A liquid is poured into a container in the shape of a right circular cone with radius 4 meters and height 16 meters, as shown in Figure 33. Express the volume $V$ of the liquid as a function of the height $h$ of the liquid.

Solution The formula for the volume of a right circular cone of radius $r$ and height $h$ is $$V=\dfrac{1}{3}\pi r^{2}h$$

The volume depends on two variables, $r$ and $h$. To express $V$ as a function of $h$ only, we use the fact that a cross section of the cone and the liquid form two similar triangles. See Figure 34.

Figure 34

Corresponding sides of similar triangles are in proportion. Since the cone's radius is 4 meters and its height is 16 meters, we have $$\begin{eqnarray*} \dfrac{r}{h} &=&\dfrac{4}{16}=\dfrac{1}{4} \\[4pt] r &=&\dfrac{1}{4}h \end{eqnarray*}$$

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Then $$\begin{array}{l} V = \frac{1}{3}\pi r^2 h\mathop = \limits_ {\color{#0066A7}{\uparrow}} \frac{1}{3}\pi \left( {\frac{1}{4}h} \right)^2 h = \frac{1}{{48}}\pi h^3 \\ \quad \quad \quad \;\;\;{\color{#0066A7}{r = \;\frac{1}{4}h}} \\ \end{array}$$

So $V=V( h) =\dfrac{1}{48}\pi h^{3}$ expresses the volume $V$ as a function of the height of the liquid. Since $h$ is measured in meters, $V$ will be expressed in cubic meters.