The shapes of infinitesimal elements. In engineering, we have to use a lot of differentiation and integration for modelling purposes. 
For example, suppose we have to mathematically model a phenomenon on a solid cuboid. The books say "at $x$, take a differential element $dxdydz$". When talking about a disc, they say "take a differential element $rd\theta dr$ at $(r,\theta)$". 
Why can't a disc be divided into infinitesimal rectangles $dxdy$? Why can't a solid  cuboid be divided into infinitesimal cylinders, or any arbitrary shape at all?
Even if they can, does the shape of an object have any relation with the shape of the infinitesimal elements we can divide those objects into?
EDIT: I know that integrating a function on a disc would be mush easier using polar coordinates than using cartesian coordinates. My question was more general in nature. For example, if we were to integrate a function on a triangle ($3$ 'sharp' points), could we divide the triangle into infinitesimal ellipses?
Thanks in advance!
 A: A disc can be divided into rectangles $dxdy$. The reason we typically divide them into $rd\theta dr$ is that these two are completely equivalent (which I'll show below), and because for most purposes (notably integration) it is more convenient to work in polar coordinates on a disc. Keeping track of the limits of integration when integrating over a disc is very difficult with $xy$-coordinates, but extremely simple in polar coordinates.
To see that $dxdy$ and $r d\theta dr$ are equivalent, note that they are related by a change of variable: set
$$x=r\cos\theta \hspace{2em}y=r\sin\theta.$$ 
Then the change of variable formula for integration tells us that $dxdy = |\det D_\Phi(r,\theta)|drd\theta$, where $\Phi(r,\theta)=(x(r,\theta),~y(r,\theta))$ is the change of variable, $D_\Phi$ is the Jacobian matrix, and $\det$ is the determinant function. The Jacobian matrix is
$$ \begin{bmatrix} x_r & x_\theta\\ y_r & y_\theta \end{bmatrix} = \begin{bmatrix} \cos\theta & r\sin\theta \\ \sin\theta & -r\cos\theta\end{bmatrix}$$
and the determinant of this matrix is $-r$. So $dxdy = |-r|drd\theta = rdrd\theta$.
So there is no difference mathematically speaking when you integrate with respect to $dxdy$ and with respect to $rdrd\theta$; but there are situations in which integrating in one coordinate system is a lot easier than integrating in the other. Typically the shape of the region you integrate over will have a lot to do with which one you prefer, but technically speaking you can choose to divide the region any way you want (and there are many ways other than rectangular or polar coordinates) so long as the change of variable map is a $C^1$ local diffeomorphism. The smart engineer/physicist/mathematician chooses a division that makes integration as easy as possible, and this choice of division will be related to the geometry of the region.
A: Sure you can.  The differential area and volume elements $dA$ and $dV$ are expressible in all kinds of fun coordinate systems.  As long as you define your integration area or volume correctly, any of them should work.
But if you're asking "Why can't a disc be divided into infinitesimal rectangles," I could just as easily answer: "Why would you want to?"
Why would you want to break a disc up into squares $dA = dx dy$ when chunks of a sector $dA = r dr d\theta$ make your life so much easier?
Why would you want to use cylindrical coordinates $dV = r dr d \theta dz$ on a square box when $dV = dx dy dz$ is right there waiting for you?
Except maybe for an exam, you don't get extra points for working harder to get the same answer that you could get much more easily.
A: The shapes you choose have to slot together.  Think of it like building a huge object using Lego (TM).  If the Lego pieces were elliptically shaped, they wouldn't slot together.  Whereas tiny rectangles, or tiny pieces of arcs of circles that are oriented properly, do fit together nicely.
