Inmersion of homogeneous straight cone into a liquid

After studying concepts about this problem, I think I could solve it through this procedure: Following these previous concepts:

Density: $rho$=$frac{m}{V}$, m=mass and V=volume

Weight: w=mg, g=gravity acceleration

Specific weight: $gamma$=$frac{w}{V}$=$frac{mg}{V}$=$rho$g

According the situation, there must be a balance between immersed part by liquid and weight of the cone.

Wcone=mg=$gamma_1$Vcone

Buoyancy Immersed part: E=$gamma$Vpart

Then, $$Wcone=E$$ $$gamma_1Vcone=gamma Vpart$$ But the implicated volumes of the cone and the immersed part are these ones: $$Vcone=frac{pi R^2h}{3}$$ $$Vpart=frac{pi y}{3} (R^2+r^2+Rr)$$ The immersed part is a cone trunk with height h=y.

Based on these volumes, the balance of forces can be rewritten $$gamma_1frac{pi R^2h}{3}=gammafrac{pi y}{3} (R^2+r^2+Rr)$$

After grouping and deleting terms... $$frac{gamma_1 R^2 h}{gamma}=y(R^2+r^2+Rr)$$ $$frac{gamma_1 h}{gamma}=y(1+frac{r^2}{R^2}+frac{r}{R})$$

Checking the picture again, I could realize there is a similarity of triangles.

$$frac{h}{R}=frac{h-y}{r}$$ It means...$$frac{r}{R}=frac{h-y}{h}$$ Then, the equation can be rewritten... $$frac{gamma_1 h}{gamma}=y(1+frac{(h-y)^2}{h^2}+frac{h-y}{h})$$ The common denominator is $h^2$, then... $$frac{gamma_1 h}{gamma}=y(frac{h^2+(h-y)^2+h(h-y)}{h^2})$$ Grouping and executing terms... $$frac{gamma_1 h^3}{gamma}=y(h^2+(h-y)^2+h(h-y))$$ $$frac{gamma_1 h^3}{gamma}=y(h^2+h^2-2hy+y^2+h^2-hy)$$ $$frac{gamma_1 h^3}{gamma}=y(3h^2-3hy+y^2)$$ $$frac{gamma_1 h^3}{gamma}=3h^2 y-3hy^2+y^3$$ Adding -$h^3$ to two sides of the equation... $$frac{gamma_1 h^3}{gamma}-h^3=3h^2 y-3hy^2+y^3-h^3$$ Multiplying -1 to two sides of the equation... $$h^3-frac{gamma_1 h^3}{gamma}=h^3-3h^2 y+3hy^2-y^3$$ Factorizing... $$h^3(1-frac{gamma_1}{gamma})=(h-y)^3$$ Cube root... $$h-y=h sqrt[3]{1-frac{gamma_1}{gamma}}$$ Finding y... $$y=h-h sqrt[3]{1-frac{gamma_1}{gamma}}$$ Finally, the height of the immersed part is $$y=h[1-sqrt[3]{1-frac{gamma_1}{gamma}}]$$

Thanks for your attention.