I need to solve the volume that's between: $$z=0,\quad x=0,\quad y=0,\quad x^2+y^2=4\quad z=12-x-y$$
1) Does it matter if I use double integrals or triple?
2) When I draw this area on the x and y plane, I have a circle with radius 2. How do I know which quadrant my volume is in?
On
1)
No. You can set this up as either: $$ \int_0^2\int_0^{\sqrt{4-x^2}}zdxdy=\int_0^2\int_0^{\sqrt{4-x^2}}(12-x-y)dydx $$ Which represents integrating the volume bounded by the cylinder $x^2+y^2=4$ bounded above by the plane $$ 12-x-y=z $$ But this is just a shortcut for $$ \int_0^2\int_0^{\sqrt{4-x^2}}\int_0^{12-x-y}dzdydx $$ 2)
You technically can't tell which octant you are in, but I think what was intended is as follows.
Assuming you are in the first octant you have $$ 0\leq z\leq12-x-y\\ 0\leq y\leq \sqrt{4-x^2}\\ 0\leq y\leq 2 $$
You'll probably want to use a triple integral here. But it doesn't really matter in this case (see qbert's answer).
You're right that the surfaces you're given do not specify one unique bounded volume.
I'd hazard a guess that your professor wants you to calculate the volume bounded by those surfaces in the first octant, but you really should bring this issue to his/ her attention.