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


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.

  • $\begingroup$ Ah, I see. Now I know that I don't have to convert at the beginning. Thank you so much. $\endgroup$ – Uchuuko Mar 31 at 20:20

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