Suppose we have a solid disc of uniform surface density $\sigma$, and radius $a$. I want to find its mass by integration. Normally we do this by the following method :
We set the origin of a polar coordinate system at the center of the disc. Then we take a small area element, at a distance $r$ from this origin $O$. Let $dm$ be the mass of this element. In polar coordinates, we know $dm=\sigma dxdy=\sigma rdrd\psi$.
Hence $$M=\int dm=\int_{0}^{2\pi}\int_{0}^{a}\sigma rdrd\psi = \sigma\pi a^2$$
However, what I want to do, is to make my origin at a point on the circumference of the disc, as follows :
Now I want to carry out the same integration. What would be the new mass element, and the limits of integration, that would get me the exact same answer ?
My guess is that the mass element would still be $dm=\sigma rdrd\psi$ however I'm not sure. The limits of integration would be very different though. Can anyone show me what these would be and how to find it, so that my final answer is exactly the same.
Another guess is that the limits of $d\psi$ are symmetric, between some $[-c,c]$ where $c$ is some constant or maybe a function of $r$ such that $c=c(r)$. However I think that it is symmetric even though I not know the form or the values.
(note : I need $d\psi$ and $dr$ to be the variables of integration, not $dx$, $dy$ or anything else. )
Any help would be highly appreciated.
You are now dealing with the disk centered at $(a,0)$ with radius $a$, that is, with the disk $\{(x,y)\in\Bbb R^2\mid(x-a)^2+y^2\leqslant a^2\}$. But\begin{array}\,(x-a)^2+y^2\leqslant a^2&\iff x^2+y^2\leqslant2ax\\&\iff r^2\leqslant2ar\cos(\theta).\end{array}So, you compute\begin{align}\int_{-\pi^/2}^{\pi/2}\int_0^{2a\cos(\theta)}\sigma r\,\mathrm dr\,\mathrm d\theta&=\int_{-\pi^/2}^{\pi/2}2a^2\sigma\cos^2(\theta)\,\mathrm d\theta\\&=\pi a^2\sigma.\end{align}