I'd like some help here because I can't get the right answer. The lamina we are working with is defined by:
$B={(x,y);x^2+y^2\le1,0\le y}$. Also the density function is "proportional to the distance of the point (x,y) to the x-axis"
I think here is where my mistake lies, for me density function is: $p(x,y)=y$
Acording to the exercise center of mass should be: $(Xc=0,Yc=\frac{3pi}{32})$
So I did this:
To find mass M I converted to polar coordinates. $x=rcos(\theta),y=rsin(\theta)$
$M=\int\int_Bp(x,y)dA => M=\int_0^{\pi}\int_0^1rsin(\theta)rdrd\theta=2/3$.
Now with the mass we can find $Xc$ and $Yc$. It's easy to note that $Xc=0$ with symmetry
$Yc = \frac{1}{M}\int\int_Byp(x,y)da = > Yc=\frac{3}{2}\int_0^\pi\int_0^1rsin(\theta)rsin(\theta)rdrd\theta=\frac{3\pi}{16}$
But Yc should be $\frac{3\pi}{32}$, so I'm close but I'm doing something wrong.
Best Answer
$X=0$ by symmetry.
$Y= {\int_0^1 2 y^2 \sqrt{1-y^2}dy \over \int_0^1 2 y \sqrt{1-y^2}dy } = {3 \pi \over 16}$.