Evaluating double integrals by inspection I'm trying to use as much information about the domain to be able to solve the integral without actually integrating. The problem is:

Evaluate $\oint\limits_C {(x\sin ({y^2}) - {y^2})dx + ({x^2}y\cos ({y^2}) +3x)dy} $ where $C$ is the counterclockwise boundary of the trapezoid with vertices $(0,-2),(1,-1),(1,1),(0,2)$. 

My attempt
It's clear that we can use Green's theorem to rewrite the integral, i.e.
$$\oint\limits_C {(x\sin ({y^2}) - {y^2})dx + ({x^2}y\cos ({y^2})+3x)dy}  = \iint\limits_T {(3 - 2y)dA}$$
The interesting thing about $T$ is that it is a trapezoid, the area which is quite easy to find without integration. I got it to be $3$. Now, based on what I've seen in the textbook, it is written as following
$$(3-2y)\times (\text{area of trapezoid})$$
If we would not have the integrand, the integral would be $3$. How should we interpret $(3-2y)$. What $y$ value is needed? The key says it is $9$. Based on this, $y=0$, but why?

The integrand should be $3+2y$. Your idea is correct. The term $3$ multiplies the area gives you the $9$. 
Now imagine integrating $y$ over the symmetric trapezoid region. $f(x,y)=y$ is a plane that is positive over the region above $x$-axis, and negative over the region below $x$-axis, and it is symmetric. Can you see what it should be then? 
