# 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.

• Can you solve the problem for the special case of a unit circle? If so, the result follows by applying the change-of-variables formula to a particularly simple coordinate change. – Travis Willse Feb 20 '17 at 16:21
• No. I don't see it through, I can't solve for the unit circle. – sigsegv Feb 20 '17 at 16:23
• If we consider the circle $x^2 + y^2 = 1 + \epsilon$ with $\epsilon$ small, we can see that the radius of the circle satisfies $r^2 = 1 + \epsilon$, so $r = 1 + \frac{1}{2} \epsilon + O(\epsilon^2)$. Thus, its area is $\pi \left(1 + \frac{1}{2} \epsilon + O(\epsilon^2)\right)^2 = \pi + \pi \epsilon + O(\epsilon^2)$, and the area of the annulus is $\pi \epsilon + O(\epsilon^2)$. – Travis Willse Feb 20 '17 at 16:32
• A uniform thickening of a noncircular ellipse is not itself an ellipse. – coffeemath Feb 20 '17 at 16:43
• I am not thickening the ellipse. I am increasing the value of c to c+dc. That is an ellipse, isn't it? – sigsegv Feb 20 '17 at 16:50

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 $\pi ab\,dc$.

• That the area of the ring, but how do I find the area of the hatched region i.e. between $x$ and $x+dx$? – sigsegv Feb 20 '17 at 18:44
• I hadn't read that comment in the picture: you'd better point that out in the main text of your question. – Intelligenti pauca Feb 20 '17 at 18:46
• Just edited my answer. – Intelligenti pauca Feb 20 '17 at 19:14

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 $$dS=ab\pi{dc}$$

Note that you just scaled up both the major and minor axes by a common factor $\sqrt c$

You want to look at

$$(\sqrt c\, y )\ d (\sqrt c x ) - y\ dx = (c-1) y\ dx .$$