Contradictory area elements in Cartesian and polar coordinates We all know that the area element $dA$ in Cartesian coordinates is given by $$dA = dx\ dy.$$ With a bit of geometry, we can also see that the area element $dA$ in polar coordinates is given by $$dA = r\ dr\ d\theta.$$ If this is the case, when why doesn't $dx\ dy = r\ dr\ d\theta$?
By implicitly differentiating the equations $x = r \cos \theta$ and $y = r \sin \theta$, we have that $$dx = dr \cos \theta - r \sin \theta\ d\theta$$ and $$dy = dr \sin \theta + r \cos \theta\ d\theta.$$ Thus, \begin{align*}
dx\ dy &= (dr \cos \theta - r \sin \theta\ d\theta)(dr \sin \theta + r \cos \theta\ d\theta)\\
&= dr^2 \cos\theta \sin\theta+r\ dr \cos^2 \theta\ d\theta-r\ dr \sin^2 \theta\ d\theta-r^2\sin\theta\cos\theta\ d\theta^2.
\end{align*}
It is not evident to me that this expression equals $r\ dr\ d\theta$, if it does at all. Why does this discrepancy exist? Is it the case that $dA$ in Cartesian coordinates is simply not the same thing as $dA$ in polar coordinates? If so, why do both correspond to area elements in the plane?
 A: In general, because we want to integrate with sign (for example, we want 
$$\int_a^b f(x)dx = - \int_b^a f(x) dx, )$$
so we require that in any coordinate system, 
$$dx dy = - dy dx$$
So 
$$\begin{align*}
dx\ dy &= (dr \cos \theta - r \sin \theta\ d\theta)(dr \sin \theta + r \cos \theta\ d\theta)\\
&=  \cos\theta \sin\theta dr\ dr +r\cos^2 \theta dr \ d\theta -r \sin^2 \theta d\theta \ dr -r^2\sin\theta\cos\theta\ d\theta\ d\theta \\
&= rcos^2 \theta dr \ d\theta - r\sin^2 \theta d\theta \ dr \\
&= r\ dr \ d\theta\ .
\end{align*}$$
A: A more elementary approach:
In Cartesian coordinates:
We have the position vector $\vec{r} = x \vec{e}_x + y \vec{e}_y $. A small area can be defined by the displacement vectors,
$$d\vec{x} = \frac{\partial \vec{r}}{\partial x} dx = dx \vec{e}_x \qquad d\vec{y} = \frac{\partial \vec{r}}{\partial y} dy = dy \vec{e}_y.$$
The area of the parallelogram swept out by these vectors is given by,
$$ dA = \vert d\vec{A} \vert = \vert  d\vec{x} \times d\vec{y}\vert = dx dy $$
In polar Coordinates:
We have the position vector $\vec{r} = \rho\cos(\theta) \vec{e}_x + \rho\sin(\theta) \vec{e}_y $. A small area can be defined by the displacement vectors,
$$d\vec{\rho} = \frac{\partial \vec{r}}{\partial \rho} d\rho =  \left( \cos(\theta) \vec{e}_x + \sin(\theta) \vec{e}_y \right)d\rho, $$ 
$$ d\vec{\theta} = \frac{\partial \vec{r}}{\partial \theta} d\theta =\left(-\rho\sin(\theta) \vec{e}_x + \rho\cos(\theta) \vec{e}_y\right)d\theta.$$
The area of the parallelogram swept out by these vectors is given by,
$$ dA = \vert d\vec{A} \vert = \vert  d\vec{\rho} \times d\vec{\theta}\vert = \rho d\rho d\theta $$

Edit: To answer the question about the general rule for constructing n-dimensional volume elements.
The important mathematical object here isn't cross product, but rather the determinant used to calculate it. Let our coordinates be denoted by $q_j$ where $j=1\dots n$. We will first examine what is going on in the two dimensional case in more detail and then generalize from there.
Our coordinate vector $\vec{r} = x(q_1,q_2) \vec{e}_x + y(q_1,q_2) \vec{e}_y$. The cross product of the two vectors $d\vec{q}_2$ and $d\vec{q}_2$ is given by,
$$d\vec{q}_1 \times d\vec{q}_2
= det \left(\begin{array} 
 _ \vec{e}_x & \vec{e}_y & \vec{e}_z\\
\frac{\partial x}{\partial q_1} & \frac{\partial y}{\partial q_1} & 0 \\
\frac{\partial x}{\partial q_2} & \frac{\partial y}{\partial q_2} & 0 
\end{array} \right) dq_1 dq_2
= det \left( \begin{array}
\ \frac{\partial x}{\partial q_1} & \frac{\partial y}{\partial q_1}  \\
\frac{\partial x}{\partial q_2} & \frac{\partial y}{\partial q_2}   
\end{array}\right) dq_1 dq_2 \vec{e}_z .
$$
The resulting $2\times 2$ matrix at the end of the calculation is called the Jacobian of the transformation. This is usually written as,
$$ \frac{\partial(x,y)}{\partial(q_1,q_2)} \equiv \left( \begin{array}
\ \frac{\partial x}{\partial q_1} & \frac{\partial y}{\partial q_1}  \\
\frac{\partial x}{\partial q_2} & \frac{\partial y}{\partial q_2}   
\end{array}\right) . $$ 
The $n$-dimensional Jacobian is,
$$ \frac{\partial(x_1,x_2,\dots,x_n)}{\partial(q_1,q_2,\dots,q_n)} = 
\left( \begin{array}
\ \frac{\partial x_1}{\partial q_1} &  \cdots & \frac{\partial x_n}{\partial q_1}  \\
 \vdots  & \ddots &  \vdots  \\
\frac{\partial x_1}{\partial q_n} & \cdots & \frac{\partial x_n}{\partial q_n}  
\end{array}\right) .$$
Notice that this is just made by making a matrix whose rows are  $\frac{\partial \vec{r}}{\partial q_j}$.
The $n$-dimensonial volume element is then given by,
$$ dV_n = \left| det \frac{\partial(x_1,x_2,\dots,x_n)}{\partial(q_1,q_2,\dots,q_n)} \right| dq_1 dq_2 \cdots dq_n $$
This formalism using determinants is related to other formalisms you may run into using the levi-cevita tensor or wedge products since all three are closely related. 
For more information you can look at,


*

*Curvilinear Coordinates

*Orthogonal Cooridnates

*Jacobians

*Wedge Products
