# Finding the volume of a cone which is partially filled with water

A cone is partially filled with water. When the base of the cone is down, the height of the water from the base $$=h_{w_1}$$. When the base of the cone is up, the height of the water from the apex $$=h_{w_2}$$.

Knowing only $$h_{w_1}$$ and $$h_{w_2}$$, can we find the volume of the cone?

What if we are given an additional information that the radius of the base of the cone and the height of the cone both are $$= a$$ units, can we find the volume of the cone in terms of $$h_{w_1},h_{w_2}$$, and $$a$$?

My thinking:

Yes we can, there is only one value of the radius of the base of the cone $$(r)$$ and only one value of the height of the cone $$(h_c)$$ so that the height of the water from the base and the height of the water from the apex are as given.

Since the volume of the water is unchanged when flipping the cone, we can somehow find $$r$$ and $$h_c$$, and then we can find the volume of the cone.

Clearly, $$h_{w_1}\le h_{w_2}$$. I do not know if the inequality is useful or no. The equality holds true if the cone is totally filled.

Useful Formulae:

• The volume of the cone with radius $$r$$ and height $$h$$, is given by $$V=\frac{\pi}{3}r^2h$$

• The volume of the conical frustum with radii $$r_1$$ and $$r_2$$ and the height between the two bases $$h$$, is given by $$V=\frac{\pi}{3}(r_1^2+r_1r_2+r_2^2)h$$.

I do not know if my thinking is right or not. Any help will be appreciated. THANKS!

Let $$r_1,r_2$$ be the radii of the upper circular water surface when the cone is base-down/base-up respectively, $$r$$ the cone's radius and $$h$$ the cone's overall height. Then we have, by similarity relations, $$\frac{h-h_{w_1}}{r_1}=\frac hr=\frac{h_{w_2}}{r_2}$$ $$\frac{r(h-h_{w_1})}h=r_1,\frac{rh_{w_2}}h=r_2\tag1$$ Since the volume of water is unchanged, $$\frac\pi3(r^2+rr_1+r_1^2)h_{w_1}=\frac\pi3r_2^2h_{w_2}$$ $$(r^2+rr_1+r_1^2)h_{w_1}=r_2^2h_{w_2}$$ Substituting the relations in $$(1)$$, $$\left(r^2+r\left(\frac{r(h-h_{w_1})}h\right)+\left(\frac{r(h-h_{w_1})}h\right)^2\right)h_{w_1}=\left(\frac{rh_{w_2}}h\right)^2h_{w_2}$$ Dividing by $$r^2$$ and then multiplying by $$h^2$$, we get a quadratic in $$h$$: $$3h_{w_1}h^2-3h_{w_1}^2h+h_{w_1}^3-h_{w_2}^3=0$$ Thus we can solve for $$h$$. If we are given only $$h_{w_1}$$ and $$h_{w_2}$$, then we are stuck; we cannot find $$r,r_1,r_2$$ even though they are now in known proportions. For example, two cones with $$h=3$$ and $$r=1,2$$ filled so that $$h_{w_1}=1$$ in both cases will have the same $$h_{w_2}$$.
If we are also given $$r=h$$, then the volume is simply $$\frac\pi3h^3$$ using the $$h$$ we have calculated. (Knowing that $$r=h=a$$ and the value of $$a$$ makes the problem trivial.)