find double integral I am trying to find double integral of function $F(x,y)=3y$ over the triangle with vertices $(-1,1), (0,0)$, and $(1,1)$.
Here is picture of this triangle:
 
I guess that we should integrate it by $x$ and then by $y$,but i am confused on region bound  ,I think  for $y$ lowest bound it $0$ and highest $1$,and for $x$  lowest is $-1$ and $1$ (or $-y$ and $y$)?am I correct or wrong? please help.
 A: Hint: $$\int\limits_{-1}^{1} \int\limits_{|x|}^{1} 3y \ dy \cdot dx = \int\limits_{-1}^{0}\int\limits_{-x}^{1} 3y \  dy \cdot dx + \int\limits_{0}^{1}\int\limits_{x}^{1} 3y \ dy \cdot dx$$
Or even you can do this: $$I = \int\limits_{0}^{1} \int\limits_{-y}^{y} 3y   \ dx \cdot dy$$
A: Because the function $F(x,y)$ is in this case symmetric across the $y$ axis, I think I would almost automatically integrate over the right half of the triangle in the picture, then double the answer. 
In this case there are various advantages, not the least of which is that we do not meet any negative numbers, which are a notorious source of error.
For any fixed $y$ with $0 \le y \le 1$, we make $x$  travel horizontally from $0$ to $y$. Then $y$ is made to travel from $0$ to $1$.
Thus our integral is equal to
$$2\int_{y=0}^{y=1}\left(\int_{x=0}^{x=y} 3y\:dx\right)dy.$$
The calculation is now easy. The inner integral is $3y^2$, so we end up calculating $2\int_{y=0}^{y=1} 3y^2\;dy$. This is $2$.  
We could also integrate first with respect to $y$.  Here again life is made easier if we integrate over the right half of the region, and then double. 
Now $y$ first travels from $x$ to $1$, and then $x$ travels from $0$ to $1$. So our integral is equal to
$$2\int_{x=0}^{x=1}\left(\int_{y=x}^{y=1} 3y\:dy\right)dx.$$
The details of evaluation are a little more messy than with the first approach. In addition, if we do not use the symmetry, the integral will have to be broken up as two integrals, as follows:
$$\int_{x=-1}^{x=0}\left(\int_{y=-x}^{y=1} 3y\:dy\right)dx +\int_{x=0}^{x=1}\left(\int_{y=x}^{y=1} 3y\:dy\right)dx.$$
Uglier, more work, much greater chance of minor errors.
Notational Comment: When we are expressing a double integral as an iterated integral, it is easy to lose track of who is going from where to where. That's why, for example, the integral in the second solution was not written as
$$2\int_{0}^{1}\left(\int_{x}^{1} 3y\:dy\right)dx.$$    
