I have the following water basin
Now imagine this basin is filled with water to the top, is there anyway to accurately calculate the volume of water stored in it using only top and bottom areas A1 and A2? The basin is basically an irregular polygon with circular rounded edges. Furthermore the edges taper down at given slope as can be seen in the section drawing (Please disregard the red line) 
Ideally the volume would be a function of y or depth. Thanks for your help.
Let's say that you know the bottom area, $A_2$, and the top area, $A_1$ at water height $d$.
Let's say that the bottom area is actually at a height $h$ above the unknown apex of the (inverted) pyramid, from which the frustum is created.
The area of the (inverted) base will scale as the square of the height from the apex, so
$$\frac{A_2}{h^2} = \frac{A_1}{(h + d)^2}.$$
By equating cross products and using the quadratic formula, this can be solved for $h$:
$$h = \left[\frac{A_2 + \sqrt{A_1A_2}}{A_2 - A_1}\right] d = Cd.$$
Now you can calculate the area at any water depth $y$:
$$\frac{A_2}{(Cd)^2} = \frac{A(y)}{(Cd + y)^2},$$
or
$$A(y) = \frac{A_2(Cd+y)^2}{(Cd)^2}.$$
Then, the volume of your frustum is:
$$V(y) = \frac{y}{3}\left(A_2 + \sqrt{A_2A(y)} + A(y)\right).$$