Integral bounds for a square with vertices $(\pm2,0), (0,\pm2)$ In exercise X §1 1(e) from "Calculus of several variables" by Serge Lang (Third Edition), we are asked to

Use Green's theorem to find the integral $\int_C y^2 \;dx+x\;dy$ When C is the following curve (taken counterclockwise) ... (e) The square with vertices $(\pm2,0), (0, \pm2)$.


I took this to mean
$$
\int_{-2}^0\int_{-2-x}^{2+x}(1-2y)dy\;dx+
\int_{0}^2\int_{x-2}^{2-x}(1-2y)dy\;dx
$$
which comes out to 0 according to my calculations. Checking in the back the answer was given as 8. I checked my work again and didn't find any error, but according to the explanation for the answer, see photograph below, the integral is just
$$
\int_{-2}^{2}\int_{-2}^{2}(1-2y)dy\;dx
$$

I think the book's answer is wrong, but I am asking for your opinion because I would like to check that I haven't missed some obvious thing.
 A: I have no idea where the book got the double integral
$$\int_{-2}^{2}\int_{-2}^{2}\left(1-2y\right)dydx$$
because it equals $16$ when you evaluate it. Their integrand is correct (and yours too) because
$$
\text{det}\begin{bmatrix}\frac{\partial}{\partial x}&\frac{\partial}{\partial y}\\y^2&x\end{bmatrix} = \frac{\partial}{\partial x}x - \frac{\partial}{\partial y}y^2 = 1-2y.
$$
And then I decided to split the square in half (diamond). Using Green's Theorem for that left-half of the square, I got $\oint_C y^2dx + xdy$ to be
$$\int_{-2}^{0}\int_{-2-x}^{2+x}\left(1-2y\right)dydx = \int_{-2}^{0}\left(2x+4\right)dx = 4.$$
Using Green's Theorem for the other half of the square, I got $\oint_C y^2dx + xdy$ to be
$$\int_{0}^{2}\int_{x-2}^{2-x}(1-2y)dydx = \int_{0}^{2}\left(-2x+4\right)dx = 4.$$
Adding them together gives us $8$, which according to you, is what the back of your book says.
I'm not sure what your calculation mistake was. Please let me know if I need to specify anything.
A: The integration is over a rectangle $R$ of the form $[a,b]\times [c,d]$, then the double integral is the form $\displaystyle \int_{a}^{b}\int_{c}^{d}f(x,y)\, {\rm d}x\, {\rm d}y$.

Remark: By Green's theorem we have$$\int_{C}P(x,y)\,{\rm d}x+Q(x,y)\,{\rm d}y=\iint_{\partial C}\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\, {\rm d}A$$
where $C$ is a simple closed curve, piecewise smoothing and $\partial C$ the boundary of $C$.

All the hypothesis it holds, so by the Green's theorem we have
$$ \int_{C}y^{2}\, {\rm d}x+x\, {\rm d}y=\iint_{\partial C}\left(1-2y \right)\, {\rm d}A=\int_{-\sqrt{2}}^{\sqrt{2}}\int_{-\sqrt{2}}^{\sqrt{2}}\left(1-2y \right)\, {\rm d}x\, {\rm d}y$$

