# Calculating mass from the integration of the surface density of a disc

If I had an equation for the surface density of a disc as a function of it's radius, such as:

$$\Sigma = A\left(\frac{r}{B}\right)^{1/2}$$

where A and B are constants, and I want to find the mass, what form does the integration take?

Would it be:

$$M = AB^{-1/2}\int{r^{1/2}} r dr$$

Do I add in that extra r since its a disc/cylinder? I get an answer for the mass in the right ballpark if I do but I'm not sure I can justify adding it in, or whether it should be a double integral.

Thanks for any help.

To give a full answer, yes you do need to do a double integral, but in this case, one of the double integrals is easy to evaluate if you switch to polar coordinates. The factor of $r$ arises when you switch to polar coordinates.
Let us set up the full problem. To get the total mass of the object, one sets up the integral: $$M=\int\int_S dm$$ where $M$ is the total mass, the integral is taken over the entire surface where there is surface density, and $dm$ is an infinitesimal mass element. So, we can transform this integral into: $$M=\int_{-\infty}^\infty\int_{-\infty}^\infty dxdy \Sigma(x,y).$$ Now, the form of $\Sigma(x,y)$ is pretty complicated in caretesian coordinates (presumably it's 0 outside some circle and equal to your expression inside some circle), but simple in polar coordinates so it makes sense for us to transform into polar coordinates. We make a coordinate change $r=\sqrt{x^2+y^2},\quad \tan\theta=y/x$. With this coordinate change, we will find that $dxdy=rdrd\theta$. I won't show the proof of that here, one can look for it in any multivariate calculus text. So now our integral has become: $$M=\int_0^{2\pi}\int_0^{\infty} rdrd\theta\Sigma(r)=2\pi A B^{-1/2}\int_0^{r_{\max}} r^{3/2}dr.$$ So in the end your equation is correct up to a factor of $2\pi$.
P.S. make sure you take care of the size of the disc - giving you the $r_{\max}$ limit in the integration. If $\Sigma$ had that form for the whole plane, then the integral obviously diverges.
There is a double integral--another one over the azimuth. Since there is no azimuthal dependence, it's trivial and you can say at radius $r$:
$$dm = (\Sigma)(2\pi)(r dr) .$$