An exercise in a book says "Prove Green's theorem for $R=[0,1]\times[0,1]$". It doesn't add any specific form, so I will assume it askes to prove $\int_R d\phi=\int_{\partial R} \phi$ for an arbitrary $\phi$.
I have encountered several problems to solve it.
(1) Since we are in $\mathbb{R}^2$ I think we can assume $\phi =Adx +Bdy.$ Therefore $d\phi = \displaystyle\frac{\partial A}{\partial y}dx\wedge dy + \displaystyle\frac{\partial B}{\partial x} dy \wedge dx = \left( \displaystyle\frac{\partial A}{\partial y} - \displaystyle\frac{\partial B}{\partial x}\right) dx \wedge dy$, The the first of the integrals would be $\displaystyle\int_0^1\int_0^1 \left(\frac{\partial A}{\partial y} - \displaystyle\frac{\partial B}{\partial x}\right)dxdy$
(2) I would parametrize the boundary of the square in the four lines: $(t, 0)$ with $t$ from $0$ to $1$, $(1,t)$ with $t$ $0$ to $1$, $(t,1)$ with $t$ $1$ to $0$ and $(0,t)$ with $t$ $1$ to $0$. Both $A$ and $B$ are function of $x$ and $y$, then the integral of the theorem would be something like $\displaystyle\int_0^1 A(t,0)dt+\int_0^1 B(1,t)dt-\int_0^1 A(t,1)dt-\int_0^1 B(0,t)dt.$
I'm not sure how to proceed to show that the latter sum of four integrals is the double integral above.
To expand on the comment: from the fundamental theorem of calculus (applied to the square bracket, we obtain $$\int_{[0,1]^2}dx\,dy\,\frac{\partial A}{\partial y} =\int_0^1 dx \left[\int_0^1 dy \frac{\partial A}{\partial y} \right] =\int_0^1dx \left[A(x,1)- A(x,0)\right].$$
Similarly,
$$\int_{[0,1]^2}dx\,dy\,\frac{\partial B}{\partial x} =\int_0^1 dy \left[\int_0^1 dx \frac{\partial B}{\partial x} \right] =\int_0^1dy \left[B(1,y)- B(0,y)\right].$$
Thus, we obtain $$\int_{[0,1]^2}dx\,dy\,\left(\frac{\partial A}{\partial y}-\frac{\partial B}{\partial x} \right) = \int_0^1 \!dt\,[ A(t,1) +B(0,t) -A(t,0)- B(1,t) ].$$