A body is in the form of solid revolution, radius of upper face of the body is  $r_0$ and volumetric density is  $\rho$ . The solid body supports a load of weight P. If the compressive stress at all cross-sections should be same, then the radius r as a function of y (see figure ) is

Consider a very small (differential) disc element at a distance y from the upper base.
Let r and e+dr denote the radii of upper and lower surfaces. Respectively and A and A+dA be the 
corresponding areas. Let w denotes the weight of the body above the elemental disc and dw that of 
the elemental disc. Normal compressive stress at the top surface of the whole body is  $\sigma \frac{P}{\pi r_0^2}\mathrm{......}\left(i\right)$
 
Now, consider the normal compressive stress acting over both the surfaces of the element. We can write 
$\frac{P+w}{A}=\frac{P+w+dw}{A+dA}=\sigma =$ constant
 $$\begin{gathered} PA+PdA+wA+wdA=PA+wA+Adw \\ dA(P+w)=Adw \\ \Rightarrow \frac{dA}{dw}=\frac{A}{P+w}=\frac{1}{\sigma } \end{gathered}$$
 
                                
The increment in area between upper and lower surfaces of the differential disc is $dA=\pi {(r+dr)}^2-\pi r^2\alpha 2\pi /dr$$dW=\pi r^{2}p\left(dy\right)$
The increment in weight is 
Using these in equation (ii) 
  $$\begin{gathered} \frac{2\pi rdr}{\pi r^2\rho dy}=\frac{1}{\sigma } \\ 2\frac{dr}{r}=\frac{\rho }{\sigma }dy \\ 2{\int }_{t0}^r\frac{1}{r}dr=\frac{\rho }{\sigma }y \\ \Rightarrow 2\ln \frac{t}{t_0}=\frac{\rho }{(P/\pi t_0^2)}y \end{gathered}$$
 
From Eq-1    $2\ln \frac{r}{r_0}=\frac{\rho }{(P/\pi r_0^2)}y$
 $$\begin{gathered} \Rightarrow \ln \frac{r}{r_0}=\frac{\pi r_0^2\rho }{2P}y \\ r=r_0{e}^{(\pi t_0^2\rho y/2P)} \end{gathered}$$