the task is to calculate line integral over vector. $ \oint_\gamma (2x+y+e^x)dx+(6y+2)dy $
$\gamma$ is $\{(x, y); |x|+|y|=4\}$
Can I use Green's theorem to solve this integral? Answer is 32 I belive, as the integral becomes $\iint_S \,dx\,dy$ where S is region bounded by $\gamma$.
Yes, you can. If we define $M(x,y)=2x+y+e^x$ and $N(x,y)=6y+2$ then Green theorem say: $\oint_\gamma Mdx+Ndy = \iint_S \left(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) dxdy$.
In our case: $\frac{\partial N}{\partial x}=0$ and $\frac{\partial M}{\partial y} = 1$, then
$$\oint_\gamma (2x+y+e^x)dx+(6y+2)dy = \iint_S - dxdy.$$
The region S has a diamon shape, so we can calculate its area as four times the area of a triangle, for example the triangle $\{(x,y) \in \mathbb{R}^2: |x|+|y| \leq 4, x \geq 0, y \geq 0 \}$. The base is $4$ (for instance we can think the base as the segment $\{ (x,0) : 0 \leq x \leq 4 \}$. On the other hand, the height is also $4$, because its $\{ (0,y) : 0 \leq y \leq 4 \}$. Finally its area is $$ \frac{4 \times 4}{2}=8, $$ so the intengral you want is
$$ \oint_\gamma (2x+y+e^x)dx+(6y+2)dy = \iint_S - dxdy = -4 \times 8 = - 32. $$