Is there any density function such that the center of mass of a semicircle about its diameter is half its radius? I am currently taking a Calculus II class in college. We recently went over how to find the center of mass of a lamina with uniform density. The professor worked through the example of a semicircle to show that the center of mass lies along the line of symmetry and at a radius of $\frac{4R}{3\pi}$ from the center. This is slightly below the midpoint of the radius, $\frac{1}{2}R$. I considered whether increasing the density as you moved further from the center could bring the center of mass to $\frac{1}{2}R$, although this was not discussed in class.
Consider the semicircle bounded by $f(x) = \sqrt{R^2-x^2}$ and the x-axis. Does there exist any function $\rho(y)$ such that $y_{CM}=\frac{1}{2}R$ ?
The generalized center of mass $y_{CM}$ can be written:
$$\frac{\int_0^R 2y\rho(y)\sqrt{R^2-y^2}\ dy}
       {\int_0^R 2\rho(y) \sqrt{R^2-y^2}\ dy} = \frac{1}{2}R$$
My intuitive first attempt was to try a linear direct proportion density $\rho(y) = ky$ cand so solve for $k$. However, this fails because:
$$\frac{2\int_0^R y(ky)\sqrt{R^2-y^2}\ dy}
       {2\int_0^R  (ky)\sqrt{R^2-y^2}\ dy} = 
  \frac{\int_0^R ky^2\sqrt{R^2-y^2}\ dy}
       {\int_0^R ky  \sqrt{R^2-y^2}\ dy}   = \frac{3\pi}{16}
$$
So the center of mass of such a lamina is always the same no matter the value of $k$. After the calculation it seems so obvious: increasing $k$ increases the moment about the axis, but it also increases the mass; there is no net effect. My next idea was to rewrite the equation for the center of mass by multiplying the bottom integral out of the denominator:
$$ 2\int_0^R y\rho(y)\sqrt{R^2-y^2}\ dy = R\int_0^R \rho(y) \sqrt{R^2-y^2}\ dy $$
and differentiate the integrals to get the integrands, but of course they are definite integrals, so their derivatives are zero. But if we ignore the bounds, I have a feeling that it is not quite right to get the integrands from the integrals as such because of the multiple variables:
$$  2y\rho(y)\sqrt{R^2-y^2} = R\rho(y)\sqrt{R^2-y^2} + \frac{dR}{dy}\int\rho(y)\sqrt{R^2-y^2}$$
and since $R$ is a constant, $\frac{dR}{dy} = 0$:
$$ 2y\rho(y)\sqrt{R^2-y^2} = R\rho(y)\sqrt{R^2-y^2} $$
$$ \rho(y) (2y-R)\sqrt{R^2-y^2} = 0 $$
$$ \rho (y) = 0\ ??$$
and I am getting nowhere to finding $\rho(y)$ outright. I think that this problem is beyond my abilities as a second-semester calculus student with no experience in multivariable calculus. Does this problem have a solution at all, and is it possible for me to understand with my basic calculus knowledge?
Thank you.
 A: If $\rho(y) = y^0 = 1$, then the center of mass is at $4R/(3\pi)$.  If $\rho(y) = y^1 = y$, then the center of mass is at $(3\pi)R/16$.  Now, let the function $g$ be defined by setting $g(\alpha)$ to the location of the center of mass when $\rho(y) = y^\alpha$.  Then $g$ is continuous (proving this is a little messy but should be intuitively obvious), and $g(0) = 4R/(3\pi), g(1) = (3\pi)R/16$, so by the intermediate value theorem there exists some $\alpha \in [0, 1]$ with $g(\alpha) = R/2$.  However this doesn't tell you what $\alpha$ is.
A: Here’s one function that works, derived from inspection of the integral for center of mass.  
We’re basically looking for some $f(y)$ such that $\frac1M\int_0^Rf(y)dy=\frac R2$. A simple choice is $f(y)=\frac M2$. Comparing this to the integral for the center of mass, we have $$
y\rho(y)\cdot2\sqrt{R^2-y^2}=\frac M2,
$$ so $$
\rho(y)=\frac M{4y\sqrt{R^2-y^2}}.
$$ This has the unfortunate property of being undefined at $y=0$ and $y=R$, but its value there doesn’t affect the integral, anyway. One can think of this mass distribution as being created by squeezing a $2R\times R$ rectangle of uniform density straight in from the sides into a semicircle.  
Clearly, for any other solution of the integral equation above, $\rho(y)=\frac{f(y)}{2y\sqrt{R^2-y^2}}$ will also place the center of mass at $y=R/2$. All of these mass distributions have the property that the density is uniform along the lines $y=\text{const}$. Removing this constraint opens the door to even more density functions that will work. For example, any function that’s symmetric about the $y$-axis will keep the center of mass at $x=0$.
