The help I'm looking for is not solving the actual problem but setting it up (from what i have heard, setting it up is the hardest part.... and from experience i found that out as well)
There seems to be a couple ways the bounds are given but i have the most trouble with this specific "set up"
$$\iint_D xy \mathrm{d}x\mathrm{d}y $$ With $D=\{(x,y) | x^2 + y^2 \le 4, x \ge 0, y\ge 0\} $
I do not understand how to make the circle with radius of 2 into the bounds.. What exactly are the x and y telling me?
Start by analyzing the region $D = \{ (x,y) \, | \, x^2 + y^2 \leq 4, x \geq 0, y \geq 0 \}$. The inequality $x^2 + y^2 \leq 4$ describes a disc of radius $2$ and center $(0,0)$. By adding the inequalities $x \geq 0, y \geq 0$, you restrict attention to the part of the disc which lies in the first quadrant. Thus, in polar coordinates, this region is described as
$$ D' = \{ (r, \theta) \, | \, 0 \leq r \leq 2, 0 \leq \theta \leq \frac{\pi}{2} \}. $$
Setting up the integral using $x = r \cos \theta, y = r \sin \theta$ and $dx \, dy = r dr \, d \theta$, we get
$$ \int_{D} xy \, dx dy = \int_{D'} r^2 \cos \theta \sin \theta \, r dr \, d \theta = \int_{D'} r^3 \cos \theta \sin \theta \, dr d \theta = \left( \int_0^2 r^3 \, dr \right) \left( \int_0^{\frac{\pi}{2}} \cos \theta \sin \theta \, d \theta \right). $$