Find the area of an infinitesimal elliptical ring. I have an ellipse given by,
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=c$$
and another ellipse that is infinitesimally bigger than the previous ellipse, i.e.
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=c+dc$$
I want to find the area enclosed by the ring from $x$ to $x+dx$ but I don't know how. Please don't solve the question, just point me in the right direction, I want to solve it myself. Here is a picture of what I want to do.

 A: As the ellipse can be derived from a circle by means of a dilation along $y$ of ratio $b/a$,
we can find the desired area by considering a circular ring with radii $a\sqrt c$ and $a\sqrt{c+dc}$ and then multiplying the result by $b/a$. A $y$ section of such a circular ring at $x$ has a width 
$$
dy=\sqrt{a^2(c+dc)-x^2}-\sqrt{a^2c-x^2}={1\over2}
{a^2\over \sqrt{a^2c-x^2}}dc
$$ 
to first order in $dc$. The area, to first order, is formed by two parallelograms of basis $dy$ and height $dx$, so we have
in the end for the desired area in the ellipse:
$$
dA = {b\over a}\,2\,dy\,dx={ab\over \sqrt{a^2c-x^2}}dc\,dx.
$$
Notice that by integrating the above for $-a\sqrt c<x<a\sqrt c$ one gets, as expected, the annulus area $\pi ab\,dc$.
A: Let $x=a\sqrt{c}\cos(\varphi)$, $y=b\sqrt{c}\sin(\varphi)$. The Jacobian is
$$J=det\begin{bmatrix}
\frac{1}{2\sqrt{c}}a\cos(\varphi) & -b\sqrt{c}\sin(\varphi) \\
\frac{1}{2\sqrt{c}}a\sin(\varphi) & b\sqrt{c}\cos(\varphi)
\end{bmatrix}=\frac{ab}{2}
$$
Hence $dS=\frac{ab}{2}dcd\varphi$, integrating over angle we get 
$$dS=ab\pi{dc}$$
A: Note that you just scaled up both the major and minor axes by a common factor $ \sqrt c$
You want to look at 
$$   (\sqrt c\, y )\ d (\sqrt c  x ) -  y\ dx = (c-1) y\ dx . $$ 
