Concept of a Volume element I am rather confused by the concept of a volume element. For example in cylindrical polar coordinates why is a 'volume element' generated by sending:
$r \rightarrow r+ dr$, $\theta \rightarrow \theta+ d\theta$, $z \rightarrow z+ dz$    ?
I understand that the volume generated by this is : $r$$ d\theta\, dr \,dz$ but I don't understand why this is a 'volume element'. 
 A: To expand on the answers of others in an attempt to also explain why the volume element gets the jacobian as a factor, but without flowing deeply into the theory of differential forms or rigorous mathematics...
Imagine you have a domain of integration $\mathcal{D}$. You have coordinate axes $x,y,z$ that  are orthogonal. You then partition $\mathcal{D}$ into small cubes with sides $dx$, $dy$ and $dz$, you sum the volume of these small cubes up weighted by the value of the function you are integrating in those cubes, idealized as constant. You make this partition finer above all limits, then you get your multivariable Riemann-integral.
If we think of $dx$, $dy$ and $dz$ as infinitesimal displacements, then we generally think of them as scalars, but why should we? Instead, imagine them as infinitesimal vector-like things.
You know that the cross product and the scalar triple product, both associated with oriented area and oriented volume are both (multi)linear and antisymmetric.
It is not hard to see, even with elementary geometry, that if you allow area/volume/hypervolume etc. to be oriented, it behaves linearly and antisymmetrically.
So think of the product in $dxdydz$ as something that is antisymmetric.
Now, if you have new coordinates, $r,\vartheta,\varphi$, you have infinitesimal displacements $dr,d\vartheta,d\varphi$ pointing along the new coordinate curves, but since this coordinate system is curved, the volume $drd\vartheta d\varphi$ depends on position in space.
Not only that, but $dxdydz$ can be viewed as a natural "infinitesimal box", because, well, the cartesian system is orthogonal, so $dxdydz$ is basically an "infinitesimal unit cube".
When you have a function $f:\mathbb{R}^3\rightarrow\mathbb{R},\ x,y,z\mapsto f(x,y,z)$, then when you integrate it as $$ \int_\mathcal{D}f(x,y,z)\ dxdydz, $$ you basically use the "infinitesimal unit cube" $dxdydz$ to measure the weighted volume of $\mathcal{D}$. So, to express this integral with coordinates $r,\vartheta,\varphi$, you also need to express $dxdydz$ with the help of $dr,d\vartheta,d\varphi$.
These are $$ dx=\frac{\partial x}{\partial r}dr+\frac{\partial x}{\partial \vartheta}d\vartheta+\frac{\partial x}{\partial \varphi}d\varphi \\ dy=\frac{\partial y}{\partial r}dr+\frac{\partial y}{\partial \vartheta}d\vartheta+\frac{\partial y}{\partial \varphi}d\varphi \\ dz=\frac{\partial z}{\partial r}dr+\frac{\partial z}{\partial \vartheta}d\vartheta+\frac{\partial z}{\partial \varphi}d\varphi.$$
Now, insert these expressions into $dxdydz$, while keeping mind that the "product" between the differentials is anticommutative and not commutative, and you see you will get the correct results.
Obviously, rigorously infinitesimals don't exist, but if you read a bit on differential forms, you will see that the general idea is what I outlined here heuristically.
A: $$r \,dr\, d\theta \, dz $$ is the volume of an almost cube with opposite corners at $$ (r, \theta, z) \text{ and } (r+dr, \theta+ d\theta, z+dz) $$ with sides $$ dr, r d\theta, \text{ and } dz$$ 
