi_2(b)} Q(b, t) \, dt - \int_{\phi_1(a)}^{\phi_2(a)} Q(a, t) \, dt \\ &= \int_{a}^{b} \frac{d}{dt} \left( \int_{\phi_1(t)}^{\phi_2(t)} Q(t, y) \, dy \right) \, dt + \int_{\phi_1(b)}^{\phi_2(b)} Q(b, t) \, dt - \int_{\phi_1(a)}^{\phi_2(a)} Q(a, t) \, dt \\ &= \int_{\phi_1(b)}^{\phi_2(b)} Q(b, y) \, dy - \int_{\phi_1(a)}^{\phi_2(a)} Q(a, y) \, dy + \int_{\phi_1(b)}^{\phi_2(b)} Q(b, t) \, dt - \int_{\phi_1(a)}^{\phi_2(a)} Q(a, t) \, dt \\ &= 2 \left( \int_{\phi_1(b)}^{\phi_2(b)} Q(b, y) \, dy - \int_{\phi_1(a)}^{\phi_2(a)} Q(a, y) \, dy \right) \end{align*} $$ At this point, I'm not convinced I haven't made a mistake, but I can't find any. How should I proceed, or does what I have done not work?