How can I calculate, how the volume element transforms under change of co-ordinates? Suppose I transform an integral $$I=\int f(x,y) \, dx \, dy$$ using polar coordinates, setting $x=r\cos\theta$ and $y=r\sin\theta$. We get
$$
\begin{split}
dx &= \cos\theta \, dr - r\sin\theta \, d\theta\\
dy &= dr\,\sin\theta + r \cos \theta \, d\theta.
\end{split}
$$
Now the volume element $dx \, dy$, should become
$$
(\cos\theta \, dr - r\sin\theta \, d\theta)(dr\,\sin\theta + r \cos \theta \, d\theta),
$$
whereas my book says it becomes $r \, dr \, d\theta$.
I understand the geometric argument but I am looking for a way to connect the result the result of my book with what should I have got had I used differentials.
 A: Multivariable transformations don't quite work out so easily in multi-d to a simple 1d change. You have to generalize the 1D change of variables carefully (see here for example).
You really have the 2D transformation of coordinates
$$
\begin{pmatrix} dx \\ dy \end{pmatrix}
= \begin{pmatrix} \cos \theta & -r \sin \theta \\
                  \sin \theta & r \cos \theta \end{pmatrix}
  \begin{pmatrix} dr \\ d\theta \end{pmatrix}
$$
so the correct change of variable factor is given by the Jacobian, which is the determinant of the transformation matrix. You get
$$
\det \begin{pmatrix} \cos \theta & -r \sin \theta \\
                  \sin \theta & r \cos \theta \end{pmatrix}
= r \cos^2 \theta + r \sin^2 \theta = r.
$$
A: The volume element $dx\,dy$ can be thought of as an infinitely small change in $x$ with $y$ not changing, times an infinitely small change in $y$ with $x$ not changing -- thus the area of an infinitely small rectangle.
In the above, if we make $\theta$ change while $r$ is fixed we move along a circle centered at $0$, and if we make $r$ change with $\theta$ fixed we move away from $0$, thus at a right angle to the other motion; hence we again have the area of an infinitely small rectangle.


*

*One side of that infinitely small rectangle has length $dr$.

*The other side has length $r\,d\theta$, since the length of an arc of a circle is the radius (in this case $r$) times the radian measure (in this case the infinitely small change $d\theta$).


Thus the area element is $(r\,d\theta)(dr)$.
For purposes of actually computing integrals, it is often convenient to change the order of multiplication and write $r\,dr\,d\theta$.
