The cross-section of a glass can be modeled by the function \(r(x) = \frac {x^4}{3}\), with units in centimeters: Follow the steps below to determine at what height would one need to place a mark indicating \(250\unit {ml}\) of fluid:

To start, we should be looking at the following cross-section:

The radius of the disk, given any value of \(y\) is given by \(\answer [given]{(3y)^{1/4}}\). Hence the volume of the infinitesimal disk is
\[ \d V = \pi \left (\answer [given]{(3y)^{1/4}}\right )^2 \d y \]
Summing these all together via integration we find:
\begin{align*} \int _0^h \pi ((3y)^{1/4})^2 \d y &= \int _0^h \pi \answer [given]{3^{1/2} y^{1/2}} \d y\\ &= \eval {\answer [given]{\frac {2\pi y^{3/2}}{\sqrt {3}}}}_0^h \textrm { by evaluating the antiderivative.} \\ &= \answer [given]{\frac {2\pi h^{3/2}}{\sqrt {3}}} \end{align*}

Now that we have a formula for volume, we need to see when it is equal to \(250\). Write with me:

\begin{align*} \frac {2\pi h^{3/2}}{\sqrt {3}} &= 250\\ h^{3/2} &= \answer {\frac {\sqrt {3} \cdot 125}{\pi }},\\ h &= \answer {\left (\frac {\sqrt {3} \cdot 125}{\pi }\right )^{2/3}} \end{align*}

Using a calculator to approximate this height to 2 decimal places, we see that we should put our mark at approximately \(16.8\) centimeters.