Center of mass of a planar lamina

I have to find the center of mass of a planar lamina bounded by $$x=0$$, $$y=1/2$$, and $$y=x$$, with the density of the lamina being $$x/(1-y^2)^{1/2}$$.

I ended up drawing a picture, and it looks like a triangle. However, I set up my integral to find the mass, and I end up getting

$$\int_0^{1/2}\int_0^y\frac{x}{\sqrt{1-y^2}}\,\rm{d}x\,\rm{d}y$$

as my double integral to find the mass. I really don't like the look of the density function. I was thinking about converting everything to cylindrical coordinates, but I am not sure how to go about doing it. When I do come up with an integral in cylindrical coordinates, I think it looks even worse. Do any of you recommend converting to cylindrical coordinates?

In addition, I remember a previous calculation that I did where I found $$(1-y^2)^{1/2}$$ to be equal to $$x$$. I do not know where that identity comes from, and so I am afraid to substitute an $$x$$ for that part of the density.

This integral can directly be performed $$\int_0^y \frac{x}{\sqrt{1-y^2}}dx=\frac{y^2}{2\sqrt{1-y^2}}$$ $$\int_0^{\frac12} \frac{y^2}{2\sqrt{1-y^2}}dy=\int_0^{\frac\pi6} \frac{\sin^2{\theta}}{2\sqrt{1-\sin^2{\theta}}} \cos{\theta} d\theta$$ $$=\int_0^{\frac\pi6} \frac{\sin^2{\theta}}{2}d\theta$$ $$=\int_0^{\frac\pi6} \frac{1-\cos{2\theta}}{4}d\theta$$ $$=[\frac14(\theta-\frac12\sin{2\theta})]_0^{\frac\pi6}$$ $$=\frac{\pi}{24}-\frac{\sqrt{3}}{16}$$ Where I have used the substitution $$y=\sin{\theta}$$ in order to calculate the second integral.