Find $M_x$ and $M_y$ for a thin wire along the semicircle $y=\sqrt{1-x^2}$. Take $\rho=1$ so $M=\text{length}=\pi$.
Here, $M_x$ and $M_y$ are moments in the x-axis and y-axis, respectively. $M$ is the mass.
I know how to find them if it were a semicircle disk: $$M_x=\int_0^12y\sqrt{1-y^2}\;dy=\frac23$$ $$M_y=\int_{-1}^1x\sqrt{1-x^2}\;dx=0$$
But now I cannot figure out what $x(y)$ and $y(x)$ are: $$M_x=\int y\;x(y)\;dy$$ $$M_y=\int x\;y(x)\;dx$$
How should I proceed?
Parametrise the wire by $(x,y)=(\cos t,\sin t)$, $0\le t\le\pi$. The $x$-moment is zero, by symmetry. The $y$ moment is got by integrating $y$ over the wire and is $$\frac1\pi\int_0^\pi y\,dt=\frac1\pi\int_0^\pi \sin t\,dt=\cdots.$$