Suppose we have a ellipse $x^2 + 4y^2 = 1$, and we want to integrate a density $f(x,y) = 3 \cdot |xy| $ over the ellipse.
I was trying to solve this problem in two different ways and i find 2 different answers.
First I called $2y = h$ and then the elipse turns to be a circumference, in this new scale: $x^2 + h^2 = 1$ and my density turns to be $f(x,h) = \frac{3}{2} |xh| $. Then i integrate this density on the parametric circumference $\alpha (t) = ( \cos t , \sin t )$ what leads me to the integral $\int_{0}^{2\pi} \frac{3}{2} |\sin t \cos t| dt = 3.$
Second approach I parameterized the elipse with $\alpha (t) = ( \cos t , 0.5 \sin t ) \implies | \alpha'(t)| = \frac{1}{2} \sqrt(1 + 3 sin^2t)$ and then the integral of $f(x,y)$ over the elipse is : $\int_{0}^{2\pi} \frac{3}{2} |\sin t \cos t| \frac{1}{2} \sqrt(1 + 3 sin^2t) dt = 7/3$ .
What approach is wrong? and what am i misunderstanding? I was suppose to think that this integral does not depend on the coordinate system.
Well, I will try to sketch what is happening with your change of coordinates so you can see the problem.
Let $A := \mathbb{R}^2\;$ be the first plane in consideration and $e:[0,1) \longrightarrow A\;$ an ellipse in that plane. Now, consider a second plane $B:= \mathbb{R}^2\;$ and a circumference $c:[0,1) \longrightarrow B\;$ in this second plane. Your change of coordinates consists in a $C^1$ invertible map $h: A \rightarrow B$ (where $h^{-1}$ is also $C^1$) such that $h \circ e = c\;$. You can click here to see: sketch.
Now, consider the initial density map $f: A \longrightarrow \mathbb{R}$. We get that the correspondent density map in $B\;$, namely $g: B \longrightarrow \mathbb{R}\;$, is actually as you have said, it is $g = f \circ h^{-1}$. So, where is the problem? Let us start by what you want to calculate and see if it reaches what you actually calculated.
$\displaystyle \int_e f \cdot ds = \int_0^1 f(e(t))\ |e'(t)|\ dt = \int_0^1 f\circ h^{-1}(c(t))\ |(h^{-1}\circ c)'(t)|\ dt = \int_0^1 g(c(t))\ |D_{c(t)}h^{-1} \cdot c'(t)|\ dt$
If we recall what you actually calculated in part 1, it was $\displaystyle \int_0^1 g(c(t))\ | c'(t)|\ dt \;$. So we can see that it was missing to take account for the action of the derivative of $h^{-1}\;$ in the velocity of the curve $c\;$.
Finally, in part 2 this problem did not appear because you computed $(h^{-1}\circ c)' \;$ directly.