I have been battling with this for a little while now, but can't really get my head around it.
The question goes: how high must you pour water into a cone before it is half-full (by volume)?
No numbers specified, so I guess what I am looking for is a generalised formula for finding the height at which the volume is halved.
Edit: I was wrong in the last paragraph, what I was looking for was not half the volume, but the height at which the value of the volume was halved. Sorry.
$\endgroup$5 Answers
$\begingroup$The volume of a cone of base $R$ and height $H$ is
$$ V(R, H) = \frac{\pi}{3}R^2 H \tag{1} $$
You want to find a new base $r$ and height $h$ such such that
$$ V(r, h) = \frac{1}{2}V(R, H) \tag{2} $$
with the constraint
$$ \frac{R}{H} = \frac{r}{h} \tag{3} $$
Putting together (1), (2) and (3)
\begin{eqnarray} \frac{1}{2}\frac{\pi}{3}R^2 H &=& \frac{\pi}{3}r^2 h \\ \frac{1}{2}R^2 H &=& \left(\frac{R^2}{H^2}h^2 \right) h \\ \frac{1}{2}H &=& \frac{h^3}{H^2} \end{eqnarray}
From this you find
$$ h = 2^{-1/3}H $$
$\endgroup$ $\begingroup$The volume of the half-full region of the cone is similar to the region of the full cone. By that I mean that a uniform scaling can be applied to one to get the other. If the volume is divided by two that means each dimension is divided by the cube root of two. So the height of the water will be $h / \sqrt[3]{2}$ where $h$ is the height of the full cone.
$\endgroup$ $\begingroup$The volume of a cone with radius $r$ and height $h$ is given by $V = \pi r^2 \frac{h}{3}$. Looking at the first diagram, we see that $\tan(\alpha) = \frac{r}{h}$, so $r = h\tan(\alpha)$. Thus $V = \pi (h \tan(\alpha))^2 \frac{h}{3} = \frac{\pi \tan(\alpha)^2}{3} h^3$.
So if we want to halve the volume of a cone, we have to divide $h$ by $\sqrt[3]{2}$. Plugging this in, we get
$$ \frac{\pi \tan(\alpha)^2}{3} (h/\sqrt[3]{2})^3 = \frac{1}{2} \cdot (\frac{\pi \tan(\alpha)^2}{3} h^3) = V/2 $$
as expected.
Note that this works because the angle $\alpha$ is the same for the small cone and the big cone. This formula doesn't work for comparing two different cones with different angles at the tip.
$\endgroup$ $\begingroup$Hint: the volume of a cone is $V = \frac{1}{3}Ah_0$, where $A$ is the base area and $h_0$ the height. What can you say about the area of the cross-section at height $h$, and can you compute the volume below that height?
$\endgroup$ $\begingroup$Integral, area of a circle is pir^2 On a cone, any circular cross section is pi(ry/h)^2 the two volumes, 0-x and x-h, should be equal Both integrals can be simplified to y^2 They work out to y^3/3 F(0) is nothing, 2F(x) is on the left, F(h) is on the right 2x^3/3=h^3/3 x^3=h^3/2 x=h2^(-1/3)
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