A "cracked glass" Riemann Sum - Double integrals For a 1-dimensional domain, you can partition a segment $(a,b)$ into uneven intervals, choose sample points in the uneven intervals, and form an integral in the limit that the uneven intervals go to $0$. 
For a 2-dimensional domain, it would be easiest to partition such a domain into squares. However, to be as crazy as possible, could you break the domain down into a union of circles, parallelograms, triangles, and trapezoids? From this crazy partition, would a double integral be attainable and would it be equal to the double integral obtained by "normal" partitions? Now lets focus on a single fragment in this "cracked glass" domain and a function $f(x,y)$ above this domain. Let the fragment be a circular fragment. Does the volume above this single circular fragment $$(f(\text{sample point}))\; (\text{area of circle fragment})$$ in the limit that the fragment goes to zero, equal the same value if the fragment was a parallelogram that went to zero [probably "yes" as both numbers by themselves are $0$. Maybe 1 tends to zero faster than the other, I don't know. But if we consider two different "cracked glass" patterns in which the only difference is one fragment is a circle while the other is a parallelogram, does the total Riemann sum remain the same]? 
 A: Yes. 
Riemannian sums are generally taught with taking a regular partition (e.g., breaking an interval into segments of equal length, partitioning a two dimensional domain into equally sized squares, etc) as this is generally the easiest to understand and by far the easiest way to compute numerical examples (e.g., the classic left/right/middle or trapezoid rules). 
For an easy example of non-rectangular pieces, consider double integrals in polar coordinates, which correspond to a Riemann sum with polar rectangles. 
In general, we can partition the domain $\mathcal{D}$ of integration (basically*) however we like by declaring a collection of subsets $\{V_i\}_{i=1}^N$ of $\mathcal{D}$ such that they combine to form $D$:
$$\mathcal{D} = \bigcup_{i=1}^n V_i$$
and the partition elements don't overlap, outside of along their boundaries:
$$\mathrm{int}(V_i) \cap \mathrm{int}(V_j) = \emptyset \quad \text{for } i \neq j.$$
There's no reason that you can't take one of the partition elements to be a circle, provided that you account for the region around the circle with the other partition elements. So we couldn't strictly do parallelograms, squares, triangles, and circles---we would also need some "anti-circle" pieces to account for pieces of the domain around each circle in the partition. We could even go crazy and take partition elements that have fractal boundaries. 
Regardless of the exact partition chosen, we get a Riemann sum upon choosing a sample point $x_i$ from each $V_i$:
$$\sum_{i=1}^N f(x_i) \mu(V_i)$$
where $\mu(V_i)$ denotes the measure of the set $V_i$---basically the length/area/volume of the set. Any such Riemann sum will provide an approximation of the integral. 
To find the integral of $f$ over domain $\mathcal{D}$, we would consider a sequence of partitions $\mathcal{P}_n$ such that the individual partition elements become "small*." If the function is Riemannian integrable, then we will get the same limit regardless of the sequence of partitions and sample points chosen. In fact, a function is Riemannian integrable only if we get the same limit for any sequence of partitions and any sample points. 

*Footnote: We are brushing many technical details under the rug here---there are a few constraints of the partition, not the least of which is that all the partition elements are measureable. Non-measureable sets do exist, but they are much much more exotic than the geometric objects you're considering. There are a few other topological and measure theoretic concerns as well; basically, we can go crazy, but not too crazy. 
